A258582 a(n) = n*(2*n + 1)*(4*n + 1)/3.

0, 5, 30, 91, 204, 385, 650, 1015, 1496, 2109, 2870, 3795, 4900, 6201, 7714, 9455, 11440, 13685, 16206, 19019, 22140, 25585, 29370, 33511, 38024, 42925, 48230, 53955, 60116, 66729, 73810, 81375, 89440, 98021, 107134, 116795, 127020, 137825, 149226, 161239, 173880

Offset: 0

Author

Ilya Gutkovskiy, Nov 06 2015

Keywords

• nonn
• easy

Comments

First bisection of the square pyramidal numbers (A000330).

Links

• Eric Weisstein's World of Mathematics, Square Pyramidal Number.
• Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Crossrefs

Cf. A000330, A001477, A005408, A016813, A053126 (partial sums), A100157.

Programs

Magma

[n*(2*n+1)*(4*n+1)/3: n in [0..50]]; // Wesley Ivan Hurt, Nov 17 2015

Maple

A258582:=n->n*(2*n + 1)*(4*n + 1)/3: seq(A258582(n), n=0..50); # Wesley Ivan Hurt, Nov 17 2015

Mathematica

Table[(1/3) n (2 n + 1) (4 n + 1), {n, 0, 45}]

PARI

vector(100, n, n--; n*(2*n+1)*(4*n+1)/3) \\ Altug Alkan, Nov 06 2015

PARI

concat(0, Vec((5*x + 10*x^2 + x^3)/(1 - x)^4 + O(x^50))) \\ Altug Alkan, Nov 06 2015

Formula

G.f.: x*(5 + 10*x + x^2)/(1 - x)^4.

a(n) = A000330(2*n).

Sum_{n>0} 1/a(n) = 3*(6 - Pi - 4*log(2)) = 0.25745587...

a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4) for n>3. - Vincenzo Librandi, Nov 18 2015

a(n) = A006918(4*n-1) = A053307(4*n-1) = A228706(4*n-1) for n>0. - Bruno Berselli, Nov 18 2015

a(n) = Sum_{k=1..2*n} k^2 (see the first comment). E.g.f. exp(x)*(5*x+ 20*x^2/2+16*x^3/3!). - Wolfdieter Lang, Mar 13 2017

Sum_{n>=1} (-1)^(n+1)/a(n) = 3*log(2) + 6*sqrt(2)*log(1+sqrt(2)) + 3*(sqrt(2)-1/2)*Pi - 18. - Amiram Eldar, Sep 17 2022