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A290900 p-INVERT of the positive integers, where p(S) = 1 - S^2 - S^3. 2
0, 1, 5, 17, 51, 149, 439, 1308, 3916, 11728, 35093, 104943, 313773, 938199, 2805439, 8389163, 25086356, 75016104, 224321012, 670787533, 2005857561, 5998122649, 17936209267, 53634716681, 160384099011, 479597177352, 1434141243492, 4288517958652 (list; graph; refs; listen; history; text; internal format)
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 (6, -14, 19, -14, 6, -1)

FORMULA

a(n) = 6*a(n-1) - 14*a(n-2) + 19*a(n-3) - 14*a(n-4) + 6*a(n-5) - a(n-6).

G.f.: x*(1 - x + x^2) / (1 - 6*x + 14*x^2 - 19*x^3 + 14*x^4 - 6*x^5 + x^6). - Colin Barker, Aug 18 2017

MATHEMATICA

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

Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)

Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A290900 *)

PROG

(PARI) concat(0, Vec(x*(1 - x + x^2) / (1 - 6*x + 14*x^2 - 19*x^3 + 14*x^4 - 6*x^5 + x^6) + O(x^40))) \\ Colin Barker, Aug 18 2017

CROSSREFS

Cf. A000027, A290890.

Sequence in context: A039783 A103685 A116521 * A137500 A146814 A034335

Adjacent sequences:  A290897 A290898 A290899 * A290901 A290902 A290903

KEYWORD

nonn,easy

AUTHOR

Clark Kimberling, Aug 17 2017

STATUS

approved

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Last modified April 5 15:53 EDT 2020. Contains 333245 sequences. (Running on oeis4.)