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Numbers n such that the pentagonal number P(n) is equal to the sum of the heptagonal numbers H(m), H(m+1) and H(m+2) for some m.
2

%I #8 May 08 2020 14:11:42

%S 1283,60266,133006667,6248482130,13790397229331,647855124125114,

%T 1429815965398472795,67170914973291570338,148246178910654059084579,

%U 6964414805612961471642122,15370460320384618188608829803,722084455808392156329506905586

%N Numbers n such that the pentagonal number P(n) is equal to the sum of the heptagonal numbers H(m), H(m+1) and H(m+2) for some m.

%C Also nonnegative integers y in the solutions to 15*x^2-3*y^2+21*x+y+16 = 0, the corresponding values of x being A252115.

%H Colin Barker, <a href="/A252116/b252116.txt">Table of n, a(n) for n = 1..398</a>

%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (1,103682,-103682,-1,1).

%F G.f.: -x*(2*x^4+57*x^3-77605*x^2+58983*x+1283) / ((x-1)*(x^2-322*x+1)*(x^2+322*x+1)).

%e 1283 is in the sequence because P(1283) = 2468492 = 819963+822829+825700 = H(573)+H(574)+H(575).

%t LinearRecurrence[{1,103682,-103682,-1,1},{1283,60266,133006667,6248482130,13790397229331},20] (* _Harvey P. Dale_, May 08 2020 *)

%o (PARI) Vec(-x*(2*x^4+57*x^3-77605*x^2+58983*x+1283)/((x-1)*(x^2-322*x+1)*(x^2+322*x+1)) + O(x^100))

%Y Cf. A000326, A000566, A252115.

%K nonn,easy

%O 1,1

%A _Colin Barker_, Dec 14 2014