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%I #7 Jun 13 2015 00:55:19
%S 17,239,25529,345647,36813809,498423743,53085488057,718726692767,
%T 76549236965393,1036403392547279,110383946618609657,
%U 1494492973326484559,159173574474798161009,2155057831133398187807,229528184008712329566329,3107591898001386860334143
%N Numbers n such that the sum of the heptagonal numbers H(n), H(n+1) and H(n+2) is equal to the sum of three consecutive squares.
%C Also nonnegative integers x in the solutions to 15*x^2-6*y^2+21*x-12*y+6 = 0, the corresponding values of y being A251770.
%H Colin Barker, <a href="/A251769/b251769.txt">Table of n, a(n) for n = 1..633</a>
%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (1,1442,-1442,-1,1).
%F a(n) = a(n-1)+1442*a(n-2)-1442*a(n-3)-a(n-4)+a(n-5).
%F G.f.: x*(x^4+6*x^3-776*x^2-222*x-17) / ((x-1)*(x^2-38*x+1)*(x^2+38*x+1)).
%e 17 is in the sequence because H(17)+H(18)+H(19) = 697+783+874 = 2354 = 729+784+841 = 27^2+28^2+29^2.
%o (PARI) Vec(x*(x^4+6*x^3-776*x^2-222*x-17)/((x-1)*(x^2-38*x+1)*(x^2+38*x+1)) + O(x^100))
%Y Cf. A000290, A000566, A251770.
%K nonn,easy
%O 1,1
%A _Colin Barker_, Dec 08 2014
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