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A355752
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a(n) = 3*(2*n - 1)*( 3*(2*n - 1)^3 + 1) / 2.
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6
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6, 369, 2820, 10815, 29538, 65901, 128544, 227835, 375870, 586473, 875196, 1259319, 1757850, 2391525, 3182808, 4155891, 5336694, 6752865, 8433780, 10410543, 12715986, 15384669, 18452880, 21958635, 25941678, 30443481, 35507244, 41177895, 47502090, 54528213, 62306376, 70888419, 80327910, 90680145, 102002148, 114352671, 127792194
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OFFSET
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1,1
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COMMENTS
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Numbers C > 0 such that A = B^3 + (B+1)^3 = C^3 - D^3 such that the difference (C - D) == 3 (mod 6), C - D = 3 (2n - 1) for n > 1, and the difference between the positive cubes C^3 - D^3 is equal to a centered cube number, C^3 - D^3 = B^3 + (B+1)^3, with C > D > B > 0, and A > 0, A = 27*t^3 *(27*t^6+1)/4 with t = 2*n-1, and where A = A352759(n), B = A355751(n), C = a(n) (this sequence), and D = A355753(n).
There are infinitely many such numbers a(n) = C in this sequence.
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LINKS
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A. Grinstein, Ramanujan and 1729, University of Melbourne Dept. of Math and Statistics Newsletter: Issue 3, 1998.
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FORMULA
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a(n) = 3*(2*n - 1)*( 3*(2*n - 1)^3 + 1) / 2.
For n > 3, a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) + 1728*(n - 2), with a(1) = 6, a(2) = 369 and a(3) = 2820.
a(n) can be extended for negative n such that a(-n) = a(n+1) - 3*(2*n + 1).
G.f.: -3*x*(2+113*x+345*x^2+115*x^3+x^4) / (x-1)^5 . - R. J. Mathar, Aug 03 2022
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EXAMPLE
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a(1) = 6 belongs to the sequence as 6^3 - 3^3 = 4^3 + 5^3 = 189 and 6 - 3 = 3 = 3 (2*1 - 1).
a(2) = 369 belongs to the sequence as 369^3 - 360^3 = 121^3 + 122^3 = 3587409 and 369 - 360 = 9 = 3*(2*2 - 1).
a(3) = 3*(2*3 - 1)*( 3*(2*3 - 1)^3 + 1) / 2 = 2820.
a(4) = 3*a(3) - 3*a(2) + a(1) + 1728*2 = 3*2820 - 3*369 + 6 + 1728*2 = 10815.
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MAPLE
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restart; for n to 20 do (1/2)* 3*(2*n - 1)*(3*(2*n - 1)^3+1); end do;
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PROG
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CROSSREFS
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Cf. A005898, A001235, A272885, A352133, A352134, A352135, A352136, A352220, A352221, A352223, A352224, A352225, A352755, A352756, A352757, A352758, A352759, A355751, A355753.
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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