A289806 p-INVERT of (1,1,2,2,3,3,...) (A008619), where p(S) = 1 - S - S^2.

1, 3, 9, 26, 74, 211, 600, 1708, 4860, 13832, 39364, 112029, 318827, 907366, 2582312, 7349121, 20915193, 59523497, 169400608, 482104856, 1372044007, 3904762096, 11112739032, 31626246588, 90006565434, 256153755080, 728999555983, 2074692805003, 5904462080604

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 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 (3, 1, -5, 2, 2, -1)

Formula

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

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

Mathematica

z = 60; s = x/((1 - x) (1 - x^2)); p = 1 - s - s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A008619 shifted *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A289806 *)

Crossrefs

Cf. A008619, A289780.

Sequence in context: A291000 A276068 A171277 * A303976 A000243 A076264

Adjacent sequences: A289803 A289804 A289805 * A289807 A289808 A289809

Keyword

nonn,easy

Author

Clark Kimberling, Aug 12 2017

Status

approved