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 (2, 5, -8, -9, 8, 5, -2, -1)
FORMULA
G.f.: (2 + x - 8 x^2 - 3 x^3 + 8 x^4 + x^5 - 2 x^6)/(1 - x - 3 x^2 + x^3 + x^4)^2.
a(n) = 2*a(n-1) + 5*a(n-2) - 8*a(n-3) - 9*a(n-4) + 8*a(n-5) + 5*a(n-6) - 2*a(n-7) - a(n-8) for n >= 9.
MATHEMATICA
PROG
(GAP)
P:=[2, 5, 12, 30, 70, 166, 382, 881];; for n in [9..10^3] do P[n]:=2*P[n-1]+5*P[n-2]-8*P[n-3]-9*P[n-4]+8*P[n-5]+5*P[n-6]-2*P[n-7]-P[n-8]; od; P; # Muniru A Asiru, Sep 03 2017
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Aug 31 2017
STATUS
approved