A353021 a(n) = Sum_{l=1..n} Sum_{k=1..l} Sum_{j=1..k} Sum_{i=1..j} (l*k*j*i)^2.

0, 1, 341, 13013, 196053, 1733303, 10787231, 52253971, 209609235, 725520510, 2230238010, 6217887390, 15973440990, 38276304066, 86383520146, 185042663146, 378620563178, 743881306623, 1409531082531, 2585397711611, 4605062303611

Offset: 0

Author

Roudy El Haddad, Apr 17 2022

Keywords

• nonn
• easy

Comments

a(n) is the sum of all products of four squares of positive integers up to n, i.e., the sum of all products of four elements from the set of squares {1^2, ..., n^2}.

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. See Theorem 4.8 for m = 4 and p = 2.
• Index entries for linear recurrences with constant coefficients, signature (13,-78,286,-715,1287,-1716,1716,-1287,715,-286,78,-13,1).

Crossrefs

Cf. A354021 (for distinct squares).

Cf. A000290 (squares), A000330 (sum of squares), A060493 (for two squares), A351105 (for three squares).

Cf. A000915 (for power 1).

Programs

PARI

{a(n) = n*(n + 1)*(n + 2)*(n + 3)*(n + 4)*(2*n + 1)*(2*n + 3)*(2*n + 5)*(2*n + 7)*(5*n - 2)*(35*n^2 - 28*n + 9)/5443200};

Python

def A353021(n): return n*(n*(n*(n*(n*(n*(n*(n*(8*n*(n*(70*n*(5*n + 84) + 40417) + 144720) + 2238855) + 2050020) + 207158) - 810600) - 58505) + 322740) + 7956) - 45360)//5443200 # Chai Wah Wu, May 14 2022

Formula

a(n) = n*(n + 1)*(n + 2)*(n + 3)*(n + 4)*(2*n + 1)*(2*n + 3)*(2*n + 5)*(2*n + 7)*(5*n - 2)*(35*n^2 - 28*n + 9)/5443200.

a(n) = binomial(2*n+8,9)*(5*n - 2)*(35*n^2 - 28*n + 9)/(5!*4).