OFFSET
0,3
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).
In the following guide to p-INVERT sequences using s = (1,3,5,7,9,...) = A005408, in some cases t(1,3,5,7,9,...) is a shifted (or differently indexed) version of the cited sequence:
p(S) *********** t(1,3,5,7,9,...)
1 - S A003946
1 - S^2 A292480
1 - S^3 (not yet in OEIS)
(1 - S)^2 (not yet in OEIS)
(1 - S)^3 (not yet in OEIS)
1 - S - S^2 A289786
1 + S - S^2 A289484
1 - S - 2 S^2 A289785
1 - S - 3 S^2 A289786
1 - S - 4 S^2 A289787
1 - S - 5 S^2 A289788
1 - S - 6 S^2 A289789
1 - S - 7 S^2 A289790
1 + S - 2 S^2 A289791
1 - S + S^2 - S^3 A289792
1 + S - 3 S^2 A289793
1 - S - S^2 - S^3 A289794
LINKS
Clark Kimberling, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (4,-5,6)
FORMULA
G.f.: x*(1 + x)^2/((1 - 3*x)*(1 - x + 2*x^2)).
a(n) = 4*a(n-1) - 5*a(n-2) + 6*a(n-3) for n >= 5.
EXAMPLE
s = (1,3,5,7,9,...), S(x) = x + 3 x^2 + 5 x^3 + 7 x^4 + ...,
p(S(x)) = 1 - ( x + 3 x^2 + 5 x^3 + 7 x^4 + ...)^2,
1/p(S(x)) = 1 + x^2 + 6 x^3 + 20 x^4 + 56 x^5 + ...
T(x) = (-1 + 1/p(S(x)))/x = x + 6 x^2 + 20 x^3 + 56 x^4 + ...
t(s) = (0,1,2,20,56,...).
MATHEMATICA
z = 60; s = x (x + 1)/(1 - x)^2; p = 1 - s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A005408 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A292480 *)
Join[{0}, LinearRecurrence[{4, -5, 6}, {1, 6, 20}, 30]] (* Vincenzo Librandi, Oct 03 2017 *)
PROG
(Magma) I:=[0, 1, 6, 20]; [n le 4 select I[n] else 4*Self(n-1)- 5*Self(n-2)+6*Self(n-3): n in [1..30]]; // Vincenzo Librandi, Oct 03 2017
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Oct 02 2017
STATUS
approved