A385029 a(n) = Sum_{-n <= a, b, c <= n} (b^2 - 4*a*c).

18, 250, 1372, 4860, 13310, 30758, 63000, 117912, 205770, 339570, 535348, 812500, 1194102, 1707230, 2383280, 3258288, 4373250, 5774442, 7513740, 9648940, 12244078, 15369750, 19103432, 23529800, 28741050, 34837218, 41926500, 50125572, 59559910, 70364110, 82682208, 96668000

Offset: 1

Author

Darío Clavijo, Jun 15 2025

Keywords

• nonn
• easy

Comments

There are (2*n + 1)^3 combinations of a, b, c.

Links

• Paolo Xausa, Table of n, a(n) for n = 1..10000
• Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

Crossrefs

Cf. A000384, A002378, A014105, A016755, A016754, A055112, A384666.

Programs

Mathematica

A385029[n_] := (n*(n + 1)*(2*n + 1)^3)/3;
Array[A385029, 50] (* Paolo Xausa, Jun 18 2025 *)

Python

a = lambda n: ((n*n+n)*((n << 1)+1)**3)//3
print([a(n) for n in range(1, 11)])

Formula

a(n) = (n*(n+1)*(2*n+1)^3)/3.

a(n) = (A055112(n)*A016754(n))/3.

a(n) = (A002378(n)*A016755(n))/3.

G.f.: 2*x*(9 + 71*x + 71*x^2 + 9*x^3)/(1 - x)^6. - Stefano Spezia, Jun 15 2025

From Amiram Eldar, Jun 18 2025; (Start)

Sum_{n>=1} 1/a(n) = 21*(1 - zeta(3)/2) - 12*log(2).

Sum_{n>=1} (-1)^(n+1)/a(n) = 3*Pi^3/8 + 3*Pi - 21. (End)