A288487 Cuboids that fit in square rings from A288486 obtaining a fifth power.

1, 8, 75, 400, 1445, 4056, 9583, 20000, 38025, 67240, 112211, 178608, 273325, 404600, 582135, 817216, 1122833, 1513800, 2006875, 2620880, 3376821, 4298008, 5410175, 6741600, 8323225, 10188776, 12374883, 14921200, 17870525, 21268920, 25165831, 29614208

Offset: 0

Author

Daniel Poveda Parrilla, Jun 11 2017

Keywords

• nonn
• easy

Comments

If we add a(n) and A288487(n) graphically we obtain a bigger cuboid which is a square of cubes (a cuboid with dimensions n^2 * n^2 * n).

a(10^n) is a palindrome in base 10.

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 (6,-15,20,-15,6,-1).

Crossrefs

Cf. A000584, A002061, A002522, A014206, A059100, A082044, A288486.

Programs

Mathematica

Table[(1 + n)*(1 + n^2)^2, {n, 0, 28}] (* or *) CoefficientList[Series[(1 + 2 x + 42 x^2 + 50 x^3 + 25 x^4)/(1 - x)^6, {x, 0, 28}], x] (* or *) LinearRecurrence[{6, -15, 20, -15, 6, -1}, {1, 8, 75, 400, 1445, 4056}, 29]

PARI

Vec((1 + 2*x + 42*x^2 + 50*x^3 + 25*x^4)/(1 - x)^6 + O(x^28))

Formula

G.f.: (1 + 2*x + 42*x^2 + 50*x^3 + 25*x^4)/(1 - x)^6.

a(n) = 6*a(n-1) - 15*a(n-2) + 20*a(n-3) - 15*a(n-4) + 6*a(n-5) - a(n-6).

a(n) = (n + 1)*(n^2 + 1)^2 = (n + 1)*(A002522(n))^2 = (n + 1)*A082044(n).

a(n) = n^5 + A002061(A002061(n+1)).

a(n) = A000584(n+1) - A288486(n)

a(n) = (n + 1)*A059100(n-1) + 4*(n^2 -1)*A014206(n-1) for n > 1.