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).
LINKS
Clark Kimberling, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (4, -1, -6, 1, 4, 1)
FORMULA
G.f.: -((1 - x - x^2 + x^3 + x^4)/(-1 + 4 x - x^2 - 6 x^3 + x^4 + 4 x^5 + x^6)).
a(n) = 4*a(n-1) - a(n-2) - 6*a(n-3) + a(n-4) + 4*a(n-5) + a(n-6) for n >= 7.
MAPLE
A292398:=proc(n) option remember:
if n=0 then 1 elif n=1 then 3 elif n=2 then 10 elif n=3 then 32 elif n=4 then 102 elif n=5 then 323 elif n>=6 then 4*procname(n-1)-procname(n-2)-6*procname(n-3)+procname(n-4)+4*procname(n-5)+procname(n-6) fi; end:
seq(A292398(n), n=0..10^2); # Muniru A Asiru, Oct 02 2017
MATHEMATICA
z = 60; s = x/(1 - x + x^2); p = 1 - s - s^2 - s^3;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A010892 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A292398 *)
LinearRecurrence[{4, -1, -6, 1, 4, 1}, {1, 3, 10, 32, 102, 323}, 30] (* Harvey P. Dale, Feb 24 2022 *)
PROG
(GAP)
a:=[1, 3, 10, 32, 102, 323];; for n in [7..10^2] do a[n]:=4*a[n-1]-a[n-2]-6*a[n-3]+a[n-4]+4*a[n-5]+a[n-6]; od; A292398:=a; # Muniru A Asiru, Oct 02 2017
(PARI) Vec(-(1 - x - x^2 + x^3 + x^4)/(-1 + 4*x - x^2 - 6*x^3 + x^4 + 4*x^5 + x^6) + O(x^20)) \\ Felix Fröhlich, Oct 02 2017
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Sep 29 2017
STATUS
approved