A101362 a(n) = (n+1)*n^4.

0, 2, 48, 324, 1280, 3750, 9072, 19208, 36864, 65610, 110000, 175692, 269568, 399854, 576240, 810000, 1114112, 1503378, 1994544, 2606420, 3360000, 4278582, 5387888, 6716184, 8294400, 10156250, 12338352, 14880348, 17825024, 21218430, 25110000, 29552672

Offset: 0

Author

Jonathan Vos Post, Dec 25 2004

Keywords

• nonn
• easy

Comments

For n>=4, a(n-1) is equal to the number of functions f:{1,2,3,4,5}->{1,2,...,n} such that for fixed, different x_1, x_2, x_3, x_4 in {1,2,3,4,5} and fixed y_1, y_2, y_3, y_ 4 in {1,2,...n} we have f(x_i)<>y_i, (i=1,2,3,4). - Milan Janjic, May 13 2007

Pierce expansion of the constant 1 - Sum_{k >= 1} (-1)^(k+1)*k^4/k!^5 = 0.48961 54584 28443 62043 ... = 1/2 - 1/(2*48) + 1/(2*48*324) - .... - Peter Bala, Feb 01 2015

Examples

			a(5) = (5+1)*5^4 = 3750 = 2 * 3 * 5^4, the sum of the divisors of which is 30008.
a(7) = 8*7^4 = 19208 = 2^3 * 7^4 = 98^2 + 98^2.
a(8) = 9*8^4 = 36864 = 2^12*3^2 = 192^2.
a(9) = 10*9^4 = 65610 = 2*3^8*5 = 243^2 + 81^2.
a(10) = 11*10^4 = 110000 = 2^4*5^4*11 = 300^2 + 100^2 + 100^2.
		

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..1000
• Milan Janjic, Enumerative Formulas for Some Functions on Finite Sets
• Katherine Kanim, Proof without Words: The Sum of Cubes: An Extension of Archimedes' Sum of Squares, Mathematics Magazine, Vol. 77, No. 4 (2004), pp. 298-299.
• Eric Weisstein's World of Mathematics, Pierce Expansion.
• Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

Crossrefs

Cf. A019583.

Programs

Magma

[n^4+n^5: n in [0..40]]; // Vincenzo Librandi, Aug 15 2016

Maple

a:= n-> (n+1)*n^4: seq(a(n), n=0..35);

Mathematica

Table[(n + 1)*n^4, {n, 0, 30}]
LinearRecurrence[{6,-15,20,-15,6,-1},{0,2,48,324,1280,3750},40] (* Harvey P. Dale, Jun 10 2019 *)

Formula

a(n) + 6*Sum_{i=1..n} i^3 + 4*Sum_{i=1..n} i^2 + Sum_{i=1..n} i = 5*Sum_{i=1..n} i^4.

G.f.: 2*x*(8*x^3+33*x^2+18*x+1) / (x-1)^6. - Colin Barker, May 06 2013

Sum_{n>=1} 1/a(n) = 0.5252003... = Pi^2/6+Pi^4/90-Zeta(3)-1. - R. J. Mathar, Oct 18 2019

Sum_{n>=1} (-1)^(n+1)/a(n) = 1 - 2*log(2) + Pi^2/12 - 3*zeta(3)/4 + 7*Pi^4/720. - Amiram Eldar, Nov 05 2020

Extensions

Corrected and extended by Ray Chandler, Dec 26 2004