A291231 - OEIS

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A291231

p-INVERT of (0,1,0,1,0,1,...), where p(S) = (1 - S)(1 - 2 S)(1 - 3 S)(1 - 4 S).

10, 65, 360, 1831, 8830, 41104, 186710, 833401, 3672840, 16034303, 69506930, 299700192, 1287010850, 5509712833, 23531008200, 100312445063, 427025152550, 1815832379312, 7714875191470, 32756357939033, 139008007848360, 589672772732671, 2500620567692890

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OFFSET

0,1

COMMENTS

Suppose s = (c(0), c(1), c(2), ...) is a sequence and p(S) is a polynomial. Let S(x) = c(0)*x + c(1)*x^2 + c(2)*x^3 + ... and T(x) = (-p(0) + 1/p(S(x)))/x. The p-INVERT of s is the sequence t(s) of coefficients in the Maclaurin series for T(x). Taking p(S) = 1 - S gives the "INVERT" transform of s, so that p-INVERT is a generalization of the "INVERT" transform (e.g., A033453).

See A291219 for a guide to related sequences.

LINKS

Clark Kimberling, Table of n, a(n) for n = 0..1000

Index entries for linear recurrences with constant coefficients, signature (10, -31, 20, 40, -20, -31, -10, -1)

FORMULA

G.f.: (10 - 35 x + 20 x^2 + 46 x^3 - 20 x^4 - 35 x^5 - 10 x^6)/(1 - 10 x + 31 x^2 - 20 x^3 - 40 x^4 + 20 x^5 + 31 x^6 + 10 x^7 + x^8).

a(n) = 10*a(n-1) - 31*a(n-2) + 20*a(n-3) + 40*a(n-4) - 20*a(n-5) -31*a(n-6) - 10*a(n-7) - a(n-8) for n >= 7.

MATHEMATICA

z = 60; s = x/(1 - x^2); p = (1 - s)(1 - 2 s)(1 - 3 s)(1 - 4 s);

Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000035 *)

Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A291231 *)

CROSSREFS