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A291221 p-INVERT of (0,1,0,1,0,1,...), where p(S) = 1 - S^3 - S^6. 2
0, 0, 1, 0, 3, 2, 6, 12, 13, 42, 42, 117, 156, 312, 531, 894, 1641, 2757, 4866, 8643, 14525, 26637, 44292, 80738, 136563, 243747, 420347, 739188, 1286250, 2252976, 3921546, 6879438, 11951510, 20993796, 36461328, 64002901, 111314775, 195060591, 339831254 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,5

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 (0, 6, 1, -15, -3, 21, 3, -15, -1, 6, 0, -1)

FORMULA

a(n) = a(n-2) + 6*a(n-2) - 15*a(n-3) - 3*a(n-4) + 21*a(n-5) + 3*a(n-6) - 15*a(n-7) - a(n-8) - a(n-10) for n >= 11.

G.f.: x^2*(1 + x - x^2)*(1 - x - x^2 + x^3 + x^4) / (1 - 6*x^2 - x^3 + 15*x^4 + 3*x^5 - 21*x^6 - 3*x^7 + 15*x^8 + x^9 - 6*x^10 + x^12). - Colin Barker, Aug 25 2017

MATHEMATICA

z = 60; s = x/(1 - x^2); p = 1 - s^3 - s^6;

Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000035 *)

Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A291221 *)

PROG

(PARI) concat(vector(2), Vec(x^2*(1 + x - x^2)*(1 - x - x^2 + x^3 + x^4) / (1 - 6*x^2 - x^3 + 15*x^4 + 3*x^5 - 21*x^6 - 3*x^7 + 15*x^8 + x^9 - 6*x^10 + x^12) + O(x^60))) \\ Colin Barker, Aug 25 2017

CROSSREFS

Cf. A000035, A291219.

Sequence in context: A274975 A188621 A175182 * A052616 A091461 A078091

Adjacent sequences: A291218 A291219 A291220 * A291222 A291223 A291224

KEYWORD

nonn,easy

AUTHOR

Clark Kimberling, Aug 24 2017

STATUS

approved

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Last modified February 3 09:09 EST 2023. Contains 360034 sequences. (Running on oeis4.)