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A252115
Numbers n such that the sum of the heptagonal numbers H(n), H(n+1) and H(n+2) is equal to the pentagonal number P(m) for some m.
2
573, 26951, 59482389, 2794406159, 6167253128301, 289729619423063, 639433138789094469, 30039746398227684383, 66297706689763639679133, 3114580985771313152847719, 6873878824368640550422845813, 322925985736701543915329589551, 712697504201891682859177859976909
OFFSET
1,1
COMMENTS
Also nonnegative integers x in the solutions to 15*x^2-3*y^2+21*x+y+16 = 0, the corresponding values of y being A252116.
FORMULA
G.f.: x*(x^4+26*x^3-45652*x^2-26378*x-573) / ((x-1)*(x^2-322*x+1)*(x^2+322*x+1)).
EXAMPLE
573 is in the sequence because H(573)+H(574)+H(575) = 819963+822829+825700 = 2468492 = P(1283).
MATHEMATICA
LinearRecurrence[{1, 103682, -103682, -1, 1}, {573, 26951, 59482389, 2794406159, 6167253128301}, 20] (* Harvey P. Dale, Jul 21 2023 *)
PROG
(PARI) Vec(x*(x^4+26*x^3-45652*x^2-26378*x-573)/((x-1)*(x^2-322*x+1)*(x^2+322*x+1)) + O(x^100))
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Colin Barker, Dec 14 2014
STATUS
approved