SOLUTION TO A PROBLEM IN RAMANUJAN'S LOST

SOLUTION TO A PROBLEM IN RAMANUJAN’S LOST NOTE BOOK A.C.Wimal Lalith De Alwis School Of Four Incalculables And Tathagatha’s Society Golden Lanka (New name for Lanka Island in Indian Ocean given by present author appeared in the history yet to be absorbed and adopt by others kept for their observation, Air Golden Lanka is the new name given by present author for Air Carriers to and from Lanka) E-Mail:[email protected] Abstract Srinivasa Ramanujan an Indian Mathematical Genius has stated three mathematical identities in his Last Note Book in Mathematics popular as Lost Note Book which was hidden in a Box after his death for more than a half century leads to sum of two cubes in two different ways a problem posed by G.H.Hardy to Ramanujan in a Hospital Bed in England which was a dull taxi cab number to Hardy but was a very interesting Number to Ramanujan is discussed with several proofs Introduction In the page 341 of Lost Note Book and other unpublished papers of S.Ramanujan a new edition published by Narosa publishing house in Bombay in the year 1993 following three identities are noted which are mathematically very interesting (i.)

1+53𝑥+9𝑥 2

= 𝑎0 + 𝑎1 𝑥 + 𝑎2 𝑥 2 + 𝑎3 𝑥 3 +

1−82𝑥−82𝑥 2 +𝑥 3

or (ii.)

2−26𝑥−12𝑥 2

𝑥

+

∝1 𝑥2

+

∝2 𝑥3

+.

= 𝑏0 + 𝑏1 𝑥 + 𝑏2 𝑥 2 + 𝑏3 𝑥 3 +..

1−82𝑥−82𝑥 2 +𝑥 3

or (iii.)

∝0

2+8𝑥−10𝑥 2 1−82𝑥−82𝑥 2 +𝑥 3

𝛽0 𝑥

+

𝛽1 𝑥2

+

𝛽2 𝑥3

+⋯

= 𝑐0 + 𝑐1 𝑥 + 𝑐2 𝑥 2 + 𝑐3 𝑥 3 +.

or

𝛾0 𝑥

+

𝛾1

𝛾2

𝑥

𝑥3

2 +

+..

then 𝑎𝑛3 + 𝑏𝑛3 = 𝑐𝑛3 + (−1)𝑛 and 𝛼𝑛3 + 𝛽𝑛3 = 𝛾𝑛3 + (−1)𝑛 examples 1353 + 1383 = 1723 -1 111613 + 114683 = 142583 + 1 7913 + 8123 = 10103 − 1 656013 + 674023 = 838023 + 1 93 + 103 = 123 + 1 63 + 83 =93 − 1 These number identities can very easily be proved by four methods except the last one 1.By cross multiplication and then by collecting coefficients of same powers of x & by comparison of both sides with further simplification very easily leads to these number identities except the last one 2.By using Taylor’s expansion with calculus 3.By expansion of denominator as a power series with factorization or without 4.More coefficients have been calculated not given by Ramanujan with use of Computer Program in Computer Language Package known as Maxima but without software it cannot be done and it is the last achievement in this problem. But still Ramanujan’s Mathematical Intelligence cannot be defeated at all even though he has expired in that birth due to sickness being in England after one year of return to India because without his birth ability these formulas will not be available in his own hand writing his wife

Janaki Ammal was only collecting and preserving his mathematical pages prior to his departure from this world which are so hard to achieve by others computer software are later developments many decades after Ramanujan they can only program in their own languages and provide further numerical results even ordinary mathematics does the job manually but those three formulas never originate from a computer only it came out of biological conscious brain of Ramanujan not by any artifact computer software Intelligence of a Robo which is only a programmed routine machine mean while Ramanujan was a conscious human it will be very difficult to get a Robotic Ramanujan. 1 + 53𝑥 + 9𝑥 2 + 0𝑥 3 = (1 − 82𝑥 − 82𝑥 2 + 𝑥 3 )(𝑎0 + 𝑎1 𝑥 + 𝑎2 𝑥 2 + 𝑎3 𝑥 3 ) =𝑎0 + (𝑎1 − 82𝑎0 )𝑥 + (−82𝑎0 − 82𝑎1 + 𝑎2 )𝑥 2 + (𝑎0 − 82𝑎1 − 82𝑎2 + 𝑎3 )𝑥 3 𝑎0 = 1, 𝑎1 − 82𝑎0 = 53, 𝑎2 − 82𝑎0 − 82𝑎1 = 9, 𝑎3 − 82(𝑎1 + 𝑎2 ) + 𝑎0 = 0 𝑎1 = 53 + 82𝑎0 = 135, 𝑎2 = 9 + 82(𝑎0 + 𝑎1 ) = 9 + 82(1 + 135) = 9 + 82.136 = 11161, 𝑎3 = 82(𝑎1 + 𝑎2 ) − 𝑎0 = 82(135 + 11161) − 1 = 82.11296 − 1 = 926272 − 1 = 926271 2 − 26𝑥 − 12𝑥 2 + 0𝑥 3 = 𝑏0 + (𝑏1 − 82𝑏0 )𝑥 + (−82𝑏0 − 82𝑏1 + 𝑏2 )𝑥 2 + (𝑏0 − 82𝑏1 − 82𝑏2 + 𝑏3 )𝑥 3 𝑏0 = 2, 𝑏1 − 82𝑏0 = −26, 𝑏1 = 82𝑏0 − 26 = 82.2 − 26 = 164 − 26 = 138, 𝑏2 = 82(𝑏0 + 𝑏1 ) − 12 = −12 + 82. (2 + 138) = −12 + 82.140 = −12 + 11480 = 11468 𝑏3 = 82(𝑏1 + 𝑏2 ) − 𝑏0 = 82. (138 + 11468) − 2 = 82.11606 − 2 = 951692 − 2 = 951690

2 + 8𝑥 − 10𝑥 2 = 𝑐0 + (𝑐1 − 82𝑐0 )𝑥 + (−82𝑐0 − 82𝑐1 + 𝑐2 )𝑥 2 + (𝑐0 − 82𝑐1 − 82𝑐2 + 𝑐3 )𝑥 3 𝑐0 = 2, 𝑐1 = 82𝑐0 + 8 = 82.2 + 8 = 164 + 8 = 172 𝑐2 = 82(𝑐0 + 𝑐1 ) − 10 = 82. (2 + 172) − 10 = 82.174 − 10 = 14268 − 10 = 14258 𝑐3 = 82. (𝑐1 + 𝑐2 ) − 𝑐0 = 82. (172 + 14258) − 2 = 82.14430 − 2 =1183260-2=1183258 𝑎𝑛3 + 𝑏𝑛3 = 𝑐𝑛3 + (−1)𝑛 𝑛 = 0, 𝑎03 + 𝑏03 = 𝑐03 + 1 13 + 23 = 23 + 1 𝑛 = 1, 𝑎13 + 𝑏13 = 𝑐13 − 1 1353 + 1383 = 1723 − 1 𝑛 = 2, 𝑎23 + 𝑏23 = 𝑐23 + 1 111613 + 114683 = 142583 + 1 𝑛 = 3, 𝑎33 + 𝑏33 = 𝑐33 − 1 9262713 + 9516903 = 11832583 − 1 The last number identity is not among Ramanujan’s list but among the mrob.com Wolfram Alpha Software Maxima Program List of Number Identities of Robotic Ramanujan a Ghost, We say raising the hand from Grave. 53

1

53

1

𝑥 2 (9 + + 2 ) 1 + 53𝑥 + 9𝑥 2 1 (9 + 𝑥 + 𝑥 2 ) 𝑥 𝑥 = = 1 − 82𝑥 − 82𝑥 2 + 𝑥 3 𝑥 3 (1 − 82 − 82 + 1 ) 𝑥 (1 − 82 − 82 + 1 ) 𝑥 𝑥2 𝑥3 𝑥 𝑥2 𝑥3 ∝0 ∝1 ∝2 ∝3 = + 2 + 3 + 4 +⋯ 𝑥 𝑥 𝑥 𝑥

9+ 1− 9+

82 𝑥

53 𝑥

−

+

1 𝑥2

82

1

𝑥

𝑥3

2 +

=∝0 +

∝1 ∝2 ∝3 + 2 + 3 +⋯ 𝑥 𝑥 𝑥

53 1 0 82 82 1 ∝1 ∝2 ∝3 + 2 + 3 = (1 − − 2 + 3 ) (∝0 + + 2 + 3 + ⋯) 𝑥 𝑥 𝑥 𝑥 𝑥 𝑥 𝑥 𝑥 𝑥 1 1 =∝0 + (∝1 − 82 ∝0 ) + (∝2 − 82 ∝1 − 82 ∝0 ) 2 𝑥 𝑥 1 + (∝3 − 82 ∝2 − 82 ∝1 +∝0 ) 3 + ⋯ 𝑥 ∝0 = 9, ∝1 − 82 ∝0 = 53, ∝2 − 82 ∝1 − 82 ∝0 = 1, ∝3 − 82 ∝2 − 82 ∝1 +∝0 = 0

∝1 = 82 ∝0 + 53 = 82.9 + 53 = 738 + 53 = 791, ∝2 = 82(∝1 +∝0 ) + 1 = 82(791 + 9) + 1 = 82.800 + 1 = 65600 + 1 = 65601, ∝3 = 82(∝2 +∝1 ) −∝0 = 82(65601 + 791) − 9 = 82.66392 − 9 = 5444144 − 9 = 5444135 26

2

26

2

𝑥 2 (−12 − + 2 ) 2 − 26𝑥 − 12𝑥 2 1 (−12 − 𝑥 + 𝑥 2 ) 𝑥 𝑥 = = 1 − 82𝑥 − 82𝑥 2 + 𝑥 3 𝑥 3 (1 − 82 − 82 + 1 ) 𝑥 (1 − 82 − 82 + 1 ) 𝑥 𝑥2 𝑥3 𝑥 𝑥2 𝑥3 𝛽0 𝛽1 𝛽2 𝛽3 = + + + +⋯ 𝑥 𝑥2 𝑥3 𝑥4 −12 − 1− −12 −

82 𝑥

26

−

+

𝑥 82 𝑥2

2 𝑥2 1

+

𝑥3

= 𝛽0 +

𝛽1 𝛽2 𝛽3 + + +⋯ 𝑥 𝑥2 𝑥3

26 2 0 82 82 1 𝛽1 𝛽2 𝛽3 + 2 + 3 = (1 − − 2 + 3 ) (𝛽0 + + 2 + 3 ) 𝑥 𝑥 𝑥 𝑥 𝑥 𝑥 𝑥 𝑥 𝑥

1 1 = 𝛽0 + (𝛽1 − 82𝛽0 ) + (𝛽2 − 82𝛽1 − 82𝛽0 ) 2 𝑥 𝑥 1 + (𝛽3 − 82𝛽2 − 82𝛽1 + 𝛽0 ) 3 + ⋯ 𝑥 𝛽0 = −12, 𝛽1 -82𝛽0 = −26, 𝛽1 = 82𝛽0 − 26 = 82. −12 − 26 = −984 − 26 = −1010, 𝛽2 − 82𝛽1 − 82𝛽0 = 2, 𝛽2 = 82(𝛽1 + 𝛽0 ) + 2 = 82. (−1010 − 12) + 2 = 82. −1022 + 2 = −83804 + 2 = −83802 𝛽3 = 82(𝛽2 + 𝛽1 ) − 𝛽0 = 82. (−83802 − 1010) + 12 = −82.84812 + 12 = −6954584 + 12 = −6954572 8

𝑥 2 (−10 + +

2

8

−10 + + 1− −10 +

82 𝑥2

−

2

𝑥 𝑥2 82 1 𝑥

+

𝑥3

= 𝛾0 +

2)

8

2

2 + 8𝑥 − 10𝑥 1 (−10 + 𝑥 + 𝑥 2 ) 𝑥 𝑥 = = 1 − 82𝑥 − 82𝑥 2 + 𝑥 3 𝑥 3 (1 − 82 − 82 + 1 ) 𝑥 (1 − 82 − 82 + 1 ) 𝑥2 𝑥 𝑥3 𝑥2 𝑥 𝑥3 𝛾0 𝛾1 𝛾2 𝛾3 = + 2+ 3+ 4+⋯ 𝑥 𝑥 𝑥 𝑥 2

𝛾1 𝛾2 𝛾3 + + +⋯ 𝑥 𝑥2 𝑥3

8 2 0 82 82 1 𝛾1 𝛾2 𝛾3 + 2 + 3 = (1 − − 2 + 3 ) (𝛾0 + + 2 + 3 ) 𝑥 𝑥 𝑥 𝑥 𝑥 𝑥 𝑥 𝑥 𝑥 1 1 = 𝛾0 + (𝛾1 − 82𝛾0 ) + (𝛾2 − 82𝛾1 − 82𝛾0 ) 2 𝑥 𝑥 1 + (𝛾3 − 82𝛾2 − 82𝛾1 + 𝛾0 ) 3 + ⋯ 𝑥

𝛾0 = −10, 𝛾1 − 82𝛾0 = 8, 𝛾1 = 82. −10 + 8 = −820 + 8 = −812, 𝛾2 − 82(𝛾1 + 𝛾0 ) = 2, 𝛾2 = 82(𝛾1 + 𝛾0 ) + 2 = 82(−812 − 10) + 2 = 82. −822 + 2 = −67404 + 2 = −67402, 𝛾3 − 82(𝛾2 + 𝛾1 ) + 𝛾0 = 0, 𝛾3 = 82(𝛾2 + 𝛾1 ) − 𝛾0 = 82(−67402 − 812)— 10 = −82.68214 + 10 = −5593548 + 10 = −5593538

𝛼𝑛3 + 𝛽𝑛3 = 𝛾𝑛3 + (−1)𝑛 𝑛 = 0, 𝛼03 + 𝛽03 = 𝛾03 + 1 93 + (−12)3 = (−10)3 + 1, 93 + 103 = 123 + 1 n=1,𝛼13 + 𝛽13 = 𝛾13 − 1 7913 + (−1010)3 = (−812)3 − 1 7913 + 8123 = 10103 − 1 n=2,𝛼23 + 𝛽23 = 𝛾23 + 1 656013 + (−83802)3 = (−67402)3 + 1 656013 + 674023 = 838023 + 1 𝑛 = 3, 𝛼33 + 𝛽33 = 𝛾33 − 1 54441353 + (−6954572)3 = (−5593538)3 − 1 54441353 + 55935383 = 69545723 − 1 The last cubic number identity is not coming under Ramanujan’ s Lost Note book number list but among the mrob.com Wolfram Alpha Maxima Program number list in which they have calculated ten such number identities by Computer program that We call as work of Robotic Ramanujan. 2

𝑛

13 + 23 + 33 + 43 + 53 + ⋯ + 𝑛3 = [ (𝑛 + 1)] , 𝑡𝑎𝑘𝑒 𝑛 = 5 𝑐𝑎𝑠𝑒 then answer 2

3

3

3

3

3

3

3

5

3

6

is, 1 + 2 + 3 + 4 + 5 = [ (5 + 1)]2 = [5. ]2 = [5.3]2 = 152 2

2 3

3

3

2 2

1 + 8 + 3 + 4 + 5 = 15 ,3 + 4 + 5 = 15 − 32 = (15 − 3)(15 + 3) = 12.18 = 6.2.6.3 = 6.6.2.3 = 6.6.6 = 63 33 + 43 + 53 = 63 33 + 43 = 63 − 53 ×23 , 23 . 33 + 23 . 43 = 23 . 63 − 23 . 53 63 + 83 = 123 − 103 = 93 − 13 = 729 − 1 = 728 63 + 83 = 93 − 1 a consistent result with the Ramanujan’s derived Number identity but unfortunately Ramanujan did not come up with another set of three formulas that can give this result directly.1729 is the G.H.Hardy’s Taxi Cab Number was a dull number to Hardy but was very interesting Ramanujan.Whether Hardy has hired a Taxi or he was driving a taxi to see Ramanujan is not clear but no record of Ramanujan ever drived any Taxies. Ramanujan was doing very high mathematics in the hospital bed during war in England. Hardy gave 100 marks to Ramanujan in Mathematics but Hardy said his own career in Math is Nil in Mathematical Apology a late novel book written by him. Hardy could have get 100

marks to Psychology where Ramanujan’s career in Psychology is Nil in Madras even in England. No evidence he did any other subject other than Math even in England only Math. Ramanujan never knew infinity even Google. With out knowing where numbers end Google talks too many nonsense on numbers. Numbers end at the incalculable & infinity is the immeasurable the distance travel by light speed in incalculable Earth years .Google’s career in Psychology is Nil. Google= 10100 , 𝐼𝑛𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑏𝑙𝑒 = 10140 𝑒𝑎𝑟𝑡ℎ 𝑦𝑒𝑎𝑟𝑠, 𝐼𝑛𝑓𝑖𝑛𝑖𝑡𝑦 = 𝐼𝑚𝑚𝑒𝑎𝑠𝑢𝑟𝑎𝑏𝑙𝑒 = 3×108 ×365.25. .×24×60×60× 10140 , 𝑓𝑜𝑢𝑛𝑑 𝑏𝑦 𝑆𝑖𝑑𝑑ℎ𝑎𝑟𝑡ℎ𝑎 𝐺𝑎𝑢𝑡𝑎𝑚𝑎, 𝐽𝑎𝑚𝑒𝑠 𝐶𝑙𝑒𝑟𝑘 𝑀𝑎𝑥𝑤𝑒𝑙𝑙 𝑎𝑛𝑑 𝑡ℎ𝑒 𝑃𝑟𝑒𝑠𝑒𝑛𝑡 𝐴𝑢𝑡ℎ𝑜𝑟. Google cannot exceed these ends. Google can change only the decimals of the earth year in days. If there are decimals in number of hours in a day Google can change that also. Google cannot change 3 in light speed it is fixed. Google is insane as Ramanujan in number game. 1

Take = 𝑡 𝑥

𝑓(𝑡) =

9+53𝑡+𝑡 2 1−82𝑡−82𝑡 2 +𝑡 3

=𝛼0 + 𝛼1 𝑡 + 𝛼2 𝑡 2 + 𝛼3 𝑡 3 + ⋯

𝑡

= 𝑓(0) + 𝑓 , (0) +𝑓 ,, (0) 𝑔(𝑡) =

1! −12−26𝑡+2𝑡 2

1−82𝑡−82𝑡 2 +𝑡 3

𝑡2 2!

𝑡

ℎ(𝑡) =

1−82𝑡−82𝑡 2 +𝑡 3

𝑡3

+⋯

3!

= 𝛽0 + 𝛽1 𝑡 + 𝛽2 𝑡 2 + 𝛽3 𝑡 3 + ⋯

=𝑔(0) + 𝑔′ (0) + 𝑔′′ (0) 1! −10+8𝑡+2𝑡 2

+ 𝑓 ,,, (0)

𝑡2

+ 𝑔′′′ (0)

2!

𝑡3

+⋯

3!

= 𝛾0 + 𝛾1 𝑡 + 𝛾2 𝑡 2 + 𝛾3 𝑡 3 + ⋯

𝑡

𝑡2

1!

2!

= ℎ(0) + ℎ, (0) + ℎ,, (0)

+ ℎ,,, (0)

𝑡3 3!

+⋯

𝑓(0) = 9 = 𝛼0 , 𝑔(0) = −12 = 𝛽0 , ℎ(0) = −10 = 𝛾0 𝛼03 + 𝛽03 = 𝛾03 + 1,𝑓(0)3 + 𝑔(0)3 = ℎ(0)3 + 1 93 + (−12)3 = (−10)3 + 1 93 + 103 = 123 + 1 (9 + 10)(92 − 9.10 + 102 ) = 19.91 = 1729 = (12 + 1)(122 − 12 + 12 ) = 13.133 𝑓 ′ (𝑡) = (9 + 53𝑡 + 𝑡 2 ).

−1(−82−2.82𝑡+3𝑡 2 ) (1−82𝑡−82𝑡 2 +𝑡 3 )2

+ (53 + 2𝑡).

1 (1−82𝑡−82𝑡 2 +𝑡 3 )

𝑓 ′ (0) = 9.82 + 53 = 738 + 53 = 791 = 𝛼1 𝑔, (𝑡) = (−12 − 26𝑡 + 2𝑡 2 )

−1(−82−2.82𝑡+3𝑡 2 ) (1−82𝑡−82𝑡 2 +𝑡 3 )2

+ (−26 + 2.2𝑡).

1 (1−82𝑡−82𝑡 2 +𝑡 3 )

𝑔, (0) = −12.82 − 26 = −984 − 26 = −1010=𝛽1 ℎ′ (𝑡) = (−10 + 8𝑡 + 2𝑡 2 )

−1(−82−2.82𝑡+3𝑡 2 ) (1−82𝑡−82𝑡 2 +𝑡 3 )2

+ (8 + 2.2𝑡).

ℎ′ (0) = −10.82 + 8 = −820 + 8 = −812=𝛾1 𝛼13 + 𝛽13 = 𝛾13 − 1 𝑓 ′ (0)3 + 𝑔′ (0)3 = ℎ′ (0)3 − 1 7913 + (−1010)3 = (−812)3 -1 7913 + −10103 = −8123 − 1 7913 + 8123 = 10103 − 1 2𝑡) 𝑡 2)

−(−82−2.82𝑡+3𝑡 2 )

+ (9 + 53𝑡 + 𝑡 2 )

(1−82𝑡−82𝑡 2 +𝑡 3 )2

−(−82−2.82𝑡+3𝑡 2 ).−2(−82−82.2𝑡+3𝑡 2 ) (1−82𝑡−82𝑡 2 +𝑡 3 )3

𝑓 ′′ (𝑡) = (53 +

−1(−2.82+3.2𝑡) (1−82𝑡−82𝑡 2 +𝑡 3 )2 2

+

1 (1−82𝑡−82𝑡 2 +𝑡 3)

1−82𝑡−82𝑡 2 +𝑡

+ (9 + 53𝑡 + −1(−82−82.2𝑡+3𝑡 2 )

+(53+2t). 3

(1−82𝑡−82𝑡 2 +𝑡 3 )2

𝑓 ′′ (0) = 53.82 + 2.9.82 + 9.82.2.82) + 2 + 53.82 = 2 + 2.53.82 + 2.9.82.83 = 2(1 + 82. (53 + 9.83)) = 2(1 + 82. (53 + 747)) = 2(1 + 82.800) = 2(1 + 65600) = 2.65601 = 131202 𝑓′′ (0) 2

= 65601 = 𝛼2

𝑓 ′′′ (𝑡) = 2. 2𝑡) 𝑡 2)

−(−82−2.82𝑡+3𝑡 2 ) (1−82𝑡−82𝑡 2 +𝑡 3 )2

+ (53 + 2𝑡)

−(−82−2.82𝑡+3𝑡 2 ).−2(−82−82.2𝑡+3𝑡 2 ) (1−82𝑡−82𝑡 2 +𝑡 3 )3 −1(3.2)

(53 + 2𝑡)

2𝑡).

+ (9 + 53𝑡 +

(1−82𝑡−82𝑡 2 +𝑡 3 )2 −(−2.82+3.2𝑡).−2(−82−82.2𝑡+3𝑡 2 ) 𝑡 2) + (1−82𝑡−82𝑡 2 +𝑡 3 )3 2 2 2.2(−82−2.82𝑡+3𝑡 )(−2.82+3.2𝑡)

+ (9 + 53𝑡 + 𝑡 )

(1−82𝑡−82𝑡 2 +𝑡 3 )3 2 3 2 2(−82−2.82𝑡+3𝑡 ) .−3

(9 + 53𝑡 + 𝑡 )

+ (53 +

(1−82𝑡−82𝑡 2 +𝑡 3 )2 −(−2.82+3.2𝑡)

+(53 + 2𝑡)

+ +(9 + 53𝑡 +

(1−82𝑡−82𝑡 2 +𝑡 3 )2 2(−82−2.82𝑡+3𝑡 2 )2

−(−2.82+3.2𝑡)

+2

(1−82𝑡−82𝑡 2 +𝑡 3 )3 −1(−82−82.2𝑡+3𝑡 2 )

+

+ (53 +

(1−82𝑡−82𝑡 2 +𝑡 3 )4 (1−82𝑡−82𝑡 2 +𝑡 3 )2 −1(−82.2+3.2𝑡) −1(−82−82.2𝑡+3𝑡 2 ) −1(−82−82.2𝑡+3𝑡 2 )2 .−2

(1−82𝑡−82𝑡 2 +𝑡 3 )2

+ 2.

(1−82𝑡−82𝑡 2 +𝑡 3 )2

+ (53 + 2𝑡).

(1−82𝑡−82𝑡 2 +𝑡 3 )3

𝑓 ′′′ (0) = 2.82 + 53.2.82 + 53.82.2.82 + 53.2.82 − 9.3.2 + 9.2.82.2.82 + 53.2.822 + 9.2. 22 . 822 + 9.2.823 . 3+2.82+53.2.82+2.82+53.2.822 =2.3.82+2.3.53.82+2.3.53.822 +2.3.2.9.822 + 2.3.9.823 -2.3.9 𝑓,,, (0) 6

=82+53.82+53.822 +2.9.822 +9.823 -9

=82+53.82.83+9.822 .84-9=82(1+53.83+9.82.84)-9 =82(54+82(53+756))-9 =82(54+82.809)-9 =5444135= 𝛼3

−(−82−2.82𝑡+3𝑡 2 )

𝑔′′ (𝑡) = (−26 + 2.2𝑡) (−12 − 26𝑡 + 2𝑡 2 )

(1−82𝑡−82𝑡 2 +𝑡 3 )2

−(−2.82+3.2𝑡)

+ (−12 − 26𝑡 + 2𝑡 2 )

(1−82𝑡−82𝑡 2 +𝑡 3 )2

+

−(−82−2.82𝑡+3𝑡 2 ).−2(−82−2.82𝑡+3𝑡 2 ) (1−82𝑡−82𝑡 2 +𝑡 3 )3

(−26 + 2.2𝑡). −(−82 − 82.2𝑡 + 3𝑡 2 ) 1 +2.2 + 1 − 82𝑡 − 82𝑡 2 + 𝑡 3 (1 − 82𝑡 − 82𝑡 2 + 𝑡 3 )2 𝑔,, (0) = −26. −(−82) + −12. −(−2.82) + (−12). −(−82). −2(−82) + 2.2 − 26. −(−82) = 2(−13.82 − 12.82 − 12.822 + 2 − 13.82) = 2[−82(38 + 12.82) + 2] = 2[−82(38 + 984) + 2] = 2[−82.1022 + 2] = 2[−83802] 𝑔′′ (0)

2.2𝑡) 2𝑡 2 )

−1(−82−82.2𝑡+3𝑡 2 ) (1−82𝑡−82𝑡 2 +𝑡 3 )2 −(−2.82+3.2𝑡)

1

ℎ′′ (𝑡) = 2.2

= −83802 = 𝛽2

2

+ (8 + 2.2𝑡)

1−82𝑡−82𝑡 2 +𝑡 3 −1(−82−2.82𝑡+3𝑡 2 )

+ (−10 + 8𝑡 +

(1−82𝑡−82𝑡 2 +𝑡 3 )2 2 2 2 −1(−82−2.82𝑡+3𝑡 ) .−2

+ (−10 + 8𝑡 + 8𝑡 )

(1−82𝑡−82𝑡 2 +𝑡 3 )2

+ (8 +

(1−82𝑡−82𝑡 2 +𝑡 3 )3

ℎ′′ (0) = 2.2 + 8.82+8.82-10.2.82-10.2.822 =2(2+8.82-10.82-10.822 ) =2(2-2.82-10.822 )=2(2-82.822)=2(-67402) ℎ′′ (0)

=-67402=𝛾2

2 3 𝛼2 + 𝛽23 𝑓′′ (0) 3

(

2

= 𝛾23 + 1

) +(

𝑔′′ (0) 3 ) 2 3

=(

ℎ′′(0) 3 ) 2 3

+ 1,656013 + (−83802)3 = (−67402)3 + 1

656013 + 67402 = 83802 + 1 𝑓(𝑡) =

9+53𝑡+𝑡 2

9+53𝑡+𝑡 2

1−82𝑡−82𝑡 2 +𝑡 3

9+53𝑡+𝑡 2

𝑖 = (1+𝑡)(1−𝑡+𝑡 2)−82𝑡(1+𝑡) = (1+𝑡)(1−83𝑡+𝑡 2 = ∑∞ 𝑖=0 𝛼𝑖 𝑡 = )

𝛼0 + 𝛼1 𝑡 + 𝛼2 𝑡 2 + 𝛼3 𝑡 3 + ⋯ + 𝛼𝑛 𝑡 𝑛 +. . =

𝐴 1+𝑡

+

𝐵𝑡+𝐶 1−83𝑡+𝑡 2

9 + 53𝑡 + 𝑡 2 = 𝐴(1 − 83𝑡 + 𝑡 2 ) + (𝐵𝑡 + 𝐶)(1 + 𝑡) = (𝐴 + 𝐶) + (−83𝐴 + 𝐵 + 𝐶)𝑡 + (𝐴 + 𝐵)𝑡 2 𝐴 + 𝐶 = 9, (−83𝐴 + 𝐵 + 𝐶) = 53, 𝐴 + 𝐵 = 1 𝐶 − 𝐵 = 8, −83(9 − 𝐶) + 𝐵 + 𝐶 = −83.9 + 84𝐶 + 𝐵 = 53,85𝐶 = 61 + 747 = 808, 𝐶 =

808 85

,𝐴 = 9 −

9+53𝑡+𝑡 2 1−82𝑡−82𝑡 2 +𝑡 3 −

43 85

1+𝑡

+

=

−43 85

1+𝑡

+

808 85

=

765−808 85

128 808 𝑡+ 85 85 1−83𝑡+𝑡 2

128 808 𝑡+ 85 85 83+9√85 83−9√85 (𝑡− )(𝑡− ) 2 2

=

−

=

43 85

1+𝑡

+

43 85

−

1+𝑡

= +

−43

43

43

, 𝐵 = 1 − (− ) = 1 + = 85 85 85

128 808 𝑡+ 85 85 83 2 83 (𝑡− ) +1−( )2 2 2

𝑃 83+9√85 𝑡− 2

+

𝑄 83−9√85 𝑡− 2

=

43 85

−

1+𝑡

=−

+

43

128 85

128 808 𝑡+ 85 85 83 2 −(±9.√85)2 (𝑡− ) + 2 22

1

85 1−(−𝑡)

+

=

𝑃 𝑃 𝑃

1

1

+𝑄

𝑡 83+9√85 1− −( ) 83+9√85 2 2

1

1

83+9√85 −( ) 2

83−9√85 2

1 85

1 85

808) ∑∞ 𝑖=0(83𝑡

(−1)𝑖 − 𝑃

85

83−9√85

𝑃(

2

83−9√85 𝑖 𝑖 ∑∞ ) 𝑡 𝑖=0( 2

43

𝑖+1

(−(−1)𝑖 43 + 83

𝛼0 = − 𝛼1 = 𝛼2 =

1 85 1 85

43 85

+

808 85

=

85

83+9√85 2

𝑖 𝑖 ∑∞ 𝑖=0(−1) 𝑡 +

85

1

𝑖 𝑖 ∑∞ 𝑖=0( 83−9√85 ) 𝑡 =−

43 85

𝑖 𝑖 ∑∞ 𝑖=0(−1) 𝑡 −

2

83+9√85 𝑖 𝑖 ∑∞ )𝑡 𝑖=0( 2

=−

43

𝑖 𝑖 ∑∞ 𝑖=0(−1) 𝑡 +

85

𝑖 ∑∞ 𝑖=0 𝛼𝑖 𝑡

2 𝑖

−𝑡 ) =

83+9√85 𝑖+1 ) 2 𝑖−1

43

=−

=−

−𝑄

43 85

83+9√85 83+9√85 𝑖 ( ) 2 2

(−1)𝑖 +

1 85

=−

43 85

(−1)𝑖 −

(128(83)𝑖−1 + 808(83)𝑖 ) =

. 67192 + 𝑐𝑜𝑟𝑟𝑒𝑐𝑡𝑖𝑜𝑛𝑠), ∀𝑖 ≥ 1

765

9.85

=

85

=9

85 67235 85

= 791 𝑖

(−43 + 83.67192 − 808) = 65601 with correction 808(−1)2 , 𝑖 =

{(64 +

43

(−1)𝑖−1 − 𝑃 (

85 83−9√85 𝑖 8√85)( ) 2 43 1

𝑖 = 1, 𝛼0 = − 𝛼0 =

1

83−9√85 −( ) 2

83−9√85 83−9√85 𝑖 ( ) 2 2

(43 + 67192) =

2, −808, 𝛼𝑖−1 = − 1

−𝑄

− 𝑄(

)

2

1

𝑖 𝑖 ∑∞ 𝑖=0( 83+9√85 ) 𝑡 + 𝑄

(128𝑡 +

𝛼𝑖 = −

1

𝑡 83−9√85 1− −( ) 83−9√85 2 2

85

−

85

83−9√85

𝑖

83+9√85 𝑖 ) 2 83+9√85 𝑖

) − 𝑄(

2

+ (64 − 8√85) (

2

83−9√85

{(64 + 8√85) (

2

=−

43 85

(−1)𝑖−1 −

) } , ∀𝑖 ≥ 1 83+9√85

) + (64 − 8√85) (

−43−{(32+4√85)(83−9√85)+(32−4√85)(83+9√85)} 85

=−

1 85

)}

2

{43 + 32.83 −

32.9√85 + 4.83√85 − 36.85 + 32.83 + 32.9√85 − 4.83√85 − 36.85} = −

1 85

{43 + 64.83 − 72.85} = −

8(664 − 765)} = −

1 85

1 85

{43 + 8(8.83 − 9.85)} = −

{43 + 8. −101} = −

1 85

1

{43 +

85 −765

{43 − 808} = −

85

= −(−9) =

+9 = 9,So that the correctness of the formula got confirmed, with Congratulations to S.Ramanujan for his great mathematical confidence , faith and devotion 𝑃 (𝑡 − 𝑄 𝑃

83−9√85

83+9√85

128

2

2

85

83+9√85 2 83−9√85 2

=

) + 𝑄 (𝑡 −

)=

808 85 83−9√85

+𝑄

2

=

83−9√85 128 2

.

85

𝑡+

808 85

,𝑃 + 𝑄 =

128 85

, −𝑃

83−9√85 2

−

808+64(83−9√85)

−𝑄9√85 =

=

85 808+64.83−576√85 6120−576√85

−

85 8√85−64 85

=

=

−𝑃

−𝑄

83+9√85

85

64−8√85 85

,𝑃 =

1

6120−576√85

9√85 85 128−64+8√85 64+8√85

=

85

=−

1 680−64√85 √85

85

=

85

83 − 9√85 808 83 + 9√85 −1 680 − 64√85 = + . 2 85 2 85 √85 808 + (83 + 9√85)(32√85 − 340) = 85√85 808 + 83.32√85 − 9.340√85 + 9.32.85 − 83.340 = 85√85 25288 − 28220 − 404√85 −2932 − 404√85 = = 85√85 85√85 2932 + 404√85 404 + 2932√85 =− =− 85 85√85 83+9√85 340−32√85

=

.

2 85 √85 28220−24480+3060√85−2656√85 85√85

𝑔(𝑡) =

, 𝑄=−

−12−26𝑡+2𝑡 2

=

83.340−9.32.85+9.340√85−83.32√85 = 85√85 3740+404√85 3740√85+34340 404+44√85

=

1+𝑡 2

+

=

85.85

85

−12−26𝑡+2𝑡 2

𝑖 2 3 = (1+𝑡)(1−83𝑡+𝑡 2 = ∑∞ 𝑖−0 𝛽𝑖 𝑡 = 𝛽0 + 𝛽1 𝑡 + 𝛽2 𝑡 + 𝛽3 𝑡 +

1−82𝑡−82𝑡 2 +𝑡 3 𝐴′ 𝐵′ 𝑡+𝐶 ′ 𝑛

⋯ + 𝛽𝑛 𝑡 +..=

=

85√85

)

1−83𝑡+𝑡 2 ′ (1

−12 − 26𝑡 + 2𝑡 = 𝐴 − 83𝑡 + 𝑡 2 ) + (𝐵′ 𝑡 + 𝐶 ′ )(1 + 𝑡) = (𝐴′ + 𝐶 ′ ) + (−83𝐴′ + 𝐵′ + 𝐶 ′ )𝑡 + (𝐴′ + 𝐵′ )𝑡 2 𝐴′ + 𝐶 ′ = −12, −83𝐴′ + 𝐵′ + 𝐶 ′ = −26, 𝐴′ + 𝐵′ = 2 𝐵′ − 𝐶 ′ = 14, −83(2 − 𝐵′ ) + 𝐵′ + 𝐶 ′ = −26,84𝐵′ + 𝐶 ′ = 83.2 − 26 = 166 − 26 = 140,85𝐵′ = 140, 𝐵′ = 16 85 −12−26𝑡+2𝑡 2 1−82𝑡−82𝑡 2 +𝑡 3 16 85

=

16 85

1+𝑡

𝑗 𝑗 ∑∞ 𝑗=0(−1) 𝑡 +

+

1 85

154

154

85

85

154 −1036 𝑡+ 85 85 1−83𝑡+𝑡 2

,𝐶 ′ =

=

16 85

1+𝑡

+

− 14 =

−1036 85

𝑃′

𝑡−

+ 83+9√85 2

,𝐴′ = 2 − 𝐵′ = 2 −

𝑄′ 83−9√85 𝑡− 2

2 𝑗 (154𝑡 − 1036) ∑∞ 𝑗=0(83𝑡 − 𝑡 )

𝑗 = ∑∞ 𝑗=0 𝛽𝑗 𝑡 =

154 85

=

𝛽0 =

16 85

−1036

+

=

85

−1020 85

= −12, 𝛽𝑗 =

1

1036(83)𝑗 ) =

(16(−1)𝑗 + 85

16

1

(−1)𝑗 +

85 (83)𝑗−1 (154

(154(83)𝑗−1 −

85

− 1036.83)) =

1 85

(16(−1)𝑗 +

(83)𝑗−1 (−85834) + 𝑐𝑜𝑟𝑟𝑒𝑐𝑡𝑖𝑜𝑛𝑠), ∀𝑗 ≥ 1 𝛽1 = 𝛽2 =

1 85 1 85

(−16 − 85834) =

−85850 85

= −1010

(16 + 831 . −85834 + 1036) =

1 85

(−7124206 + 1036) = 𝑗

−83802 𝑤𝑖𝑡ℎ 𝑡ℎ𝑒 𝑐𝑜𝑟𝑟𝑒𝑐𝑡𝑖𝑜𝑛 − 1036(−1)2 , 𝑗 = 2 83𝑡 − 𝑡 2 < 1, 𝑡 2 − 83𝑡 + 1 > 0, (𝑡 − 1

83 2 ) 2

+1−

832

1

> 0, 𝑡 > ± {√832 − 22 + 2

22

83} = ± {9√85 + 83} should be satisfied for the expansion but t was taken 2 arbitrarily and mean while S. Ramanujan did not provide a general expression for the sequence except first few fitting numbers of the number sequence for his famous cubic identity these are subsequent discoveries by the present author in the analysis of his contributions in Mathematics under the supervision of G. H. 𝑗 𝑗 Hardy. (83𝑡 − 𝑡 2 )𝑗 = ∑𝑟=0 𝑗−𝑟𝐶(83𝑡)𝑗−𝑟 (−𝑡 2 )𝑟 = ∑𝑗𝑟=0 𝑗−𝑟𝑗𝐶(83)𝑗−𝑟 (−1)𝑟 𝑡 𝑗−𝑟+2𝑟 = ∑𝑗𝑟=0 𝑗−𝑟𝑗𝐶(83)𝑗−𝑟 (−1)𝑟 𝑡 𝑗+𝑟 , 𝑗−𝑟𝑗𝐶 = 𝛽𝑗−1 = 𝑃′ (𝑡 − 𝑄′

16

2 83+9√85

2

2

=

−𝑃′ 9√85=

) + 𝑄′ (𝑡 −

1036 85

1036

−

, 𝑃′

83−9√85 2

154 83+9√85

85 85 5355√85+77.9.85

83+9√85 𝑗 ) 2 154 1036

𝑟!𝑗−𝑟!

) − 𝑄′ (

85 83−9√85

83+9√85 2

83−9√85

(−1)𝑗−1 − 𝑃′ (

𝑗

𝑗!

)=

+( =

154 85

−

85

𝑡−

85 83+9 85 √ 𝑃′ ) 2

1036−77.83−77.9√85

2 7√85+77

, 𝑃′ + 𝑄′ = =

=

154 85

, 𝑃′

83−9√85 2

+

1036 85 −5355−77.9√85

85 85 154−7 85−77 77−7√85 √ 𝑃′ = = , 𝑄′ = = 9.852 85 85 85 16 1 83−9 85 √ 𝛽𝑗−1 = (−1)𝑗−1 − ((77 + 7√85)( )𝑗 + (77 − 85 85 2 83+9√85 𝑗 7√85)( ) ), ∀𝑗 ≥ 1 2

𝑓𝑜𝑟 𝑗 = 1, 𝛽0 1 16 − (77.83 − 77.9√85 + 7.83√85 − 63.85 + 77.83 + 77.9√85 − 7.83√85 − 63.85) 2 = 85 =

1 2

16− (2.77.83−2.63.85) 85

=

16−77.83+63.85 16−6391+5355 85

=

85

=

16−1036 85

=−

1020 85

= −12

The formula got confirmed .With congratulations to Ramanujan for his great courage. ℎ(𝑡) =

−10+8𝑡+2𝑡 2

−10+8𝑡+2𝑡 2

1−82𝑡−82𝑡 2 +𝑡 3

𝑖 = (1+𝑡)(1−83𝑡+𝑡 2) = ∑∞ 𝑖=0 𝛾𝑖 𝑡

= 𝛾0 + 𝛾1 𝑡 + 𝛾2 𝑡 2 + 𝛾3 𝑡 3 + ⋯ + 𝛾𝑛 𝑡 𝑛 +..= 2

𝐴′′

1+𝑡 (𝐵′′

186 85

−10+8𝑡+2𝑡 2

85

−16

𝛾0 = 𝛾1 =

85 −16 85 16

1+𝑡

2) =

(−1)𝑗 + 1

+

+

85

𝛾2 = −

85

16 85

186

186−1020

1−83𝑡+𝑡 2 ′′ )(1

85

+

− 12 =

186 −834 𝑡+ 85 85 1−83𝑡+𝑡 2

=

16 85

−

1+𝑡

85

−834

=

85

𝑃′′

+

83+9√85 𝑡− 2

, 𝐴′′ = 2 −

+

𝑄′′

=

83−9√85 𝑡− 2

186 85

−16 85

=−

16 85

𝑗 𝑗 ∑∞ 𝑗=0(−1) 𝑡 +

∞ 2 𝑗 𝑗 (186𝑡 − 834) ∑∞ 𝑗=0(83𝑡 − 𝑡 ) = ∑𝑗=0 𝛾𝑗 𝑡

𝛾𝑗 =

1

−

=

1−82𝑡−82𝑡 2 +𝑡 3 1

,𝐶 ′′ =

2)

𝐵′′ 𝑡+𝐶 ′′

−10 + 8𝑡 + 2𝑡 = 𝐴 − 83𝑡 + 𝑡 + 𝑡+𝐶 + 𝑡) = (𝐴′′ + 𝐶 ′′ ) + (−83𝐴′′ + 𝐵′′ + 𝐶 ′′ )𝑡 + (𝐴′′ + 𝐵′′ )𝑡 2 𝐴′′ + 𝐶 ′′ = −10, −83𝐴′′ + 𝐵′′ + 𝐶 ′′ = 8, 𝐴′′ + 𝐵′′ = 2,𝐵′′ − 𝐶 ′′ = 12 -83(2 − 𝐵′′ ) + 𝐵′′ + 𝐶 ′′ = 8,84𝐵′′ + 𝐶 ′′ = 166 + 8 = 174,85𝐵′′ = 186, 𝐵′′ =

′′ (1

+

16 85

85 1

(186(83)𝑗−1 -834(83)𝑗 + 𝑐𝑜𝑟𝑟𝑒𝑐𝑡𝑖𝑜𝑛𝑠), ∀𝑗 ≥ 1

. −834 = −

850

= −10

85

(186 − 834.83) =

85

+

1 85

16 85

1

+

85

(−69036) = −

69020 85

= −812 𝑗

1

(186.83 − 834.83.83 + 834, 𝑤𝑖𝑡ℎ 𝑐𝑜𝑟𝑟𝑒𝑐𝑡𝑖𝑜𝑛 − 834(−1)2 , 𝑗 = 85

1

1

(−16 + 15438 + 834(−6888)) = 85 (15422 − 5744592) = 85

(−5729170)=-67402

𝛾𝑘−1 = − 𝑃′′ (𝑡 −

16

85 83−9√85 2

(𝑃′′ + 𝑄′′ ) = 83−9√85 𝑃′′ ( ) 2

−𝑃′′ 9√85 = 𝑃′′ =

83−9√85 𝑘 ) 2 83+9√85

(−1)𝑘−1 − 𝑃′′ ( ) + 𝑄′′ (𝑡 −

186

2

)=

83−9√85

, 𝑃′′ (

85

2

186

83+9√85 𝑘 ) 2 186 834

− 𝑄′′ ( 85

𝑡−

83+9√85

)+𝑄′′ (

83+9√85

2 834

)=

85 834 85

− 𝑃′′ ) ( ) = 85 85 2 834 186 83+9√85 834−93.83−93.9√85 +(

−

(

)=

85 85 2 6885√85+93.9.85 9√85+93 ′′ = ,𝑄 9.85.85 85

=

85 186−9√85−93 85

=

=

−6885−93.9√85

85 93−9√85 85

𝛾𝑘−1 = −

16 85

(−1)𝑘−1 −

83+9√85

9√85) (

1 85

((9√85 + 93)(

83−9√85 𝑘 ) 2

+ (93 −

𝑘

) ), ∀𝑘 ≥ 1

2

𝑓𝑜𝑟 𝑘 = 1, 𝛾0 1 −16 − (9.83√85 − 81.85 + 93.83 − 9.93√85 + 93.83 + 9.93√85 − 9.83√85 − 81.85) 2 = 85 1 2

−16− (2.93.83−2.81.85)

=

−16−834 85

=

85 −850 85

=

−16−(93.83−81.85) 85

=

−16−93.83+81.85 85

=

−16−7719+6885 85

=

= −10,the formula got confirmed. Congratulations to Ramanujan for greater aspirations. Solutions to the Ramanujan indicial equation 𝑎𝑛+3 = 82𝑎𝑛+2 + 82𝑎𝑛+1 − 𝑎𝑛 𝑛 𝑛+3 𝑎𝑛 = 𝜇 , 𝜇 = 82𝜇𝑛+2 + 82𝜇𝑛+1 − 𝜇𝑛 ÷ 𝜇𝑛 , 𝜇3 − 82𝜇2 − 82𝜇1 + 1 = 0 (𝜇 + 1)(𝜇2 − 𝜇 + 1 − 82𝜇) = 0 83 9 𝜇 = −1, 𝜇2 − 83𝜇 + 1 = 0, (𝜇 − )2 = (± √85)2 2

1

2

𝜇 = (83 ± 9√85) 𝑛

1

2

1

𝑛

𝑛

𝑎𝑛 = 𝑝(−1) + 𝑞( (83 + 9√85)) + 𝑟( (83 − 9√85)) is the general solution 2

2

𝑎0 = 𝑝 + 𝑞 + 𝑟 = 1 1

1

𝑎1 = −𝑝 + 𝑞 ( (83 + 9√85)) + 𝑟 ( (83 − 9√85)) = 135 1

2

1

2

2

𝑎2 = 𝑝 + 𝑞( (83 + 9√85)) + 𝑟( (83 − 9√85))2 =11161 2 2 So that we can find p, q, r such that the sequence is a superposition of three solutions given initial boundary values of S.Ramanujan for a. Similarly we can write down b, c, alpha, beta and gamma using their initial boundary values given by Ramanujan then they must automatically satisfy Ramanujan’s cubic number relations for any n. 1

1

𝑞 ( (83 + 9√85) + 1) + 𝑟( (83 − 9√85) + 1 )=136 2 2 1

1

𝑞 (85 + 9√85) + 𝑟 (85 − 9√85) = 136 2 2 1

1

1

1

𝑞( (83 + 9√85))( (85 + 9√85)) + 𝑟 ( (83 − 9√85)) ( (85 − 9√85)) = 2 2 2 2 11296

1

1

1

1

𝑞 ( (83 + 9√85) (85 + 9√85)) + 𝑟 (83 + 9√85) (85 − 9√85) = 2 2 2 2 1

136( (83 + 9√85)) 2 1

𝑟(9√85) (85 − 9√85) = 2 68.83 + 68.9√85 − 11296 = 5644 − 11296 + 612√85 = −5652 + 612√85 = 1 2 9(−628 + 68√85) = 𝑟(9.85) (√85 − 9) = 𝑟(9.85) 2

𝑟=

1 85

√85+9

1

(9 + √85)(−314 + 34√85) = 85 (−314.9 + 9.34√85 − 314√85 +

34.85) = 1

1

1

(−2826 + 2890 + 306√85 − 314√85) = 85 (64 − 8√85) 85 1

1

𝑞 85(4) = 𝑞2(85) = (85 − 9√85)(136 − (85 − 9√85) (64 − 8√85)) 2 2 85 𝑞= 1 852

1

4(85)2

{(85 − 9√85)(136.2.85 − 85.64 − 72.85 + 85.8√85 + 9.64√85)} =

{(85 − 9√85)(85(68 − 16 − 18) + (170 + 144)√85 )}

𝑞=

1

2

852

(85 − 9√85)(85.34 + 314√85) = 852 (85.85.17 − 9.157.85 −

9.17.85√85 + 85.157√85) =

2

85

(85.17 − 9.157 − 9.17√85 + 157√85) =

2

2

(1445 − 1413 − 153√85 + 157√85) = 85 (32 + 4√85) = 85 (𝑞 + 𝑟) = 1 − 𝑎𝑛 =

128 85

=

85−128 85

=

−43

1 (−1)𝑛 + {(64 + 85 85 83−9√85 𝑛

8√85) (

−43

64+8√85 85

,𝑝 = 1 −

85

8√85)(

83+9√85 𝑛 ) 2

+ (64 −

) } , ∀𝑛 𝑖𝑛𝑐𝑙𝑢𝑑𝑖𝑛𝑔 𝑧𝑒𝑟𝑜

2

b, c, alpha ,beta, gamma can also be derived using their three initial boundary values given by Ramanujan 1

1

𝑎𝑛 = 𝑝(−1)𝑛 + 𝑞( (83 + 9√85))𝑛 + 𝑟( (83 − 9√85))𝑛 2 2 −43

𝑎𝑛−1 = ( (−

85

)

404+44√85

) (−1)𝑛 − ( 85 1

85 43

−1

𝑝=−

85

85

𝑛

1

) (2 (83 + 9√85)) +

1

404+44√85

−(

+ (−

404−44√85

)( (83 − 9√85))𝑛 2

85 −43

𝑎𝑛 = (

(−1)𝑛−1

404−44√85 1 )( (83 85 2 1

1

+ 9√85))( (83 + 9√85 ))𝑛 + 2

) (2 (83 − 9√85))(2 (83 − 9√85 ))𝑛

,𝑞 =

(202 − 22√85)(83 + 9√85) = 85

22.83√85 + 202.9√85) = −

1

85

−1 85

(202.83 − 22.9.85 −

(16766 − 16830 − 1826√85 + 1818√85) =

−1

(−64 − 8√85), 𝑟 =

85

−1 85

1

(202 + 22√85 )(83 − 9√85 )= − 85 (202.83 −

22.9.85 − 202.9√85 + 22.83√85) = 𝑎𝑛 =

43 85

(−1)𝑛+1 +

1 85

−1

85 1

(−64 + 8√85) 1

[(64 + 8√85)( (83 + 9√85 ))𝑛 + (64 − 8√85 )( (83 − 2

2

𝑛

9√85 )) ] 𝑓𝑜𝑟 𝑛 = 0, 𝑎0 =

43 85

(−1) +

128 85

=

128−43 85

=

85 85

=

1, 𝑠𝑜 𝑡ℎ𝑎𝑡 𝑓𝑜𝑟𝑚𝑢𝑙𝑎 𝑔𝑜𝑡 𝑐𝑜𝑛𝑓𝑖𝑟𝑚𝑒𝑑, 𝐶𝑜𝑛𝑔𝑟𝑎𝑡𝑢𝑙𝑎𝑡𝑖𝑜𝑛𝑠 𝑡𝑜 𝑅𝑎𝑚𝑎𝑛𝑢𝑗𝑎𝑛 Similarly b, c, alpha, beta and gamma can be worked out for any n to calculate but they are not appearing by Ramanujan himself at the surface level in his note books or published papers but he provided boundary values but some others came up with similar results but they are very hard to achieve after Ramanujan but prior to him similar results should be searched with Euler king of mathematics. People of mathematics compare Ramanujan only with Euler. 43

𝐹𝑜𝑟 𝑛 = 4, 𝑎3 = − (− ) + (− 85 (−

404+44√85 85

404−44√85

4

1

) (2 (83 + 9√85)) +

85

1

) (2 (83 − 9√85 )4

1

1

( (83 − 9√85))3 = (833 + 3.832 . −9√85 + 3.83. (−9√85 )2 + (−9√85)3 ) = 2 8 1 8

(833 − 27.832 √85 + 243.81.85 − 729.85√85) 1

1

= (571787 + 1673055 − 186003√85 − 61965√85) = (2244842 − 8 8 1

247968√85)= . 1122421 − 30996√85 4

1

1

( (83 + 9√85))3 = (833 + 3.832 . 9√85 + 3.83. (9√85 )2 + (9√85 )3 ) = 1 8

2

8

3

2

(83 + 27.83 √85 + 243.83.85 + 729.85√85) 1

= (83(832 + 243.85) + 9(3.832 + 81.85)√85 ) 8

1+53𝑥+9𝑥 2 1−82𝑥−82𝑥 2 +𝑥 3

𝑖 = ∑∞ 𝑖=0 𝑎𝑖 𝑥 =

𝑖 𝑖 𝐴 ∑∞ 𝑖=0(−1) 𝑥 −

𝐵 (

83+9√85 ) 2

𝐴 1−(−𝑥)

+

𝐵 83+9√85 𝑥−( ) 2

𝑥

𝑖 ∑∞ 𝑖=0( 83+9√85 ) − 2

+

𝐶 83−9√85 ) 2

(

𝐶 83−9√85 𝑥−( ) 2

=

𝑥

𝑖 ∑∞ 𝑖=0( 83−9√85 ) 2

83+9√85 𝑖 83−9√85 𝑖 𝑎𝑖−1 = 𝐴(−1)𝑖−1 − 𝐵( ) − 𝐶( ) , ∀𝑖 ≥ 1 2 2 −43 1 83−9√85 𝑖 𝑎𝑖−1 = (−1)𝑖−1 − {(404 + 44√85)( ) + (404 − 85 85 2 83+9√85 𝑖 44√85)( ) } ,∀𝑖 ≥ 1, 2

for i=1,𝑎0 = 9√85)} = − −43+128 85

=

85 85

−43 85 43 85

+

+

−1

85.2 −1 85

{(404 + 44√85)(83 − 9√85) + (404 − 44√85)(83 +

{404.83 − 22.18.85} =

−43 85

+−

1 85

(33532 − 33660) =

= 1,the formula confirmed. Congratulations to S. Ramanujan for a

remarkable discovery 1

1 + 53𝑥 + 9𝑥 2 = 𝐴(𝑥 2 − 83𝑥 + 1) + 𝐵(1 + 𝑥) (𝑥 − (83 − 9√85)) + 2 1

1

𝐶(1 + 𝑥)(𝑥 − (83 + 9√85)) = (𝐴 + 𝐵 + 𝐶)𝑥 2 + (−83𝐴 + (1 − (83 − 2 2 1

1

1

9√85)) 𝐵 +C(1- (83 + 9√85)))𝑥 + (𝐴 − (83 − 9√85)𝐵 − (83 + 9√85)𝐶) 2

2

1

53

1

(−81 + 9√85)𝐵 − 2.83 (81 + 9√85)𝐶=83 ,A+2 (−83 + 2.83

A+B+C=9,-A+ 1

2 1

9√85)𝐵 − (83 + 9√85)𝐶 = 1 2 (85+9√85)𝐵 + (85 − 9√85)𝐶 = 2(9.83 + 53) = 2(747 + 53) = 2(800) = 1600,(85-9√85)𝐵 + (85 + 9√85)𝐶 =8.2=16 {(85-9√85)2 − (85 + 9√85 )2 }C= 1600(85 − 9√85) − 16(85 + 9√85) = 1584.85 − 1616.9√85 = −36.85√85C,1584.85√85 − 1616.9.85 = 1

4

−36.85.85𝐶, −1584√85+1616.9=36.85C,C= {404 − 44√85} = {101 − 85 85 11√85},85.4B=16(85 + 9√85) − 1

4

85

(101 − 11√85)(85 + 9√85 )2 1

B= {4(85 + 9√85) − (101 − 11√85)(166 + 18√85)} = {340 + 36√85 − 85 85 101.166 + 11.166√85 − 101.18√85 + 11.18.85} = 16830 + (36 + 1826 − 1818)√85} = 4

1 85

𝐴 = 9 − (𝐵 + 𝐶) = 9 −

808 85

=

765−808 85

85

{340 − 16766 + 1

{340 + 64 + 44√85}= {404 + 85

44√85}= {101 + 11√85} 85

1

=−

43 85

2−26𝑥−12𝑥 2

𝑗 = ∑∞ 𝑗=0 𝑏𝑗 𝑥 =

1−82𝑥−82𝑥 2 +𝑥 3

𝐴′ 1−(−𝑥)

83−9√85 𝑗 ) 2

𝑏𝑗−1 = 𝐴′ (−1)𝑗−1 − 𝐵′ (

+

𝐵′ 83+9√85 ) 2

𝑥−(

+

𝐶′ 83−9√85 ) 2

𝑥−(

83+9√85 𝑗 ) , ∀𝑗 2

− 𝐶 ′(

≥1 1

2 − 26𝑥 − 12𝑥 2 = 𝐴′ (𝑥 2 − 83𝑥 + 1) + 𝐵′ (1 + 𝑥) (𝑥 − (83 − 9√85)) + 2 1

𝐶 ′ (1 + 𝑥)(𝑥 − (83 + 9√85)) 2 𝐴′ + 𝐵′ + 𝐶 ′ = −12 1 −𝐴′ + (−81 + 9√85)𝐵′ − ′

2.83

1

1

(81 + 9√85)𝐶 ′ = 2.83

1

′

−26 83

′

𝐴 + (−83 + 9√85)𝐵 − (83 + 9√85)𝐶 = 2 2 2 (85 + 9√85)𝐵′ + (85 − 9√85)𝐶 ′ =

−12.83−26 83

. 2.83 = −(24.83 + 52) = −2044

(85 − 9√85)𝐵′ + (85 + 9√85)𝐶 ′ = −14.2 = −28 {(85 − 9√85)2 − (85 + 9√85)2 }𝐶 ′ = −2044(85 − 9√85) + 28(85 + 9√85) = −2016.85 + 2072.9√85 = (−2.9√85)(2.85)𝐶 ′ , 𝐶 ′ = 7

1

1 85

(−518 + 56√85) =

1

(−74 − 8√85) = 85 (518 − 56√85),𝐵′ = 85−9√85 {−28 − (85 + 85 9√85)

1

(−518 + 56√85)} = 85

9.518√85 − 9.56.85} = 9.259)√85) = 8√85) =

1 85

𝑏𝑗−1 =

2.85.85

4.85

85+9√85 2.85

(85(−14 + 259 − 252) + (−28.85 +

(−7.85 − 49√85) = −

1036 85

(−1)

, 𝐴′ = −12 +

𝑗−1

−

1

1036 85

{(−518 +

=

7(√85+9)(√85+7)

85

−

2.85

−1020+1036

=−

7 85

(74 +

16+42994−42840

=

16

(−518 −

{−518.83 + 518.9√85 − 56.83√85 + 56.9.85 −

16+154 85

=

− 9√85) + (−518 + 56√85)(83 +

518.83 − 518.9√85 + 56.83√85 + 56.9.85} = 85

2.85

85 85 83−9√85 𝑗 56√85)( ) + 2

85 85 83+9√85 𝑗 56√85)( ) , ∀𝑗 ≥ 1 2 16 1 for 𝑏0 = − { (-518-56√85)(83 85 2.85 −2056 1

9√85)} =

{−28.85 + 518.85 − 85.56√85 +

(−518 − 56√85)

𝐵′ + 𝐶 ′ = − 16

85+9√85

85+9√85

=

170 85

=2

16+518.83−56.9.85 85

=

2+8𝑥−10𝑥 2 1−82𝑥−82𝑥 2 +𝑥 3

𝑘 = ∑∞ 𝑘=0 𝑐𝑘 𝑥 =

𝑐𝑘−1 = 𝐴′′ (−1)𝑘−1 − 𝐵′′ (

𝐴′′ 1−(−𝑥)

83−9√85 𝑘 ) 2

+

𝐵′′

+ 83+9√85

𝑥−(

C′′ 83−9√85 ) 2

) x−(

2

83+9√85 𝑘 ) , ∀𝑘 2

− 𝐶 ′′ (

≥1 1

2 + 8𝑥 − 10𝑥 2 = 𝐴′′ (𝑥 2 − 83𝑥 + 1) + 𝐵′′ (1 + 𝑥) (𝑥 − (83 − 9√85)) + 2 1

𝐶 ′′ (1 + 𝑥)(𝑥 − (83 + 9√85)) 2

𝐴′′ + 𝐵′′ + 𝐶 ′′ = −10 1 −𝐴′′ + (−81 + 9√85)𝐵′′ − 2.83

1

′′

1

1

8

(81 + 9√85)𝐶 ′′ = 83 2.83

𝐴 + (−83 + 9√85)𝐵′′ − (83 + 9√85)𝐶 ′′ = 2 2 2 (85 + 9√85)𝐵′′ + (85 − 9√85)𝐶 ′′ =

−830+8 83

. 2.83 = −822.2 = −1644

(85 − 9√85)𝐵′′ + (85 + 9√85)𝐶 ′′ = −12.2 = −24 {(85 − 9√85)2 − (85 + 9√85)2 }𝐶 ′′ = −1644(85 − 9√85) + 24(85 + 9√85)=85(24-1644)+9√85(24 + 1644)=1620.85+9(1668)√85=(−2.9√85(2.85)𝐶 ′′ =-4.9.85√85𝐶 ′′ 𝐶 ′′ =

−1 85

(417 − 45√85),𝐵′′ =

45√85)} = 85+9√85 4.85.85 85+9√85 4.85.85

85+9√85 85.4.85

3

85 −834 85

{85(−24 + 417 − 405) + 9(417 − 425 )√85} = 1 85.85

,𝐴′′ = −10 +

45√85

−1 85

(417 −

{−24.85 + (85 + 9√85)(417 − 45√85)} =

3

−16

85+9√85 4.85.85

{−85.12 −

(85 + 9√85)(3.85 + 9.2√85) = − 85 (√85 + 9)(√85 + 6) =

(85 + 54 + 15√85) = −

𝑐𝑘−1 =

{−24 − (85 + 9√85) 85−9√85

{−24.85 + 85.417 − 45.85√85 + 9.417√85 − 9.45.85} =

9.8√85} = − −

1

834 85

(−1)𝑘−1 +

= 1

3(139+15√85)

−850+834 85

85

=−

=

16

85

(417 + 45√85),𝐵′′ + 𝐶 ′′ =

85 83−9√85 𝑘 ) 2

{(417 + 45√85)(

85 85 83+9√85 𝑘 )( ) }, ∀𝑘 ≥ 2

−1

+ (417 −

1 :Maxima Programmer’s Formula to come up with

any k by Robotic Ramanujan under Wolfram software for an another sequence for Sloan, we have six such sequences for a, b, c, alpha, beta and gamma with Ramanujan’ s initial values, raising and waving both the hands from the grave a Ghost with infinity of numbers in the sequence for Hardy’s Mathematical Fans interested in 1729 Taxi Cab Number.

for 𝑘 = 1 𝑐0 =

−16

85 83+9√85

45√85) (

2

+

1

83−9√85

{(417 + 45√85) ( 85

) + (417 −

2

1

)} = 2.85 {−32 + 417.83 − 417.9√85 + 45.83√85 − 45.9.85 + 1

417.83 + 417.9√85 − 45.83√85 − 45.9.85}= {−16 + 417.83 − 45.9.85} = 1 85

{−16 + 34611 − 34425} =

186−16 85

=

170 85

85

= 2,the formula got confirmed with

Congratulations S.Ramanujan for the prediction for many centuries in mathematics. Even without any computer program one can keep on substituting values for k=1,2,3,4,5,6,7,8,9,10,11…….and reproduced the relevant number sequence, but with proper working computer program computer does the mathematical labor work that Srinivasa Ramanujan was having hard time on various aspects in his type of Number Theory even present author does the work you cannot always expect that much genius as Ramanujan was in that birth others also do the work some times more than Ramanujan because world has evolved with binary discoveries in computer age computer do more efficient work than a human no body can always expect a biological Ramanujan that is why we suggested one day technology in it’s own advancement come up with a Robotic Ramanujan. Our hard core mathematical work related to this problem has ended. In Dickson & Hardy and Wright you can observe the following formula (9𝑞4 −3𝑞)3 + (9𝑞3 − 1)3 = (9𝑞4 )3 − 1 𝑓𝑜𝑟 𝑞 = 1, (9 − 3)3 + (9 − 1)3 = 93 − 1 63 + 83 = 93 − 1 the cubic number relation not derivable under the above project of three polynomial ratios of Ramanujan .It derives an another sequence. 𝑞 = 2, (9. 24 − 3.2)3 + (9. 23 − 1)3 = (9. 24 )3 − 1 1383 + 713 = 1443 − 1 (9. 34 − 3.3)3 + (9. 33 − 1)3 = (9. 34 )3 − 1 7203 + 2423 = 7293 − 1 (9. 44 − 3.4)3 + (9. 43 − 1)3 = (9. 44 )3 − 1 2292 3 + 5753 = 23043 − 1 ……………………………… • This sequence of cubic relations do not have (−1)𝑛 but instead -1. If write n instead of q the cubic number relation becomes 1 + (9𝑛4 − 3𝑛)3 + (9𝑛3 − 1)3 = (9𝑛4 )3 1 + 𝑎𝑛3 + 𝑏𝑛3 = 𝑐𝑛3 , 𝑎𝑛 = 9𝑛4 − 3𝑛, 𝑏𝑛 = 9𝑛3 − 1, 𝑐𝑛 = 9𝑛4 𝑓𝑜𝑟 𝑛 = 0, 𝑎0 = 0, 𝑏0 = −1, 𝑐𝑛 = 0,1 + (−1)3 = 1 − 1 = 0

1 originate by factor 1-x in the denominator instead 1+x of Ramanujan of his cubic number relations. Even though second cubic number relation of this sequence appears in his hand written notes of lost note book he has not provided out of which polynomial ratios it originate.The denominator of the polynomial ratio of the form D=(1-x)(1-mx+x2)=1-mx+x2-x+mx2-x3=1-(m+1)x+(m+1)x2 -x3 As in the case of Ramanujan if we take m=83 then D=1-84x+84x2-x3 ∞ 𝑖 4 𝑛 0 1 2 3 4 N1/D1=∑∞ 𝑖=0 𝑎𝑖 𝑥 = ∑𝑛=0(9𝑛 − 3𝑛)𝑥 =0x +6x +138x +720x +2292x +. ∞ 𝑖 3 𝑛 0 1 2 3 4 N2/D2=∑∞ 𝑖=0 𝑏𝑖 𝑥 = ∑𝑛=0(9𝑛 − 1)𝑥 = -1x +8x +71x +242x +575x +… ∞ 𝑖 4 𝑛 0 1 2 3 4 N3/D3=∑∞ 𝑖=0 𝑐𝑖 𝑥 = ∑𝑖=0(9𝑛 )𝑥 =0x +9x +144x +729x +2304x +… This is re formalism of S. Ramanujan with Dickson, Hardy and Wright 𝑛 .Mean while 1/1-x=1+x1+x2+x3+x4+…..=∑∞ 𝑛=0(1)𝑥 ,so that 1 is reproduced without difficulty. Reproducing the rest is difficult that is what Ramanujan has done for 1729 the Hardy’s Taxi Cab Number. But this time Taxi Cab Number has changed from 1729 to 728 we say Wright’s Taxi Cab Number in the case he also has visited Ramanujan by Taxi Cab with 728 Number assuming such a case with Wright for Ramanujan. This is where Ramanujan is a Genius but this case he has not handled but only listed.Ramanujan’ s role is clear in the case of 1729. x2-mx+1=0,(x-m/2)2+1-m2/4=0,x={m±(m2-4)1/2}/2 𝑎𝑛 = 𝑃(1) − {𝑄((

𝑚−√𝑚2 −4 2

))𝑛+1 + 𝑅 ((

𝑚+√𝑚2 −4 2

))𝑛+1 } , ∀𝑛 ≥ 1

for b, c we have prime and double prime over P,Q and R This is the style of the solution but we have to reproduced coefficients of x to powers numerically. By preserving m=83 as it is we define a new sequence 𝜎𝑛 = 1

1

{(

9√85

83+9√85 𝑛 ) 2

(36.83√85) =

36√85 1 )3 }= {(833 72√85 2

83−9√85 𝑛 ) } ,𝜎0 = 0, 𝜎1 = 1, 𝜎2 = 2 1 83 𝜎3 = {(83 + 9√85 )3 − (83 − 72√85

−(

+ 3. 832 . 9√85 + 3.83(9√85)2 + 93 (√85 )3 ) − (833 −

3.83 9√85 + 3.83(9√85)2 − 93 (√85)3 } = 3

3

9 (√85) }=

1 4

9√85

(3.832

3

1 72√85

{2(3.832 . 9√85 +

+ 27.85) = (6889 + 2995) = 4

3

(9184)=3(2298)=6894

4

1 83+9√85 2 𝑛 83+9√85 83−9√85 83−9√85 2 𝑛 {(( ) ) -2(( )( ) )𝑛 + (( ) ) } 81.85 2 2 2 2 1 83+9√85 2𝑛 83−9√85 2𝑛 {( ) +( ) − 2} 81.85 2 2 9 83+9√85 4𝑛 83−9√85 4𝑛 83+9√85 2𝑛 9𝜎𝑛4 = 2 2 {( ) +( ) − 4( ) − 81 85 2 2 2 83−9√85 2𝑛 1 83+9√85 3𝑛 83−9√85 𝑛 4( ) + 6} 9𝜎𝑛3 = {( ) + 3( ) − 2 81.85√85 2 2

𝜎𝑛2 =

3(

83+9√85 𝑛 ) 2

−(

83−9√85 3𝑛 ) } 2

−3𝜎𝑛 =

−1 3√85

{(

83+9√85 𝑛 ) 2

−(

=

83−9√85 𝑛 ) } 2

-1 9𝜎24 − 3𝜎2 = 9(83)4 − 3(83) = 3(83)(3(83)3 − 1) = 249(1715360) = 427124640,9𝜎23 − 1 = 9(83)3 − 1 = 5146082,9𝜎24 = 9(83)4 = 427124889 ((𝑥 − 9√𝑥) − 1) ((𝑥 + 9√𝑥) − 1) = (𝑥 − 9√𝑥)(𝑥 + 9√𝑥) − (𝑥 − 9√𝑥) − (𝑥 + 9√𝑥) + 1 = 𝑥 2 − 81𝑥 − 2𝑥 + 1 = 𝑥 2 − 83𝑥 + 1 𝐷(𝑥) = (1 + 𝑥)(1 − 83𝑥 + 𝑥 2 ) = (1 + 𝑥)(𝑥 − 9√𝑥 − 1)(𝑥 + 9√𝑥 − 1) = 1 − 82𝑥 − 82𝑥 2 + 𝑥 3 the denominator of Ramanujan 𝐷(𝑥) = (1 − 𝑥)(1 − 83𝑥 + 𝑥 2 ) = (1 − 𝑥)(𝑥 − 9√𝑥 − 1)(𝑥 + 9√𝑥 − 1) = 1 − 84𝑥 + 84𝑥 2 − 𝑥 3 the new denominator given for Ramanujan 𝑎𝑛+3 = 82𝑎𝑛+2 + 82𝑎𝑛+1 − 𝑎𝑛 the indicial equation of Ramanujan 𝑎𝑛+3 = 84𝑎𝑛+2 − 84𝑎𝑛+1 + 𝑎𝑛 the new indicial equation given for Ramanujan 𝑎𝑛 = 𝜇𝑛 the trial solution then 𝜇3 − 82𝜇2 − 82𝜇 + 1 = 𝜇3 + 1 − 82𝜇(1 + 𝜇) = (𝜇 + 1)(𝜇2 − 𝜇 + 1) − 82𝜇(1 + 𝜇) = (𝜇 + 1)(𝜇2 − 83𝜇 + 1) = (𝜇 + 1)(𝜇 − 9√𝜇 − 1)(±) = 9 (𝜇 + 1)((√𝜇)2 − 9√𝜇 − 1)((√𝜇)2 + 9√𝜇 − 1) = (𝜇 − (−1))((√𝜇 − )2 − 2 9 2 85 9+ 85 √ √ + ) − ( )2 ) = (𝜇 − (−1))(√𝜇 − ( ))(√𝜇 − 2 2 2 −9+√85 −9−√85 ))(√𝜇 − ( 2 ))(√𝜇 − ( 2 )) = 0 the equation of Ramanujan 2 3 2 (𝜇

√85 ( )2 )((√𝜇 2 9−√85

(

2 3

𝜇 − 84𝜇 + 84𝜇 − 1 = 𝜇 − 1 − 84𝜇(𝜇 − 1) = − 1)(𝜇 + 𝜇 + 1) − 84𝜇(𝜇 − 1) = (𝜇 − 1)(𝜇2 − 83𝜇 + 1) = (𝜇 − 1)(𝜇 − 9√𝜇 − 1)(𝜇 + 9√𝜇 − 1) = (𝜇 − 1)((√𝜇)2 − 9√𝜇 − 1)((√𝜇)2 + 9√𝜇 − 1) =

9+√85 )) (√𝜇 2

(𝜇 − 1) (√𝜇 − ( −9−√85 )) 2

(

−(

9−√85 )(√𝜇 2

−9+√85 ))(√𝜇 2

−(

−

= 0 the new equation given for Ramanujan

Here 𝜇 = −1 𝑜𝑟 + 1 𝑎𝑛𝑑 (√𝜇)2 ∓ 9√𝜇 − 1 = 0, (√𝜇 )2 = ±9√𝜇 + 1 𝑡𝑎𝑘𝑒 √𝜇 = 𝜀 then 𝜀 2 = ±9𝜖 + 1, 𝑡ℎ𝑒 𝑐𝑜𝑟𝑟𝑒𝑠𝑝𝑜𝑛𝑑𝑖𝑛𝑔 𝑖𝑛𝑑𝑖𝑐𝑖𝑎𝑙 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛𝑠 𝑎𝑟𝑒 𝜏𝑛+2 = ±9𝜏𝑛+1 + 𝜏𝑛 which are primitive indicial equations for root 𝜇 used in literature by Hirschhorn, M. Craig and Laughlin further by G. Andrews & B. Berndt in their own edition of Ramanujan’ s Lost note book(part-iv) whose origin was totally unknown till this factorization how they came out and what was it’ s role we have to square their solutions to get the solutions for 𝜇. 𝜏𝑛+ = 𝑛

9−√85 𝑛 −(−1)𝑛 9+√85 𝑛 9−√85 𝑛 ) },𝜏𝑛− = {( ) −( ) }= √85 2 2 √85 2 2 1 83+9√85 𝑛 83−9√85 𝑛 −(−1)𝑛 𝜏𝑛+ ,(𝜏𝑛± )2 = {( ) +( ) − 2(−1)𝑛 } 85 2 2 1 6887+747√85 𝑛 6887−747√85 𝑛 83+9√85 𝑛 (𝜏𝑛± )4 = 2 {( ) +( ) − 4(−1)𝑛 ( ) − 85 2 2 2 83−9√85 𝑛 4(−1)𝑛 ( ) + 6} 2 1 756+82√85 𝑛 9+√85 𝑛 9−√85 𝑛 (𝜏𝑛± )2 (𝜏𝑛+ ) = {( ) − 3(−1)𝑛 ( ) + 3(−1)𝑛 ( ) 3 (√85) 2 2 2 756−82√85 𝑛 ( ) },(𝜏𝑛± )2 (𝜏𝑛− ) = −(−1)𝑛 (𝜏𝑛± )2 𝜏𝑛+ 2 1

9+√85

{(

) −(

-1 𝑛 − ∑∞ 𝑛=0 𝑥 =

−1 1−𝑥

+ 𝑛 ∑∞ 𝑛=0 𝜏𝑛 𝑥 =

1

{ √85

± 4 𝑛 ∑∞ 𝑛=0(𝜏𝑛 ) 𝑥 = 1 83−9√85 1+( )𝑥 2

852

1−(

1 852

9+√85 )𝑥 2

−

1 9−√85 )𝑥 2

1−(

1

{

6887+747√85 1−( )𝑥 2

1

𝑥

+

2−6887𝑥

} = 1−9𝑥+𝑥 2 1

6887−747√85 1−( )𝑥 2

− 4(

1

83+9√85 1+( )𝑥 2

2+83𝑥

) + 6}=852 {1−6887𝑥+𝑥 2 − 4 (1+83𝑥+𝑥 2 ) + 6}

± 4 𝑛 ∑∞ 𝑛=0 9(𝜏𝑛 ) 𝑥 = 9

1

9

{

2−6887𝑥

1−6887𝑥+𝑥 2

−(

8+332𝑥

) + 6} =

1+83𝑥+𝑥 2 2 (2−6887𝑥)(1+83𝑥+𝑥 )−(8+332𝑥)(1−6887𝑥+𝑥 2 )+6(1−6887𝑥+𝑥 2 )(1+83𝑥+𝑥 2 )

{

852

(1−6887𝑥+𝑥 2 )(1+83𝑥+𝑥 2 )

}

+

−

= 2+166𝑥+2𝑥 2 −6887𝑥−6887.83𝑥 2 −6887𝑥 3 −8+8.6887𝑥−8𝑥 2 −332𝑥+332.6887𝑥 2 −332𝑥 3 +6−6.6887𝑥+6𝑥 2 + 6.83𝑥−6.6887.83𝑥 2 +6.83𝑥 3 +6𝑥 2 −6.6887𝑥 3 +6𝑥 4 { } 2 85 1+83𝑥+𝑥 2 −6887𝑥−6887.83𝑥 2 −6887𝑥 3 +𝑥 2 +83𝑥 3 +𝑥 4 9𝑥 7219−1714857𝑥 1 −48043𝑥 2 +6𝑥 3 9

=

± 4 𝑛 ∑∞ 𝑛=0 𝑥 (9(𝜏𝑛 ) −

=

{

}

852 1−6804𝑥−571619𝑥 2 − 6804𝑥 3 +𝑥 4 9 2−6887𝑥 8+332𝑥 3(𝜏𝑛+ )) = 2 { − 2 85 1−6887𝑥+𝑥 1+83𝑥+𝑥 2

+ 6} −

3𝑥 1−9𝑥+𝑥 2

( 7219−1714857𝑥−48043𝑥 2 +6𝑥 3 )(3−27𝑥+3𝑥 2 )−7225(1−6804𝑥−571619𝑥 2 −6804𝑥 3 +𝑥 4 )

3𝑥

{ 2

(1−6804𝑥−571619𝑥 2 −6804𝑥 3 +𝑥 4 )(1−9𝑥+𝑥 2 ) 85 − 𝑛 𝑛 ∞ + 𝑛 ∑∞ 𝑛=0 𝜏𝑛 𝑥 = −(−1) ∑𝑛=0 𝜏𝑛 𝑥 1 1 1 1 ± 2 + 𝑛 ∑∞ − + 3( 9−√85 𝑛=0(𝜏𝑛 ) (𝜏𝑛 ) 𝑥 = (√85)3 { 756+82√85 756−82√85 1−( )𝑥 1−( )𝑥 1+( )𝑥 2

1 9+√85 1+( )𝑥 2

)} =

1

{

82𝑥

85 1−756𝑥−𝑥 2

+(

3𝑥 1+9𝑥−𝑥 2

−

2

1

2 2 2) )+3𝑥(1−756𝑥−𝑥 82𝑥(1+9𝑥−𝑥

)}=85{

(1−756𝑥−𝑥 2 )(1+9𝑥−𝑥 2 )

85𝑥+(82.9−756.3)𝑥 2 −85𝑥 3

1

}

}

𝑥(1−18𝑥−𝑥 2 )

= { } = 1−747𝑥−6806𝑥 2 +747𝑥 3+𝑥 4 85 1+9𝑥−𝑥 2 −756𝑥−756.9𝑥 2 +756𝑥 3 −𝑥 2 −9𝑥 3 +𝑥 4 ± 2 + 𝑛 ∑∞ 𝑛=0(9(𝜏𝑛 ) (𝜏𝑛 ) − 1)𝑥 =

=

9

82𝑥

3𝑥

1

+( { )} − 1−𝑥 85 1−756𝑥−𝑥 2 1+9𝑥−𝑥 2

9𝑥(1−𝑥)(1−18𝑥−𝑥 2 )−(1−747𝑥−6806𝑥 2 +747𝑥 3 +𝑥 4 )

=

(1−747𝑥−6806𝑥 2 +747𝑥 3 +𝑥 4 )(1−𝑥) 2 (9𝑥−9𝑥 )−9.18𝑥 2 +9.18𝑥 3 −9𝑥 3 +9𝑥 4 −1+747𝑥+6806𝑥 2 −747𝑥 3 −𝑥 4

=

1−747𝑥−6806𝑥 2 +747𝑥 3 +𝑥 4 −𝑥+747𝑥 2 +6806𝑥 3 −747𝑥 4 −𝑥 5 −1+756𝑥+6635𝑥 2 −594𝑥 3 +8𝑥 4

1−748𝑥−6059𝑥 2 +7553𝑥 3 −746𝑥 4 −𝑥 5 ± 2 − 𝑛 ± 2 + 𝑛 𝑛 ∞ ∑∞ 𝑛=0(𝜏𝑛 ) (𝜏𝑛 )𝑥 = −(−1) ∑𝑛=0(𝜏𝑛 ) (𝜏𝑛 )𝑥 1 9+√85 𝑛+1 9−√85 𝑛+1 (−1)𝑛 9+√85 𝑛+1 + − 𝜏𝑛+1 = {( ) −( ) } , 𝜏𝑛+1 = {( ) √85 2 2 2 √85 1 83+9√85 𝑛+1 83−9√85 𝑛+1 ± (𝜏𝑛+1 )2 = {( ) +( ) + 2(−1)𝑛 } 85 2 2

9+√85 83+9√85 𝑛 9−√85 83−9√85 𝑛 )( ) − 9(−1)𝑛 + ( )( ) } 85 2 2 2 2 + − 𝜏𝑛− 𝜏𝑛+1 = −𝜏𝑛+ 𝜏𝑛+1 ± + − 𝜏𝑛− 𝜏𝑛+1 + 𝜏𝑛+ 𝜏𝑛+1 =0 (𝜏𝑛± )2 + (𝜏𝑛+1 )2 = 1 85+9√85 83+9√85 𝑛 85−9√85 83−9√85 𝑛 ± {( )( ) +( )( ) } 9(𝜏𝑛+1 )2 85 2 2 2 2 1 765+83√85 83+9√85 𝑛 765−83√85 83−9√85 𝑛 + ) 2(𝜏𝑛+ 𝜏𝑛+1 = {( )( ) +( )( ) } 85 2 2 2 2 + 𝜏𝑛+ 𝜏𝑛+1 =

1

9−√85 𝑛+1 ) }, 2

−(

{(

+ ) 2(𝜏𝑛+ 𝜏𝑛+1 − 9(𝜏𝑛± )2 =

9(𝜏𝑛± )4 = 9(

1 √85

{(

1 85

9+√85 𝑛 ) 2

{(

9+2√85 83+9√85 𝑛 )( ) 2 2 9−√85 𝑛 4

−(

2

) })

+(

9−2√85 2

)(

82−9√85 𝑛 ) } 2

+

−3𝜏𝑛+ = −3( (−1)𝑛 3𝜏𝑛+ 9(𝜏𝑛+ )3 = 9(

1 √85 1

√85

9+√85 𝑛 ) 2

−(

9+√85 𝑛 ) 2

−(

{(

{(

9−√85 𝑛 ) } )3 2

,−3𝜏𝑛− = −(−1)𝑛 (−3𝜏𝑛+ ) =

9−√85 𝑛 3 ) }) 2

-1 𝜏0± = 0, 𝜏1± = 1, 𝜏2+ = 9, 𝜏2− = −9, 𝜏1± =1,reproduces 6,8,9 number triplet so things must be all right. Hirschhorn was keep on squaring the solutions to these primitive indicial equations to get the solutions to indicial equation of Ramanujan and thought full of how S.Ramanujan arrived these polynomial ratios corresponding to taxi cab number. Laughlin found similar polynomial ratios inspired by Ramanujan. This is Laughlin of Mathematics there is a Laughlin in Physics. Dickson has highlighted the discoveries made by Ramanujan under G.H.Hardy in Cambridge in his History of Theory of Numbers Part-ii very valuable book available under Google search. Ken Ono once Professor of Letters be able to scan the valuable pages of Dickson on Ramanujan Mathematics can make easily accessible to one and all. Hirschhorn was inspired by a lecture given by Dorothy Zeilberger addressing American Mathematical Society but not easily accessible, Professor Ken Ono has to make that also accessible to others meanwhile Zeilberger is a great lecturer in mathematics coming under You Tube. Hirschhorn used simple h notation in his indicial equations which are primitive achieving amazing identities of Ramanujan. Dickson mention on same cubic number relations as Ramanujan in his historical book referring to another name before Ramanujan that author also has given a strange story on those number relations that his ability to provide infinity of them but he is not telling how & there is a paragraph on it Dickson published his book in the same year Srinivasa Ramanujan has expired that is in 1920.Was that story teller on Ramanujan Numbers in Dickson’s Book was a former birth of Ramanujan. Dickson’ s volume-ii must be the most wonder full book in mathematics. There is a third volume also in it accessible via Google. There are so many things on numbers in Dickson un explicabaly interesting. Ramanujan did not pass Physiology Science of Human Body pioneered by Leonardo Da Vinci a Medical Subject given by a former Genius who lived before Ramanujan in the history of Science and Leonardo was a Master of All Arts and Sciences. Ramanujan became a Master only in Mathematics lost interest in other subjects doing excessive mathematics with too much obsession in it. Ramanujan failed Physiology twice in Bachelor of Arts Degree first year exam at two colleges

had to give up getting B.A. When he do not pass first year B.A how he get into second and third year he has to pass all subjects to get the degree in this case no body can excuse young Ramanujan he lost Scholarship. At young age he could have interest in every subject not only Mathematics & Mathematics is only one branch of science and art. His married life with Janaki his wife made him sick had to go under a surgery & got recovered. He was sick even in Madras. If he concentrated only on college mathematics with in the curriculum of first year B.A and work out all other subjects he could have pass B.A first year at one college either at Kumbakonam or Pachayappa. At young age Ramanujan had to pass the degree. It is not only Ramanujan who fail there are many others. When he was a kid he passed Geography, History, English, Mathematics and became first and won prizes. He could not continue the tradition that he had set up as a kid at early ages of his life. If Ramanujan did the sketches of human body as Leonardo da Vinci and made contributions in Physiology of Medicine than dull Mathematics that could have been better for his life. He wanted ugly mathematics that has no place in true world. Too much mathematics for what? one should do all other subjects at preliminary stages. In Dickson’s Book it is stated G. Osborn gave Young’s identity (𝑥 2 − 7𝑥𝑦 + 63𝑦 2 )3 + (8𝑥 2 − 20𝑥𝑦 − 42𝑦 2 )3 + (6𝑥 2 + 20𝑥𝑦 − 56𝑦 2 )3 = (9𝑥 2 − 7𝑥𝑦 + 7𝑦 2 )3 𝑥 = 1, 𝑦 = 0, 13 + 63 + 83 = 93 𝑥 = 1, 𝑦 = 1, 573 + (−54)3 + (−30)3 = 93 ,93 + 303 + 543 = 573 33 + 103 + 183 = 193 For further investigation on this type. 𝑁 = 𝑟𝑠 = 𝑥 3 + 𝑦 3 = (𝑥 + 𝑦)(𝑥 2 − 𝑥𝑦 + 𝑦 2 ) = (𝑥 + 𝑦)((𝑥 + 𝑦)2 − 3𝑥𝑦) = (𝑥 + 𝑦)3 − 3𝑥𝑦(𝑥 + 𝑦) , 𝑥 + 𝑦 = 𝑟, 𝑥 2 − 𝑥𝑦 + 𝑦 2 = 𝑠 = 𝑟 2 − 3𝑥𝑦, 𝑥𝑦 = (𝑟 2 − 𝑠)/3 ,𝑥𝑦 + 𝑦 2 = 𝑟𝑦, 𝑦 2 − 𝑟𝑦 + √ √

4𝑠−𝑟 2 3

4𝑠−𝑟 2 3

1 2

3

})3 ,4𝑠 ≥ 𝑟 2 , 𝑠 ≥ 1 2

3

4𝑠−𝑟 2

},𝑥 = 𝑟 − 𝑦 = {𝑟 ∓ √

𝑁 = 𝑟𝑠 = ( {𝑟 ∓ √

(𝑟 2 −𝑠)

𝑟2 4

𝑟 2 −𝑠

2

3

1

4𝑠−𝑟 2

2

3

},𝑁 = 𝑟𝑠 = ( {𝑟 ∓ √

𝑟

𝑟

2

√𝑠

, 𝑠 ≥ ( )2 , 2√𝑠 ≥ 𝑟 ,

(2√𝑠−𝑟)(2√𝑠+𝑟) 3 }) 3

If N=1729=19.91=13.133 Case 1.,r=19,s=91

1

= 0, (𝑦 − 𝑟)2 +

1

𝑟2 4

1

= 0, 𝑦 = {𝑟 ± 2

1

})3 + ( {𝑟 ± 2

≤2

+ ( {𝑟 ± √ 2

−

(2√𝑠−𝑟)(2√𝑠+𝑟) 3 }) , 2√𝑠 3

≥𝑟

1

4.91−192

2

3

N=19.91=( {19 ∓ √ √

364−361 3

1

364−361

2

3

})3 + ( {19 ± √

1

4.91−192

2

3

})3 + ( {19 ± √

1

})3 = ( {19 ∓ 2

1

1

2

2

})3 = ( {19 ∓ 1})3 + ( {19 ± 1})3 = 93 +

103 Case 2.,r=13,s=133 1

4.133−132

2

3

𝑁 = 13.133 = ( {13 ∓ √ 1

532−169

2

3

= ( {13 ∓ √ 1

363

2

3

( {13 ± √

1

4.133−132

2

3

})3 + ( {13 ± √

1

532−169

2

3

})3 + ( {13 ± √ 1

1

2

2

})3

1

363

2

3

})3 = ( {13 ∓ √

})3 +

})3 = ( {13 ∓ 11})3 + ( {13 ± 11})3 = 13 + 123

1729 = 93 + 103 = 13 + 123 Congratulations to S. Ramanujan for a remarkable prediction given to G.H.Hardy. But it is not clear whether Ramanujan was already aware of this number from the mathematical literature from the time of Frenicle a French mathematician who lived in the sixteenth century or he himself has analyzed it prior to Hardy’s dull quiz to him at Hospital bed in Putney. Frenicle was competing with Wallis another mathematician. Mathematicians are playing games with fancy numbers Hardy’s taxi cab number was such a game to Ramanujan. How Hardy was unaware of the literature coming from Frenicle is also not clear. At his time Frenicle was one of the biggest competitor in mathematics in France doing many miracles as S. Ramanujan was. That is why we kept a question whether Ramanujan was a reincarnation of Frenicle in an another birth. Every human has many births after birth. According to Dickson, Ramanujan is only an one noted figure in mathematics playing excessively only in mathematics had thirty hour working habit and then twenty hour sleeping habit. He could not pass B.A first year with physiology over a period of six years from 1903 to 1909 what a great misery? .But no body has ever recorded what were the contents in physiology syllabus of Ramanujan whether there were Practical also as part of that particular subject. Did he took that in Tamil or in English. But he was writing three Note Books in Mathematics with dedication at young age got married to Janaki Ammal during this hard period all the time running in trains in Tamil Nadu state of India never passed the exam in two trials. But his performance in mathematics is excellent & brilliant it is said his English was also not up to the required standard because he

must have tamil pronounciation mean while his English hand writing was elegant with evidence in his note books. If 𝑁 = 728= 2.364=8.91 Case-1, N=2.364,r=2,s=364 Case-2 ,N=8.91,r=8,s=91 1

Case-1, N=2.364={ (2 ∓ √ 2 {1 ± √

363 3 } ={1∓11}3 3

4.364−22 3

1

)}3 + {2 (2 ± √

4.364−22 3

)}3 = {1 ∓ √

363 3 } 3

+

+ {1 ± 11}3 = (−10)3 + (12)3 = 123 − 103

1

4.91−82

2

3

Case-2 , N=8.91={ (8 ∓ √

1

4.91−82

2

3

) }3 +{ (8 ± √

)}3 ={4∓5}3 + {4 ± 5}3 =

(−1)3 + 93 = 93 − 13 728=123 − 103 = 93 − 13 =23 (63 − 53 ) = 23 (33 + 43 ) = 63 + 83 1

(𝑟 ∓ √ 2

4𝑠−𝑟 2 3

1

) = 6, 2 (𝑟 ± √

4𝑠−𝑟 2 3

) = 8, 𝑟 = 14,4𝑠 − 𝑟 2 = 4.3 = 12 = 4𝑠 −

142 , 4𝑠 = 12 + 4.49, 𝑠 = 3 + 49 = 52,14.52 = 728 1

(𝑟 ∓ √ 2

4𝑠−𝑟 2 3

1

) = 6, 2 (𝑟 ± √

4𝑠−𝑟 2 3

) = −5, 𝑟 = 1,4𝑠 − 𝑟 2 = 121.3 = 363,4𝑠 =

364, 𝑠 = 91,1.91 = 91 = 63 + (−5)3 1

(𝑟 ∓ √ 2

4𝑠−𝑟 2 3

1

) = 3, 2 (𝑟 ± √

4𝑠−𝑟 2 3

) = 4, 𝑟 = 7,4𝑠 − 𝑟 2 = 1.3 = 3 = 4𝑠 −

72 , 4𝑠 = 49 + 3 = 52, 𝑠 = 13,7.13 = 91 = 33 + 43 = 63 − 53 = 1.91 Congratulations to Ramanujan again for listing it at last N=4104=2.2052=4.1026=8.513=8.3.171=8.3.3.57=24.171=8.3.3.3.19=2.9.12.19 =18.228 =8.27.19 =216.19= 93 + 153 = 23 + 163 given by Frenicle 1

(𝑟 ∓ √ 2

4𝑠−𝑟 2 3

1

) = 9, 2 (𝑟 ± √

4𝑠−𝑟 2 3

) = 15,𝑟 = 24, √

4𝑠−𝑟 2 3

= 6,4𝑠 − 𝑟 2 = 36.3 =

108,4𝑠 = 108 + 242 , 𝑠 = 27 + 144 = 171, 𝑁 = 4104 = 24.171 Congratulations to Frenicle for a remarkable prediction 1

(𝑟 ∓ √ 2

4𝑠−𝑟 2 3

1

) = 2, 2 (𝑟 ± √

4𝑠−𝑟 2 3

)=16,r=18,√

4𝑠−𝑟 2 3

=14,

4s-r2=196.3,4s=196.3+182,s=49.3+81=147+81=228,4104=18.228 Congratulations to Frenicle again being up to the target for the discovery he had made in 1657 published in 1693 in Oxford in a Wallis’s document. The method of

calculation must have been given by Francois Viete 1646 who found the relation 33 + 43 + 53 = 63 Ramanujan stated the following relation in his note book (3𝑎2 + 5𝑎𝑏 − 5𝑏 2 )3 + (4𝑎2 − 4𝑎𝑏 + 6𝑏 2 )3 + (5𝑎2 − 5𝑎𝑏 − 3𝑏 2 )3 = (6𝑎2 − 4𝑎𝑏 + 4𝑏 2 )3 𝑓𝑜𝑟, 𝑎 = 1, 𝑏 = 0, 33 + 43 + 53 = 63 𝑓𝑜𝑟 𝑎 = √2, 𝑏 = 0, 63 + 83 + 103 = 123 Frenicle has stated 𝑚2 + 𝑚𝑛 + 𝑛2 = 3𝑎2 𝑏, (𝑚 + 𝑎𝑏 2 )3 + (𝑏𝑛 + 𝑎)3 = (𝑏𝑚 + 𝑎)3 + (𝑛 + 𝑎𝑏 2 )3 , 𝑚 = 3, 𝑛 = 0, 𝑎 = 1 𝑎𝑛𝑑 𝑏 = 3, (3 + 32 )3 + (1)3 = (3.3 + 1)3 + (0 + 1. 32 )3 = 123 + 13 = 103 + 93 = 1729 Far more generalized relation was given by Frenicle 𝑚2 + 𝑚𝑛 + 𝑛2 = 3𝑎2 𝑏𝑐, ((𝑚 + 𝑛)𝑐 + 𝑎𝑏 2 )3 + (−(𝑚 + 𝑛)𝑏 − 𝑎𝑐 2 )3 = (−𝑚𝑐 + 𝑎𝑏 2 )3 + (𝑚𝑏 − 𝑎𝑐 2 )3 = (−𝑛𝑐 + 𝑎𝑏 2 )3 + (𝑛𝑏 − 𝑎𝑐 2 )3 𝑓𝑜𝑟 𝑐 = 1, 𝑚2 + 𝑚𝑛 + 𝑛2 = 3𝑎2 𝑏, (𝑚 + 𝑛 + 𝑎𝑏 2 )3 + (−(𝑚 + 𝑛)𝑏 − 𝑎)3 = (−𝑚 + 𝑎𝑏 2 )3 + (𝑚𝑏 − 𝑎)3 = (−𝑛 + 𝑎𝑏 2 )3 + (𝑛𝑏 − 𝑎)3 𝑓𝑜𝑟 𝑚 = 0, 𝑛 = 3, 𝑎 = 1 & 𝑏 = 3,123 − 103 = 93 − 13 = 63 + 83 It looks Frenicle is far better at his time than Ramanujan but Frenicle looks dull in appearance mean while Ramanujan looks brilliant & Frenicle played too much with magic squares and cubes & Ramanujan also in magic squares. It is said Frenicle played in Combinatorics at his time as Ramanujan was in his time. Ramanujan has stated a similar identity in his note books 𝐼𝑓 𝛼 2 + 𝛼𝛽 + 𝛽 2 = 3𝛿𝛾 2 , 𝑡ℎ𝑒𝑛 (𝛼 + 𝛿 2 𝛾)3 + (𝛿𝛽 + 𝛾)3 = (𝛿𝛼 + 𝛾)3 + (𝛽 + 𝛿 2 𝛾)3 and instead of delta Ramanujan has used the notation lambda. 𝛼 = 3, 𝛽 = 0, 𝛿 = 3, 𝛾 = 1, (3 + 32 )3 + (1)3 = (3.3 + 1)3 + (0 + 32 . 1)3 = 123 + 13 = 103 + 93 =1729 This must have reached up to Ramanujan time could be from Frenicle era of mathematics. Frenicle must be primary source & Ramanujan was a secondary source. One of the former birth of Hardy must be Wallis. But Hardy must have forgotten everything. Hardy apologized the death of Ramanujan. Hardy said Ramanujan discovered himself and here himself must be Frenicle. Ramanujan found Hardy and Hardy must be the Wallis. Originally discovered by Bernhard Frenicle de Bessey(1604-1674)in the year 1657 in France prior to Srinivasa Ramanujan(1887-1920)of India.But Hardy was a British mathematician from Trinity College at Cambridge was unaware of this discovery because it was mainly in French Language. Similar formulas are appearing in Ramanujan’s Note Books. It looks Ramanujan must have had this

information. Frenicle worked on Magic Squares, Magic Cubes, Combinatorics and Cubic Number relations as Ramanujan did later. It is suspected Ramanujan’s friends must have borrowed a book on Mathematics to him from a library in Kumbakonam where Frenicle’s achievements were recorded. Ramanujan must have used such a literature prior to Hardy. Sometimes Taxi Cab problems in mathematics must have been there Hardy having knowing these problems has hidden the information and given a quiz to check Ramanujan’s health conditions as well his knowledge and capability in the Hospital in Putney. His statement of 1729 as a dull number must have been the way to check whether Ramanujan is dull on it. But Ramanujan proved he is a Genius as well as a Crank. It is further suspected whether Ramanujan was becoming a medical doctor at the length of six year duration with a B.A in Tamil Nadu. What he really failed was that practical s of Physiology. Ramanujan’ s Pronunciations in English must have been in Tamil Style not much satisfactory to British. Ramanujan had a Carriage a Dhola he must have been a King of Tamil Nadu. He must have travelled to England due to Skin sickness coming from seasonal hotness in Tamil state of British India to enjoy cold seasons of England. In 1920 in the month of April he died due to seasonal hotness where too much water pouring is a must in bodily cleaning & need lot of water in thirst. Pale beyond the porch and portal Crowned with calm leaves, she stands Who gathers all things immortal with cold immortal hands Ramanujan’s passing away Was recorded in this way In life there is no permanent stay Mathematics was his fun of the day His taste in other subjects Already left from his thoughts But mathematics remained Till his last breath

From one single man born to this world There were so many mathematical thoughts They blossomed like wild flowers There were hardly few to appreciate Their sweetness vanished in the air his Life became like a soap bubble did not much remained in the air You can still enjoy the mathematics that he has so much scribbled Life span of Ramanujan Was short and concise But his thoughts were broad and wide so hard to find What a brilliant mathematical thoughts Written by his own hand Rhythm of the poem written Move with the band He kept written so many ideas For many generations to come Fragrance of his flowers Wisdom and courage Dedication at young days No words to appreciate Ships that he had travelled By the steam they propelled with success wish achieved by the ocean he travelled

Bird s will sing What a lovely Ramanujan Great thinker for many centuries The treasure of lovely thoughts Formulas written by ink pen By his own hand Note books he has written At the beginning on the sand On the slate at young days Rubbed by hand Witnessed the Genius Marks on the hand Problems he cranked By his own hand Mathematics remained Genius what a rank His His His His

lovely mother was Komalatha Ammal lovely wife was Janaki Ammal adopted son was W. Narayanan father was Srinivasa Iyyangar

Note Books written by his own hand Were preserved with scan Mathematics remained Genius what a rank Why Tata Institute of Fundamental Research, TIFR published Ramanujan’ s Note Books in the year 1957.Commermoration of Frenicle also must be there his discovery of famous cubic identities in 1657 exactly after 300 years an another important point. We now call them Frenicle-Ramanujan cubic number identities.

Considering the solutions a and b 1

1

𝑎𝑛 = ( (83 + 9√85))𝑛 , 𝑏𝑛 = ( (83 − 9√85))𝑛 2 2 𝑎𝑛3 + 𝑏𝑛3 =

1

23𝑛 1

9√85)3𝑛 }= 1 8𝑛

8𝑛

{(83 + 9√85 )3𝑛 + (83 −

3𝑛 𝑟 3𝑛−𝑟 {∑3𝑛 + 83𝑟 (−9√85 )3𝑛−𝑟 ]}= 𝑟=0 𝑟 𝑐 [83 (9√85 )

3𝑛 𝑟 3𝑛−𝑟 {∑3𝑛 [1+(-1)3𝑛−𝑟 ]} 𝑟=0 𝑟 𝑐83 (9√85 )

Ken Ono’s

recent investigations and his comments on Ramanujan’ s Lost Note Book with Ramanujan’ s hand written scan copy of one page inspired to do this work if Ken Ono didn’t kept the scan copy of an important page of Ramanujan then this work of the present author could have been missed by mathematics. Further an another page of noted dedicated hard worker mathematician from University of New South Wales at Sydney Australia gave many inspiring inputs for further work without them this work of the present author could have been missed by mathematics. After Ramanujan that mathematician from Sydney has done very genuine work and they are already appearing in the Lost Book edition by Berndt and Andrews. One of the assistant of the present author Jayantha Neel Senadheera from open university Nawala has worked under Ken Ono in his Ph.D publications in USA & it is worth of mentioning and he has a copy of the lost note book of Ramanujan gifted by the present author in 1993 of Narosa publications for others observation.Ken Ono in 2016 spoke on Ramanujan in ICTP Ramanujan Prize awarded to a Chinese Mathematician in Trieste & made a attractive presentation in presence of M.Virasoro. Chinese Mathematician 2 2 2 2 3 showed 𝑋 = 𝑈 𝑉, 𝑌 = 𝑉 𝑈, 𝑋𝑌 = 𝑈 𝑉𝑉 𝑈 = (𝑈𝑉) = 𝑍 3 , 𝑍 = 𝑋𝑌.Further he discussed on Genus 1,2 toruses where interiors are of negative curvature and exteriors are of positive curvature in Algebraic topology and he received a bust of Ramanujan with a certificate. But it looks he has no contributions in Ramanujan Mathematics even though he is reputed in Algebraic topology. Hirschchorn , Sloan & Wolfram are the main workers after Ramanujan on cubic number relations. Further worked out mathematical details are presented in this revised version with lengthy and hard mathematical labor work & others may come up with very valuable things appearing in this work. We acknowledge W. Nadun, Susitha. S, Yasith. J We thank the organizers of Chemical Engineering Conference in Chicago, USA, October 2017 for the invitation extended to present the work.

References 1.Ramanujan Lost Note Book and other unpublished papers, Narosa publications from New Bombay 2.Papers by Ken Ono 3.Ramanujan Lost Note Book Part-iv, Edited by George Andrews and Bruce Berndt 4.Hirschchorn’s papers and related papers on Ramanujan and a Novel book on Ramanujan from Bangalore, M. C. Craig’s Papers, Laughlin’s papers, Dorothy Zeilberg’s Lectures, Some you tube presentations on Ramanujan mathematics and films on Ramanujan, Don Zagier’s papers are having important mathematical treasure inputs worth of mentioning 5.mrob.com website to study maxima computer program of Wolfram on further numbers satisfying cubic number relations of Ramanujan and Wolfram is expected to program the work in the present paper .Wolfram and his company are the pioneers of Mathematica and Maxima Computer Languages and it’s software which handle mathematics. Sloan’s affiliated sites give more program pieces and numbers. 6.Introduction to Number Theory by Hardy and Wright 7.The History of Theory of Numbers-part i, ii by Dickson,1920,published in the year Ramanujan has expired in India