A291407 p-INVERT of (1,1,0,0,0,0,...), where p(S) = 1 - S^3 - S^6.

0, 0, 1, 3, 3, 3, 12, 30, 43, 57, 120, 259, 438, 708, 1360, 2703, 4827, 8276, 15114, 28488, 51769, 92031, 167334, 309341, 564237, 1016487, 1844115, 3374343, 6151563, 11151098, 20253876, 36931695, 67280308, 122243430, 222174201, 404488108, 736378968

Offset

0,4

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 A291382 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 (0, 0, 1, 3, 3, 2, 6, 15, 20, 15, 6, 1)

Formula

G.f.: -((x^2 (1 + x)^3 (1 + x + x^2) (1 - x + 2 x^3 + x^4))/(-1 + x^3 + 3 x^4 + 3 x^5 + 2 x^6 + 6 x^7 + 15 x^8 + 20 x^9 + 15 x^10 + 6 x^11 + x^12)).

a(n) = a(n-3) + 3*a(n-4) + 3*a(n-5) + 2*a(n-6) + 6*a(n-7) + 15*a(n-8) + 20*a(n-9) + 15*a(n-10) + 6*a(n-11) + a(n-12) for n >= 13.

Mathematica

z = 60; s = x + x^2; p = 1 - s^3 - s^6;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A019590 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291407 *)

Crossrefs

Cf. A019590, A291382.

Sequence in context: A170851 A052900 A024947 * A147823 A341211 A335518

Adjacent sequences: A291404 A291405 A291406 * A291408 A291409 A291410

Keyword

nonn,easy

Author

Clark Kimberling, Sep 07 2017

Status

approved