A292492 p-INVERT of the odd positive integers, where p(S) = 1 - S + S^2 - S^3.

1, 3, 5, 8, 22, 100, 444, 1680, 5496, 16096, 43936, 117360, 323056, 946288, 2930320, 9287792, 29222800, 89856944, 269619792, 795460592, 2334102160, 6882700336, 20508738256, 61728245104, 186833742864, 565643533232, 1706639551568, 5125652284144, 15338915301264

Offset: 0

Author

Clark Kimberling, Oct 03 2017

Keywords

• nonn
• easy

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 A292480 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 (7, -19, 23, -8, 6)

Crossrefs

Cf. A005408, A292480.

Programs

Mathematica

z = 60; s = x (x + 1)/(1 - x)^2; p = 1 - s + s^2 - s^3;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A005408 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A292492 *)

PARI

x='x+O('x^99); Vec(((1+x)*(1-5*x+8*x^2-x^3+x^4))/((1-3*x)*(1-4*x+7*x^2-2*x^3+2*x^4))) \\ Altug Alkan, Oct 03 2017

Formula

G.f.: -(((1 + x) (1 - 5 x + 8 x^2 - x^3 + x^4))/((-1 + 3 x) (1 - 4 x + 7 x^2 - 2 x^3 + 2 x^4))).

a(n) = 7*a(n-1) - 19*a(n-2) + 23*a(n-3) - 8*a(n-4) + 6*a(n-5) for n >= 7.