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%I #12 Mar 03 2016 11:10:54
%S 94,5908,366282,22703656,1407260470,87227445564,5406694364578,
%T 335127823158352,20772518341453326,1287561009346947940,
%U 79808010061169319034,4946809062783150832248,306622353882494182280422,19005639131651856150553996,1178043003808532587152067410
%N Numbers n such that the sum of the pentagonal numbers P(n), P(n+1), P(n+2) and P(n+3) is equal to the heptagonal number H(m) for some m.
%C Also positive integers x in the solutions to 12*x^2-5*y^2+32*x+3*y+36 = 0, the corresponding values of y being A252770.
%H Colin Barker, <a href="/A252769/b252769.txt">Table of n, a(n) for n = 1..557</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (63,-63,1).
%F a(n) = 63*a(n-1)-63*a(n-2)+a(n-3).
%F G.f.: 2*x*(7*x-47) / ((x-1)*(x^2-62*x+1)).
%F a(n) = 2*(-2/3+1/240*(31+8*sqrt(15))^(-n)*(80-27*sqrt(15)+(31+8*sqrt(15))^(2*n)*(80+27*sqrt(15)))). - _Colin Barker_, Mar 03 2016
%e 94 is in the sequence because P(94)+P(95)+P(96)+P(97) = 13207+13490+13776+14065 = 54538 = H(148).
%t LinearRecurrence[{63,-63,1},{94,5908,366282},30] (* _Harvey P. Dale_, Mar 04 2015 *)
%o (PARI) Vec(2*x*(7*x-47)/((x-1)*(x^2-62*x+1)) + O(x^100))
%Y Cf. A000326, A000566, A252770.
%K nonn,easy
%O 1,1
%A _Colin Barker_, Dec 21 2014
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