A290899 p-INVERT of the positive integers, where p(S) = 1 - S^2 - S^4.

0, 1, 4, 12, 36, 110, 332, 983, 2876, 8380, 24428, 71357, 208868, 612178, 1795228, 5264684, 15436060, 45248195, 132616392, 388652536, 1138993032, 3338020181, 9782903524, 28671786116, 84032220964, 246284956558, 721820483900, 2115530739035, 6200240318564

Offset

0,3

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 A290890 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 (8, -27, 52, -63, 52, -27, 8, -1)

Formula

a(n) = 8*a(n-1) - 27*a(n-2) + 52*a(n-3) - 63*a(n-4) + 52*a(n-5) - 27*a(n-6) + 8*a(n-7) - a(n-8).

G.f.: x*(1 - 4*x + 7*x^2 - 4*x^3 + x^4) / (1 - 8*x + 27*x^2 - 52*x^3 + 63*x^4 - 52*x^5 + 27*x^6 - 8*x^7 + x^8). - Colin Barker, Aug 18 2017

Mathematica

z = 60; s = x/(1 - x)^2; p = 1 - s^2 - s^4;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A290899 *)

Prog

(PARI) concat(0, Vec(x*(1 - 4*x + 7*x^2 - 4*x^3 + x^4) / (1 - 8*x + 27*x^2 - 52*x^3 + 63*x^4 - 52*x^5 + 27*x^6 - 8*x^7 + x^8) + O(x^40))) \\ Colin Barker, Aug 18 2017

Crossrefs

Cf. A000027, A290890.

Sequence in context: A170637 A170685 A177881 * A290905 A000781 A192205

Adjacent sequences: A290896 A290897 A290898 * A290900 A290901 A290902

Keyword

nonn,easy

Author

Clark Kimberling, Aug 17 2017

Status

approved