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 A252630 Numbers n such that the sum of the hexagonal numbers X(n), X(n+1), X(n+2) and X(n+3) is equal to the heptagonal number H(m) for some m. 2
 50, 16503, 5314316, 1711193649, 550999041062, 177419980028715, 57128682570205568, 18395258367626164581, 5923216065693054789914, 1907257177894796016188127, 614130888066058624157787380, 197748238700092982182791348633, 63674318730541874204234656472846 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 COMMENTS Also positive integers x in the solutions to 16*x^2-5*y^2+40*x+3*y+44 = 0, the corresponding values of y being A252631. LINKS Colin Barker, Table of n, a(n) for n = 1..398 Index entries for linear recurrences with constant coefficients, signature (323,-323,1). FORMULA a(n) = 323*a(n-1)-323*a(n-2)+a(n-3). G.f.: x*(3*x^2-353*x-50) / ((x-1)*(x^2-322*x+1)). a(n) =-5/4+1/80*(161+72*sqrt(5))^(-n)*(-70-37*sqrt(5)+(-70+37*sqrt(5))*(161+72*sqrt(5))^(2*n)). - Colin Barker, Mar 03 2016 EXAMPLE 50 is in the sequence because X(50)+X(51)+X(52)+X(53) = 4950+5151+5356+5565 = 21022 = H(92). PROG (PARI) Vec(x*(3*x^2-353*x-50)/((x-1)*(x^2-322*x+1)) + O(x^100)) CROSSREFS Cf. A000384, A000566, A252631. Sequence in context: A152258 A203097 A028465 * A034205 A173170 A146518 Adjacent sequences:  A252627 A252628 A252629 * A252631 A252632 A252633 KEYWORD nonn,easy AUTHOR Colin Barker, Dec 19 2014 STATUS approved

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Last modified November 29 14:52 EST 2020. Contains 338769 sequences. (Running on oeis4.)