

A276917


Numbers obtained by alternatively adding centered pentagonal layers of 5*(2^n1) and 5*(3^n1) elements.


1



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
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OFFSET

0,2


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



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(n1) + 5*((2+((n+1) mod 2))^((n+(n mod 2))/2)  1) for n>0.
G.f.: (1+4*x15*x^3+6*x^46*x^5)/((1+x)^2*(15*x^2+6*x^4)).
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)


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]


PROG

(PARI) Vec((1+4*x15*x^3+6*x^46*x^5) / ((1+x)^2*(15*x^2+6*x^4)) + O(x^40)) \\ Colin Barker, Dec 30 2016


CROSSREFS



KEYWORD

nonn,easy


AUTHOR



STATUS

approved



