A355753 a(n) = 3*(2*n - 1)*( 3*(2*n - 1)^3 - 1) / 2 for n > 0.

3, 360, 2805, 10794, 29511, 65868, 128505, 227790, 375819, 586416, 875133, 1259250, 1757775, 2391444, 3182721, 4155798, 5336595, 6752760, 8433669, 10410426, 12715863, 15384540, 18452745, 21958494, 25941531, 30443328, 35507085, 41177730, 47501919, 54528036, 62306193, 70888230, 80327715, 90679944, 102001941, 114352458, 127791975

Offset: 1

Author

Vladimir Pletser, Jul 15 2022

Keywords

• nonn
• easy

Comments

Numbers D > 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*(2*n - 1) for n > 1, and the difference of the positive cubes C^3 - D^3 is equal to centered cube numbers, 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 = A355752(n), and D = a(n) (this sequence).

There are infinitely many such numbers a(n) = D in this sequence.

Subsequence of A352136 and of A352223.

Examples

			a(1) = 3 belongs to the sequence as 6^3 - 3^3 = 4^3 + 5^3 = 189 and 6 - 3 = 3 = 3*(2*1 - 1).
a(2) = 360 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 = 2805.
a(4) = 3*a(3) - 3*a(2) + a(1) + 1728*2 = 3*2805 - 3*360 + 3 + 1728*2  = 10794.
		

Links

• Vladimir Pletser, Table of n, a(n) for n = 1..10000
• A. Grinstein, Ramanujan and 1729, University of Melbourne Dept. of Math and Statistics Newsletter: Issue 3, 1998.
• Vladimir Pletser, Euler's and the Taxi-Cab relations and other numbers that can be written twice as sums of two cubed integers, submitted. Preprint available on ResearchGate, 2022.
• Eric Weisstein's World of Mathematics, Centered Cube Number
• Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Crossrefs

Cf. A005898, A001235, A272885, A352133, A352134, A352135, A352136, A352220, A352221, A352222, A352224, A352225, A352755, A352756, A352757, A352759, A352759, A355751, A355752.

Programs

Maple

restart; for n to 20 do (1/2)* 3*(2*n - 1)*(3*(2*n - 1)^3-1); end do;

Formula

A355752(n)^3 - a(n)^3 = A355751(n)^3 + (A355751(n) + 1)^3 = A352759(n) and A355752(n) - a(n) = 3*(2*n - 1).

a(n) = 3*(2*n - 1)*( 3*(2*n - 1)^3 - 1) / 2 for n > 0.

For n > 3, a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) + 1728*(n - 2), with a(1) = 3, a(2) = 360 and a(3) = 2805.

a(n) can be extended for negative n such that a(-n) = a(n+1) + (2n + 1).

G.f.: -3*x*(1+115*x+345*x^2+113*x^3+2*x^4) / (x-1)^5 . - R. J. Mathar, Aug 03 2022