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A253826 Indices of centered octagonal numbers (A016754) which are also triangular numbers (A000217).

%I A253826 #15 Jun 13 2015 00:55:22
%S A253826 1,18,595,20196,686053,23305590,791703991,26894630088,913625718985,
%T A253826 31036379815386,1054323288004123,35815955412324780,
%U A253826 1216688160731038381,41331581509442980158,1404057083160330286975,47696609245941786776976,1620280657278860420130193
%N A253826 Indices of centered octagonal numbers (A016754) which are also triangular numbers (A000217).
%C A253826 Also positive integers y in the solutions to x^2 - 8*y^2 + x + 8*y - 2 = 0, the corresponding values of x being A008843.
%C A253826 Also the indices of centered octagonal numbers (A016754) which are also hexagonal numbers (A000384). Also positive numbers y in the solutions to 4x^2-8y^2-2x+8y-2=0. - _Colin Barker_, Jan 25 2015
%H A253826 Colin Barker, <a href="/A253826/b253826.txt">Table of n, a(n) for n = 1..654</a>
%H A253826 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (35,-35,1).
%F A253826 a(n) = 35*a(n-1)-35*a(n-2)+a(n-3).
%F A253826 G.f.: x*(17*x-1) / ((x-1)*(x^2-34*x+1)).
%F A253826 a(n) = sqrt((-2-(17-12*sqrt(2))^n-(17+12*sqrt(2))^n)*(2-(17-12*sqrt(2))^(1+n)-(17+12*sqrt(2))^(1+n)))/(8*sqrt(2)). - _Gerry Martens_, Jun 04 2015
%e A253826 18 is in the sequence because the 18th centered octagonal number is 1225, which is also the 49th triangular number.
%e A253826 18 is in the sequence because the 18th centered octagonal number 1225 is also the 25th hexagonal number. - _Colin Barker_, Jan 25 2015
%o A253826 (PARI) Vec(x*(17*x-1)/((x-1)*(x^2-34*x+1)) + O(x^100))
%Y A253826 Cf. A000217, A008843, A016754, A046177.
%K A253826 nonn,easy
%O A253826 1,2
%A A253826 _Colin Barker_, Jan 16 2015