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%I #73 Jun 13 2015 00:55:05
%S 1,2,57,166,5561,16242,544897,1591526,53394321,155953282,5232098537,
%T 15281830086,512692262281,1497463395122,50238609604977,
%U 146736130891846,4922871049025441,14378643364005762,482391124194888217,1408960313541672806,47269407300050019801
%N Numbers n such that the hexagonal number H(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^2-4*y^2+4*x+2*y+2 = 0, the corresponding values of x being A122513.
%H Colin Barker, <a href="/A245783/b245783.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*(6*x^4+11*x^3-43*x^2+x+1) / ((x-1)*(x^2-10*x+1)*(x^2+10*x+1)).
%e 57 is in the sequence because H(57) = 6441 = 3151+3290 = P(46)+P(47).
%o (PARI) Vec(-x*(6*x^4+11*x^3-43*x^2+x+1)/((x-1)*(x^2-10*x+1)*(x^2+10*x+1)) + O(x^100))
%Y Cf. A000326, A000384, A122513.
%K nonn,easy
%O 1,2
%A _Colin Barker_, Dec 15 2014
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