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 A251769 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. 2

%I

%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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Last modified April 9 06:59 EDT 2020. Contains 333344 sequences. (Running on oeis4.)