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A276917 Numbers obtained by alternatively adding centered pentagonal layers of 5*(2^n-1) and 5*(3^n-1) 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 (list; graph; refs; listen; history; text; internal format)
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
Daniel Poveda Parrilla, Table of n, a(n) for n = 0..1000
Daniel Poveda Parrilla, Illustration of initial terms
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)
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*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
CROSSREFS
Cf. A005891.
Sequence in context: A092286 A301723 A288113 * A097118 A369548 A296957
KEYWORD
nonn,easy
AUTHOR
STATUS
approved

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Last modified June 16 11:31 EDT 2024. Contains 373429 sequences. (Running on oeis4.)