A202107 a(n) = n^4*(n+1)^4/8.

2, 162, 2592, 20000, 101250, 388962, 1229312, 3359232, 8201250, 18301250, 37949472, 74030112, 137149922, 243101250, 414720000, 684204032, 1095962562, 1710072162, 2606420000, 3889620000, 5694792642, 8194304162, 11605565952, 16200000000, 22313281250, 30356972802

Offset: 1

Author

Martin Renner, Dec 11 2011

Keywords

• nonn
• easy

Comments

A relation between fourth powers and the sum of fifth and seventh powers. See the first formula, which is from Beiler.

References

• Albert H. Beiler, Recreations in the theory of numbers, New York, Dover, (2nd ed.) 1966, p. 161.

Links

• Temple Rice Hollcroft, On sums of powers of n consecutive integers, Bulletin of the American Mathematical Society 59 (1953), nr. 6, p. 526 (574t).
• Index entries for linear recurrences with constant coefficients, signature (9,-36,84,-126,126,-84,36,-9,1).

Crossrefs

Cf. A000217, A000539, A000541, A002117, A059977.

Programs

Maple

A202107:=n->(n^4)*(n+1)^4/8; seq(A202107(n), n=1..100); # Wesley Ivan Hurt, Nov 12 2013

Mathematica

Table[n^4 (n+1)^4/8, {n, 100}] (* Wesley Ivan Hurt, Nov 12 2013 *)

Formula

a(n) = 2*(Sum_{k=1..n} k)^4 = Sum_{k=1..n} (k^5 + k^7).

a(n) = 2*A059977(n-1).

a(n) = A000539(n) + A000541(n).

G.f.: -2*x*(1+72*x+603*x^2+1168*x^3+603*x^4+72*x^5+x^6) / (x-1)^9. - R. J. Mathar, Dec 13 2011

a(n) = 2*(A000217(n)^4). - Zak Seidov, Jan 21 2012

From Amiram Eldar, Apr 09 2024: (Start)

Sum_{n>=1} 1/a(n) = 8*Pi^4/45 + 80*Pi^2/3 - 280.

Sum_{n>=1} (-1)^(n+1)/a(n) = 280 - 320*log(2) - 48*zeta(3). (End)