OFFSET
0,2
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 A292480 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 (5, -2, 13, 5)
FORMULA
G.f.: -(((1 + x) (1 + 3 x + 6 x^2))/(-1 + 5 x - 2 x^2 + 13 x^3 + 5 x^4)).
a(n) = 5*a(n-1) - 2*a(n-2) + 13*a(n-3) + 5*a(n-4) for n >= 5.
MATHEMATICA
PROG
(PARI) x='x+O('x^99); Vec(((1+x)*(1+3*x+6*x^2))/(1-5*x+2*x^2-13*x^3-5*x^4)) \\ Altug Alkan, Oct 03 2017
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Oct 03 2017
STATUS
approved