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 A291382 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, 3, 1, -2, -3, -1)
FORMULA
G.f.: -(((-1 + x) (1 + x) (2 + x) (1 + x + x^2))/(1 - 2 x - 3 x^2 - x^3 + 2 x^4 + 3 x^5 + x^6)).
a(n) = 2*a(n-1) + 3*a(n-2) + a(n-3) - 2*a(n-4) - 3*a(n-5) - a(n-6) for n >= 7.
MATHEMATICA
PROG
(GAP)
a:=[2, 7, 21, 63, 189, 567];; for n in [7..10^2] do a[n]:=2*a[n-1]+3*a[n-2]+a[n-3]-2*a[n-4]-3*a[n-5]-a[n-6]; od; a; # Muniru A Asiru, Sep 12 2017
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Sep 07 2017
STATUS
approved