A351760 - OEIS

%I #25 Oct 04 2024 08:53:04

%S 0,0,16,1393,26481,247731,1516515,6978790,26131686,83684778,237014778,

%T 607915231,1436816095,3170754405,6600189141,13064343516,24750198748,

%U 45116627556,79482515700,135826148445,225852708445,366397514791,581244702423,903454469346,1378306878690,2066986566190

%N a(n) = Sum_{1 <= i < j <= n} (i*j)^4.

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

%H Roudy El Haddad, <a href="https://arxiv.org/abs/2102.00821">Multiple Sums and Partition Identities</a>, arXiv:2102.00821 [math.CO], 2021.

%H Roudy El Haddad, <a href="https://doi.org/10.7546/nntdm.2022.28.2.200-233">A generalization of multiple zeta value. Part 2: Multiple sums</a>. Notes on Number Theory and Discrete Mathematics, 28(2), 2022, 200-233, DOI: 10.7546/nntdm.2022.28.2.200-233.

%H <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 a(n) = Sum_{j=2..n} Sum_{i=1..j-1} j^4*i^4.

%F a(n) = n*(n - 1)*(n + 1)*(2*n - 1)*(2*n + 1)*(9*n^5 + 20*n^4 - 15*n^3 - 50*n^2 + n + 30)/1800.

%F a(n) = binomial(2*n+2, 5)*(9*n^5 + 20*n^4 - 15*n^3 - 50*n^2 + n + 30)/5!.

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

%o (PARI) {a(n) = n*(n-1)*(n+1)*(2*n-1)*(2*n+1)*(9*n^5+20*n^4-15*n^3-50*n^2+n+30)/1800};

%o (PARI) a(n) = sum(j=2, n, sum(i=1, j-1, i^4*j^4));

%o (Python)

%o def A351760(n): return n*(n*(n*(n*(n*(n*(n*(n*(n*(9*n+20<<2)-105)-300)+88)+390)-20)-200)+1)+30)//1800 # _Chai Wah Wu_, Oct 03 2024

%Y Cf. A000217 (for power 0), A000914 (for power 1), A000596 (for squares), A347107 (for cubes).

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

%K nonn,easy

%O 0,3

%A _Roudy El Haddad_, Feb 18 2022