A355752 a(n) = 3*(2*n - 1)*( 3*(2*n - 1)^3 + 1) / 2.

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

Offset: 1

Author

Vladimir Pletser, Jul 15 2022

Keywords

• nonn
• easy

Comments

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.

Examples

			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.
		

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, A352223, A352224, A352225, A352755, A352756, A352757, A352758, A352759, A355751, A355753.

Programs

Maple

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

PARI

a(n)=3*(2*n-1)*(3*(2*n-1)^3+1)/2 \\ Charles R Greathouse IV, Oct 21 2022

Formula

a(n)^3 - A355753(n)^3 = A355751(n)^3 + (A355751(n) + 1)^3 = A352759(n) and a(n) - A355753(n) = 3*(2*n - 1).

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