cp's OEIS Frontend

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

%I A351766 #22 May 31 2022 22:28:25
%S A351766 0,1,273,8211,98835,710710,3659110,14886186,50816298,151416111,
%T A351766 404746111,990005445,2248888005,4798557036,9703780828,18730825828,
%U A351766 34711648356,62053834605,107439683325,180766879111,296393439111,474761104818,744484165986,1145004918190,1729932641710,2571200219835
%N A351766 a(n) = Sum_{j=1..n} Sum_{i=1..j} (i*j)^4.
%C A351766 a(n) is the sum of all products of two elements from the set {1^4, ..., n^4}.
%H A351766 Roudy El Haddad, <a href="https://arxiv.org/abs/2101.09089">Recurrent Sums and Partition Identities</a>, arXiv:2101.09089 [math.NT], 2021.
%H A351766 Roudy El Haddad, <a href="https://doi.org/10.7546/nntdm.2022.28.2.167-199">A generalization of multiple zeta value. Part 1: Recurrent sums</a>. Notes on Number Theory and Discrete Mathematics, 28(2), 2022, 167-199, DOI: 10.7546/nntdm.2022.28.2.167-199.
%H A351766 <a href="/index/Rec#order_11">Index entries for linear recurrences with constant coefficients</a>, signature (11,-55,165,-330,462,-462,330,-165,55,-11,1).
%F A351766 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.
%F A351766 a(n) = binomial(2*n+4,5) * (9*n^5 + 25*n^4 - 5*n^3 - 25*n^2 + 21*n - 5)/5!.
%F A351766 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
%o A351766 (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};
%o A351766 (PARI) a(n) = sum(j=1, n, sum(i=1, j, i^4*j^4));
%Y A351766 Cf. A000217 (for power 0), A001296 (for power 1), A060493 (for squares), A346642 (for cubes).
%Y A351766 Cf. A000583 (fourth powers), A000538 (sum of fourth powers).
%K A351766 nonn,easy
%O A351766 0,3
%A A351766 _Roudy El Haddad_, Feb 18 2022