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Numbers n such that the sum of the octagonal numbers N(n), N(n+1) and N(n+2) is equal to the pentagonal number P(m) for some m.
2

%I #7 Jun 13 2015 00:55:19

%S 36,34503186,32463979328256,30545293221963537966,

%T 28740005301926584966432476,27041413508541574648524420892746,

%U 25443211887331010498345403984177120696,23939467178338931702363652343255760088359526,22524596789139300949003224751966312751633124800916

%N Numbers n such that the sum of the octagonal numbers N(n), N(n+1) and N(n+2) is equal to the pentagonal number P(m) for some m.

%C Also nonnegative integers x in the solutions to 18*x^2-3*y^2+24*x+y+18 = 0, the corresponding values of y being A252093.

%H Colin Barker, <a href="/A252092/b252092.txt">Table of n, a(n) for n = 1..168</a>

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (940899,-940899,1).

%F a(n) = 940899*a(n-1)-940899*a(n-2)+a(n-3).

%F G.f.: 6*x*(599*x^2-105137*x-6) / ((x-1)*(x^2-940898*x+1)).

%e 36 is in the sequence because N(36)+N(37)+N(38) = 3816+4033+4256 = 12105 = P(90).

%o (PARI) Vec(6*x*(599*x^2-105137*x-6)/((x-1)*(x^2-940898*x+1)) + O(x^100))

%Y Cf. A000326, A000567, A252093.

%K nonn,easy

%O 1,1

%A _Colin Barker_, Dec 14 2014