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a(n) = 3*(2*n - 1)*( 3*(2*n - 1)^3 + 1) / 2.
6

%I #17 Feb 16 2025 08:34:03

%S 6,369,2820,10815,29538,65901,128544,227835,375870,586473,875196,

%T 1259319,1757850,2391525,3182808,4155891,5336694,6752865,8433780,

%U 10410543,12715986,15384669,18452880,21958635,25941678,30443481,35507244,41177895,47502090,54528213,62306376,70888419,80327910,90680145,102002148,114352671,127792194

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

%C 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).

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

%H Vladimir Pletser, <a href="/A355752/b355752.txt">Table of n, a(n) for n = 1..10000</a>

%H A. Grinstein, <a href="https://web.archive.org/web/20040320144821/http://zadok.org/mattandloraine/1729.html">Ramanujan and 1729</a>, University of Melbourne Dept. of Math and Statistics Newsletter: Issue 3, 1998.

%H Vladimir Pletser, <a href="https://www.researchgate.net/publication/359706361">Euler's and the Taxi-Cab relations and other numbers that can be written twice as sums of two cubed integers</a>, submitted. Preprint available on ResearchGate, 2022.

%H Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/CenteredCubeNumber.html">Centered Cube Number</a>

%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (5,-10,10,-5,1).

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

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

%F 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.

%F a(n) can be extended for negative n such that a(-n) = a(n+1) - 3*(2*n + 1).

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

%e 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).

%e 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).

%e a(3) = 3*(2*3 - 1)*( 3*(2*3 - 1)^3 + 1) / 2 = 2820.

%e a(4) = 3*a(3) - 3*a(2) + a(1) + 1728*2 = 3*2820 - 3*369 + 6 + 1728*2 = 10815.

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

%o (PARI) a(n)=3*(2*n-1)*(3*(2*n-1)^3+1)/2 \\ _Charles R Greathouse IV_, Oct 21 2022

%Y Cf. A005898, A001235, A272885, A352133, A352134, A352135, A352136, A352220, A352221, A352223, A352224, A352225, A352755, A352756, A352757, A352758, A352759, A355751, A355753.

%K nonn,easy,changed

%O 1,1

%A _Vladimir Pletser_, Jul 15 2022