A291226 p-INVERT of (0,1,0,1,0,1,...), where p(S) = (1 - S)^6.

6, 21, 62, 168, 426, 1029, 2394, 5403, 11892, 25626, 54228, 112958, 232056, 470904, 945152, 1878351, 3699666, 7227807, 14015538, 26991978, 51654946, 98275461, 185958162, 350093468, 655988730, 1223722623, 2273327418, 4206691146, 7755620994, 14248825833

Offset

0,1

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 (6, -9, -10, 30, 6, -41, -6, 30, 10, -9, -6, -1)

Formula

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

a(n) = 6*a(n-1) - 9*a(n-2) - 10*a(n-3) + 30*a(n-4) + 6*a(n-5) - 41*a(n-6) - 6*a(n-7) + 30*a(n-8) + 10*a(n-9) - 9*a(n-10) - 6*a(n-11) - a(n-12) for n >= 13.

Mathematica

z = 60; s = x/(1 - x^2); p = (1 - s)^6;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000035 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291226 *)

Crossrefs

Cf. A000035, A291219.

Sequence in context: A132130 A022571 A321947 * A228704 A320855 A319613

Adjacent sequences: A291223 A291224 A291225 * A291227 A291228 A291229

Keyword

nonn,easy

Author

Clark Kimberling, Aug 28 2017

Status

approved