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%I #11 Sep 08 2022 08:46:11
%S 1,56,297,26752,142913,12894168,68883529,6214961984,33201717825,
%T 2995598781880,16003159107881,1443872397903936,7713489488280577,
%U 695943500190915032,3717885930192129993,335443323219623141248,1792013304863118375809,161682985848358163166264
%N Indices of centered pentagonal numbers (A005891) which are also octagonal numbers (A000567).
%C Also positive integers y in the solutions to 6*x^2 - 5*y^2 - 4*x + 5*y - 2 = 0, the corresponding values of x being A253921.
%H Colin Barker, <a href="/A253922/b253922.txt">Table of n, a(n) for n = 1..745</a>
%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (1,482,-482,-1,1).
%F a(n) = a(n-1)+482*a(n-2)-482*a(n-3)-a(n-4)+a(n-5).
%F G.f.: x*(55*x^3+241*x^2-55*x-1) / ((x-1)*(x^2-22*x+1)*(x^2+22*x+1)).
%e 56 is in the sequence because the 56th centered pentagonal is 7701, which is also the number 51st octagonal number.
%t CoefficientList[Series[(55 x^3 + 241 x^2 - 55 x - 1)/((x - 1)(x^2 - 22 x + 1) (x^2 + 22 x + 1)), {x, 0, 30}], x] (* _Vincenzo Librandi_, Jan 20 2015 *)
%o (PARI) Vec(x*(55*x^3+241*x^2-55*x-1)/((x-1)*(x^2-22*x+1)*(x^2+22*x+1)) + O(x^100))
%o (Magma) I:=[1,56,297,26752,142913]; [n le 5 select I[n] else Self(n-1)+482*Self(n-2)-482*Self(n-3)-Self(n-4)+Self(n-5): n in [1..25]]; // _Vincenzo Librandi_, Jan 20 2015
%Y Cf. A000567, A005891, A253921, A253923.
%K nonn,easy
%O 1,2
%A _Colin Barker_, Jan 19 2015
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