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A254962 Indices of hexagonal numbers (A000384) that are also centered pentagonal numbers (A005891).

%I A254962 #15 Jun 08 2020 11:10:39
%S A254962 1,2,12,31,211,552,3782,9901,67861,177662,1217712,3188011,21850951,
%T A254962 57206532,392099402,1026529561,7035938281,18420325562,126254789652,
%U A254962 330539330551,2265550275451,5931287624352,40653650168462,106432637907781,729500152756861
%N A254962 Indices of hexagonal numbers (A000384) that are also centered pentagonal numbers (A005891).
%C A254962 Also positive integers x in the solutions to 4*x^2 - 5*y^2 - 2*x + 5*y - 2 = 0, the corresponding values of y being A254627.
%H A254962 Colin Barker, <a href="/A254962/b254962.txt">Table of n, a(n) for n = 1..1000</a>
%H A254962 <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (1,18,-18,-1,1).
%F A254962 a(n) = a(n-1)+18*a(n-2)-18*a(n-3)-a(n-4)+a(n-5).
%F A254962 G.f.: -x*(x^4+x^3-8*x^2+x+1) / ((x-1)*(x^2-4*x-1)*(x^2+4*x-1)).
%F A254962 a(n) = (2 + (2-r)^n - (-2-r)^n*(-2+r) + 2*(-2+r)^n + r*(-2+r)^n + (2+r)^n)/8 where r = sqrt(5). - _Colin Barker_, Nov 25 2016
%F A254962 a(n+2) - a(n) = A000032(3*n + 2) if n is odd, A000032(3*n + 1) if n is even. - _Diego Rattaggi_, May 11 2020
%e A254962 12 is in the sequence because the 12th hexagonal number is 276, which is also the 11th centered pentagonal number.
%o A254962 (PARI) Vec(-x*(x^4+x^3-8*x^2+x+1)/((x-1)*(x^2-4*x-1)*(x^2+4*x-1)) + O(x^100))
%Y A254962 Cf. A000032 (Lucas numbers), A000384, A005891, A254627, A254628.
%K A254962 nonn,easy
%O A254962 1,2
%A A254962 _Colin Barker_, Feb 11 2015