A289809 p-INVERT of (1,2,1,3,1,4,1,5,...) (A133622), where p(S) = 1 - S - S^2.

1, 4, 12, 38, 114, 354, 1076, 3311, 10120, 31043, 95044, 291284, 892242, 2733804, 8375092, 25659298, 78610859, 240840496, 737856017, 2260561368, 6925635380, 21217961710, 65005083598, 199154984626, 610147638720, 1869298875531, 5726938575936, 17545523113507

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 + ^2c(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 A289780 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 (1, 7, 1, -9, -3, 5, 1, -1)

Formula

G.f.: (1 + 3 x + x^2 - 3 x^3 - 3 x^4 + x^5 + x^6)/(1 - x - 7 x^2 - x^3 +

9 x^4 + 3 x^5 - 5 x^6 - x^7 + x^8).

a(n) = a(n-1) + 7*a(n-2) + a(n-3) - 9*a(n-4) - 3*a(n-5) + 5*a(n-6) + a(n-7) - a(n-8)..

Mathematica

z = 60; s = x (1 + 2 x - x^2 - x^3)/(1 - x^2)^2; p = 1 - s - s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A133622 *)
u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A289809 *)

Crossrefs

Cf. A133622, A289780.

Sequence in context: A189499 A183159 A369682 * A014345 A006192 A354341

Adjacent sequences: A289806 A289807 A289808 * A289810 A289811 A289812

Keyword

nonn,easy

Author

Clark Kimberling, Aug 12 2017

Status

approved