|
%I #7 Jun 13 2015 00:55:20
%S 4,257,1772,123729,853956,59636977,411604876,28744899041,198392696132,
%T 13854981700641,95624867930604,6678072434809777,46090987949854852,
%U 3218817058596611729,22215760566962107916,1551463144171132043457,10707950502287786160516
%N Numbers n such that the heptagonal number H(n) is equal to the sum of the pentagonal numbers P(m) and P(m+1) for some m.
%C Also positive integers y in the solutions to 6*x^2-5*y^2+4*x+3*y+2 = 0, the corresponding values of x being A252585.
%H Colin Barker, <a href="/A252586/b252586.txt">Table of n, a(n) for n = 1..745</a>
%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (1,482,-482,-1,1).
%F a(n) = a(n-1)+482*a(n-2)-482*a(n-3)-a(n-4)+a(n-5).
%F G.f.: -x*(x^4+11*x^3-413*x^2+253*x+4) / ((x-1)*(x^2-22*x+1)*(x^2+22*x+1)).
%e 4 is in the sequence because H(4) = 23 = 12+22 = P(3)+P(4).
%o (PARI) Vec(-x*(x^4+11*x^3-413*x^2+253*x+4)/((x-1)*(x^2-22*x+1)*(x^2+22*x+1)) + O(x^100))
%Y Cf. A000326, A000566, A252585.
%K nonn,easy
%O 1,1
%A _Colin Barker_, Dec 18 2014
| 