A347107 a(n) = Sum_{1 <= i < j <= n} j^3*i^3.
0, 0, 8, 251, 2555, 15055, 63655, 214918, 616326, 1561110, 3586110, 7612385, 15139553, 28506101, 51229165, 88438540, 147420940, 238291788, 374813076, 575377095, 864177095, 1272587195, 1840775123, 2619572626, 3672629650, 5078879650, 6935344650, 9360309933
Offset: 0
Examples
For n=3, a(3) = (2*1)^3+(3*1)^3+(3*2)^3 = 251.
Links
- Roudy El Haddad, Multiple Sums and Partition Identities, arXiv:2102.00821 [math.CO], 2021.
- Roudy El Haddad, A generalization of multiple zeta value. Part 2: Multiple sums. Notes on Number Theory and Discrete Mathematics, 28(2), 2022, 200-233, DOI: 10.7546/nntdm.2022.28.2.200-233.
- Index entries for linear recurrences with constant coefficients, signature (9,-36,84,-126,126,-84,36,-9,1).
Crossrefs
Programs
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Mathematica
CoefficientList[Series[-(x^5 + 64 x^4 + 424 x^3 + 584 x^2 + 179 x + 8) x^2/(x - 1)^9, {x, 0, 27}], x] (* Michael De Vlieger, Feb 04 2022 *) LinearRecurrence[{9,-36,84,-126,126,-84,36,-9,1},{0,0,8,251,2555,15055,63655,214918,616326},30] (* Harvey P. Dale, Jul 07 2025 *)
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PARI
a(n) = sum(i=2, n, sum(j=1, i-1, i^3*j^3));
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PARI
{a(n) = n*(n+1)*(n-1)*(21*n^5+36*n^4-21*n^3-48*n^2+8)/672};
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Python
def A347107(n): return n*(n**2*(n*(n*(n*(n*(21*n + 36) - 42) - 84) + 21) + 56) - 8)//672 # Chai Wah Wu, Feb 17 2022
Formula
a(n) = Sum_{j=2..n} Sum_{i=1..j-1} j^3*i^3.
a(n) = n*(n+1)*(n-1)*(21*n^5+36*n^4-21*n^3-48*n^2+8)/672 (from the generalized form of Faulhaber's formula).
From Alois P. Heinz, Jan 27 2022: (Start)
G.f.: -(x^5+64*x^4+424*x^3+584*x^2+179*x+8)*x^2/(x-1)^9. (End)
Comments