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
Index entries for linear recurrences with constant coefficients, signature (2,4,-10,-1,12,-6).
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
KEYWORD
nonn,easy
AUTHOR
Daniel Poveda Parrilla, Dec 29 2016
STATUS
approved