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.
All terms = 0 mod 5. - Muniru A Asiru, Sep 07 2017
LINKS
Clark Kimberling, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (10, -15, -50, -25)
FORMULA
G.f.: -((5 (1 + x) (-2 + 5 x + 5 x^2))/(-1 + 5 x + 5 x^2)^2).
a(n) = 10*a(n-1) - 15*a(n-2) - 50*a(n-3) - 25*a(n-4) for n >= 5.
MATHEMATICA
PROG
(GAP)
a:=5*[2, 17, 130, 940];; for n in [5..10^2] do a[n]:=10*a[n-1]-15*a[n-2]-50*a[n-3]-25*a[n-4]; od; a; # Muniru A Asiru, Sep 07 2017
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Sep 04 2017
STATUS
approved