A352980 a(n) = Sum_{1 <= i < j < k <= n} (k*j*i)^3.
0, 0, 0, 216, 16280, 335655, 3587535, 25421200, 135459216, 584760870, 2145870870, 6918983280, 20073184560, 53334782501, 131555523645, 304453955520, 666698215360, 1390977293580, 2780695001196, 5351537889480, 9954554649480, 17957698726275
Offset: 0
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. See Theorem 5.1 for m = 3 and p = 3.
- Index entries for linear recurrences with constant coefficients, signature (13,-78,286,-715,1287,-1716,1716,-1287,715,-286,78,-13,1). [Typo corrected by _Georg Fischer_, Sep 30 2022]
Crossrefs
Programs
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PARI
{a(n) = n^2 * (n + 1)^2 * (n - 1) * (n - 2) * (35*n^6 + 5*n^5 - 237*n^4 - 77*n^3 + 502*n^2 + 148*n -336)/13440};
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Python
def A352980(n): return n**2*(n*(n*(n*(n*(n*(n*(n*(n*(n*(35*n - 30) - 347) + 180) + 1365) - 350) - 2541) + 240) + 2160) - 40) - 672)//13440 # Chai Wah Wu, May 15 2022
Formula
a(n) = Sum_{k=3..n} Sum_{j=2..k-1} Sum_{i=1..j-1} k^3*j^3*i^3.
a(n) = n^2 * (n + 1)^2 * (n - 1) * (n - 2) * (35*n^6 + 5*n^5 - 237*n^4 - 77*n^3 + 502*n^2 + 148*n -336)/13440.
a(n) = binomial(n+1,4)*binomial(n+1,2)*(35*n^6 + 5*n^5 - 237*n^4 - 77*n^3 + 502*n^2 + 148*n -336)/280.
Comments