A291249 p-INVERT of (0,1,0,1,0,1,...), where p(S) = 1 - S - S^2 - S^3 + S^4 + S^5.

1, 2, 5, 10, 23, 47, 102, 214, 452, 955, 2003, 4223, 8854, 18610, 39032, 81896, 171752, 360103, 754985, 1582497, 3316978, 6951684, 14568692, 30530311, 63977107, 134063288, 280920507, 588643384, 1233430247, 2584481968, 5415381139, 11347029572, 23775710791

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).

See A291219 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, 6, -3, -14, 3, 14, -3, -6, 1, 1)

Formula

G.f.: -((1 + x - 3 x^2 - 4 x^3 + 3 x^4 + 4 x^5 - 3 x^6 - x^7 + x^8)/((-1 + x + x^2) (1 + x - x^2 - x^3 + x^4) (1 - x - 3 x^2 + x^3 + x^4))).

a(n) = a(n-1) + 6*a(n-2) - 3*a(n-3) - 14*a(n-4) + 3*a(n-5) + 14*a(n-6) - 3*a(n-7) - 6*a(n-8) + a(n-9) + a(n-10) for n >= 11.

Mathematica

z = 60; s = x/(1 - x^2); p = 1 - s - s^2 - s^3 + s^4 + s^5;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000035 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291249 *)
LinearRecurrence[{1, 6, -3, -14, 3, 14, -3, -6, 1, 1}, {1, 2, 5, 10, 23, 47, 102, 214, 452, 955}, 40] (* Harvey P. Dale, Jul 21 2018 *)

Crossrefs

Cf. A000035, A291219.

Sequence in context: A018344 A284181 A365376 * A365441 A260744 A317535

Adjacent sequences: A291246 A291247 A291248 * A291250 A291251 A291252

Keyword

nonn,easy

Author

Clark Kimberling, Aug 29 2017

Status

approved