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%I #14 Jun 13 2015 00:55:17
%S 1,3,113,331,11121,32483,1089793,3183051,106788641,311906563,
%T 10464197073,30563660171,1025384524561,2994926790243,100477219209953,
%U 293472261783691,9845742098050881,28757286728011523,964782248389776433,2817920627083345611,94538814600100039601
%N Numbers n such that the triangular number T(n) is equal to the sum of the pentagonal numbers P(m) and P(m+1) for some m.
%C Also nonnegative integers y in the solutions to 6*x^2-y^2+4*x-y+2 = 0, the corresponding values of x being A122513.
%H Colin Barker, <a href="/A249164/b249164.txt">Table of n, a(n) for n = 1..1000</a>
%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (1,98,-98,-1,1).
%F a(n) = a(n-1)+98*a(n-2)-98*a(n-3)-a(n-4)+a(n-5).
%F G.f.: -x*(x+1)^2*(11*x^2+1) / ((x-1)*(x^2-10*x+1)*(x^2+10*x+1)).
%e 113 is in the sequence because T(113) = 6441 = 3151+3290 = P(46)+P(47).
%o (PARI) Vec(-x*(x+1)^2*(11*x^2+1)/((x-1)*(x^2-10*x+1)*(x^2+10*x+1)) + O(x^100))
%Y Cf. A000217, A000326, A122513.
%K nonn,easy
%O 1,2
%A _Colin Barker_, Dec 15 2014
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