A276917 Numbers obtained by alternatively adding centered pentagonal layers of 5*(2^n-1) and 5*(3^n-1) elements.

1, 6, 16, 31, 71, 106, 236, 311, 711, 866, 2076, 2391, 6031, 6666, 17596, 18871, 51671, 54226, 152636, 157751, 452991, 463226, 1348956, 1369431, 4026631, 4067586, 12039196, 12121111, 36035951, 36199786, 107944316, 108271991, 323505591, 324160946, 969861756

Offset: 0

Author

Daniel Poveda Parrilla, Dec 29 2016

Keywords

• nonn
• easy

Comments

a(0), a(1), a(2) and a(3) are the first four centered pentagonal numbers, as they match the same pattern. From a(4) onwards all terms are a different kind of centered pentagonal numbers, as the number of elements in subsequent layers doesn't increase uniformly.

a(13) is the first palindromic number in the sequence. a(19) is the second one.

First prime terms are a(3), a(4), a(7), a(31), a(100) and a(115).

Links

• Daniel Poveda Parrilla, Table of n, a(n) for n = 0..1000
• Daniel Poveda Parrilla, Illustration of initial terms
• Index entries for linear recurrences with constant coefficients, signature (2,4,-10,-1,12,-6).

Crossrefs

Cf. A005891.

Programs

Mathematica

Table[5 (Sum[2^i, {i, 0, ((n + Mod[n, 2])/2)}] + Sum[3^j, {j, 0, ((n - Mod[n, 2])/2)}]) - 5 n - 9, {n, 0, 28}] (* or *)
CoefficientList[Series[(1 + 4 x - 15 x^3 + 6 x^4 - 6 x^5)/((-1 + x)^2 (1 - 5 x^2 + 6 x^4)), {x, 0, 28}], x] (* or *)
LinearRecurrence[{2, 4, -10, -1, 12, -6}, {1, 6, 16, 31, 71, 106}, 29]

PARI

Vec((1+4*x-15*x^3+6*x^4-6*x^5) / ((-1+x)^2*(1-5*x^2+6*x^4)) + O(x^40)) \\ Colin Barker, Dec 30 2016

Formula

a(n) = 5*(Sum_{i=0..((n+(n mod 2))/2)} 2^i + Sum_{j=0..((n-(n mod 2))/2)} 3^j) - 5*n - 9.

a(n) = a(n-1) + 5*((2+((n+1) mod 2))^((n+(n mod 2))/2) - 1) for n>0.

G.f.: (1+4*x-15*x^3+6*x^4-6*x^5)/((-1+x)^2*(1-5*x^2+6*x^4)).

From Colin Barker, Dec 30 2016: (Start)

a(n) = (-10*n + 5*3^(n/2+1) + 5*2^(n/2+2) - 33)/2 for n even.

a(n) = (-10*n + 5*3^(n/2+1/2) + 5*2^(n/2+5/2) - 33)/2 for n odd.

(End)