A351766 a(n) = Sum_{j=1..n} Sum_{i=1..j} (i*j)^4.

0, 1, 273, 8211, 98835, 710710, 3659110, 14886186, 50816298, 151416111, 404746111, 990005445, 2248888005, 4798557036, 9703780828, 18730825828, 34711648356, 62053834605, 107439683325, 180766879111, 296393439111, 474761104818, 744484165986, 1145004918190, 1729932641710, 2571200219835

Offset: 0

Author

Roudy El Haddad, Feb 18 2022

Keywords

• nonn
• easy

Comments

a(n) is the sum of all products of two elements from the set {1^4, ..., n^4}.

Links

• Roudy El Haddad, Recurrent Sums and Partition Identities, arXiv:2101.09089 [math.NT], 2021.
• Roudy El Haddad, A generalization of multiple zeta value. Part 1: Recurrent sums. Notes on Number Theory and Discrete Mathematics, 28(2), 2022, 167-199, DOI: 10.7546/nntdm.2022.28.2.167-199.
• Index entries for linear recurrences with constant coefficients, signature (11,-55,165,-330,462,-462,330,-165,55,-11,1).

Crossrefs

Cf. A000217 (for power 0), A001296 (for power 1), A060493 (for squares), A346642 (for cubes).

Cf. A000583 (fourth powers), A000538 (sum of fourth powers).

Programs

PARI

{a(n) = n*(n+1)*(n+2)*(2*n+1)*(2*n+3)*(9*n^5+25*n^4-5*n^3-25*n^2+21*n-5)/1800};

PARI

a(n) = sum(j=1, n, sum(i=1, j, i^4*j^4));

Formula

a(n) = n*(n + 1)*(n + 2)*(2*n + 1)*(2*n + 3)*(9*n^5 + 25*n^4 - 5*n^3 - 25*n^2 + 21*n - 5)/1800.

a(n) = binomial(2*n+4,5) * (9*n^5 + 25*n^4 - 5*n^3 - 25*n^2 + 21*n - 5)/5!.

G.f.: x*(16*x^7 + 1217*x^6 + 12038*x^5 + 30415*x^4 + 23364*x^3 + 5263*x^2 + 262*x + 1)/(1 - x)^11. - Alois P. Heinz, Feb 18 2022