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%I #10 Apr 26 2016 14:34:15
%S 158,76252,36753402,17715063608,8538623905750,4115599007507988,
%T 1983710182994944562,956144192604555770992,460859517125212886673678,
%U 222133331110160006820941900,107067804735579998074807322218,51606459749218448912050308367272
%N Numbers n such that the sum of the heptagonal numbers H(n) and H(n+1) is equal to the pentagonal number P(m) for some m.
%C Also positive integers x in the solutions to 10*x^2-3*y^2+4*x+y+2 = 0, the corresponding values of y being A252360.
%H Colin Barker, <a href="/A252359/b252359.txt">Table of n, a(n) for n = 1..372</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (483,-483,1).
%F a(n) = 483*a(n-1)-483*a(n-2)+a(n-3).
%F G.f.: 2*x*(31*x-79) / ((x-1)*(x^2-482*x+1)).
%F a(n) = (-24+(12-5*sqrt(30))*(241+44*sqrt(30))^(-n)+(12+5*sqrt(30))*(241+44*sqrt(30))^n)/120. - _Colin Barker_, Apr 26 2016
%e 158 is in the sequence because H(158)+H(159) = 62173+62964 = 125137 = P(289).
%o (PARI) Vec(2*x*(31*x-79)/((x-1)*(x^2-482*x+1)) + O(x^100))
%Y Cf. A000326, A000566, A252360.
%K nonn,easy
%O 1,1
%A _Colin Barker_, Dec 17 2014
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