A289844 p-INVERT of A175676 (starting at n=3), where p(S) = 1 - S - S^2.

1, 2, 3, 7, 16, 31, 64, 134, 274, 567, 1168, 2405, 4967, 10232, 21094, 43505, 89672, 184892, 381203, 785886, 1620327, 3340606, 6887304, 14199737, 29275538, 60357622, 124439898, 256558196, 528948160, 1090536002, 2248364880, 4635470266, 9556979689, 19703689739

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) + c(1)*x + c(2)*x^2 + ... 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 A289780 for a guide to related sequences.

Links

Table of n, a(n) for n=0..33.

Index entries for linear recurrences with constant coefficients, signature (1, 1, 4, -2, 0, -6, 1, 0, 4, 0, 0, -1)

Formula

a(n) = a(n-1) + a(n-2) + 4*a(n-3) - 2*a(n-4) - 6*a(n-7) + a(n-8) + 4*a(n-10) - a(n-13).

G.f.: (1 + x - 2*x^3 + x^6) / (1 - x - x^2 - 4*x^3 + 2*x^4 + 6*x^6 - x^7 - 4*x^9 + x^12). - Colin Barker, Aug 13 2017

Mathematica

z = 60; s = x/((x - 1)^2*(1 + x + x^2)^2); p = 1 - s - s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A175676, shifted *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A289844 *)

Prog

(PARI) Vec((1 + x - 2*x^3 + x^6) / (1 - x - x^2 - 4*x^3 + 2*x^4 + 6*x^6 - x^7 - 4*x^9 + x^12) + O(x^60)) \\ Colin Barker, Aug 13 2017

Crossrefs

Cf. A175686, A289780.

Sequence in context: A250193 A004782 A049956 * A153056 A235112 A081207

Adjacent sequences: A289841 A289842 A289843 * A289845 A289846 A289847

Keyword

nonn,easy

Author

Clark Kimberling, Aug 12 2017

Status

approved