A288486 Square rings obtained by adding four identical cuboids from A169938, a(n) = 4*n*(n+1)*(n*(n+1)+1).

0, 24, 168, 624, 1680, 3720, 7224, 12768, 21024, 32760, 48840, 70224, 97968, 133224, 177240, 231360, 297024, 375768, 469224, 579120, 707280, 855624, 1026168, 1221024, 1442400, 1692600, 1974024, 2289168, 2640624, 3031080, 3463320, 3940224, 4464768, 5040024

Offset: 0

Author

Daniel Poveda Parrilla, Jun 10 2017

Keywords

• nonn
• easy

Comments

If we fill the empty space with A288487(n) cubes, we get a solid cuboid with (n+1)^5 cubes (A000584(n+1)).

Links

• Daniel Poveda Parrilla, Table of n, a(n) for n = 0..10000
• Daniel Poveda Parrilla, Illustration of initial terms
• Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Crossrefs

Cf. A000584, A001844, A002061, A002378, A007204, A033996, A169938, A288487.

Programs

Mathematica

Table[4 n (n + 1) (n^2 + n + 1), {n, 0, 28}] (* or *) CoefficientList[Series[24 (x + 2 x^2 + x^3)/(1 - x)^5, {x, 0, 28}], x] (* or *) LinearRecurrence[{5, -10, 10, -5, 1}, {0, 24, 168, 624, 1680}, 29]

PARI

Vec(24*(x + 2*x^2 + x^3)/(1 - x)^5 + O(x^28))

Formula

G.f.: 24*(x + 2*x^2 + x^3) / (1 - x)^5.

a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5).

a(n) = A007204(n) - 1 = (A001844(n))^2 - 1.

a(n) = 4*A169938(n) = 4*A002378(n)*A002061(n+1) = A033996(n)*A002061(n+1).