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A291031 p-INVERT of the positive integers, where p(S) = 1 - 3*S + 2*S^3. 2
3, 15, 70, 321, 1461, 6624, 29967, 135399, 611318, 2758881, 12447753, 56154744, 253306119, 1142572767, 5153589754, 23244956169, 104843981505, 472885383744, 2132882300571, 9620044596687, 43389716584682, 195702453488433, 882684641446989, 3981207177094608 (list; graph; refs; listen; history; text; internal format)
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 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 (9, -27, 36, -27, 9, -1)

FORMULA

G.f.: (3 - 12 x + 16 x^2 - 12 x^3 + 3 x^4)/(1 - 9 x + 27 x^2 - 36 x^3 + 27 x^4 - 9 x^5 + x^6).

a(n) = 9*a(n-1) - 27*a(n-2) + 36*a(n-3) - 27*a(n-4) + 90*a(n-5) - a(n-6).

MATHEMATICA

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

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

Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291031 *)

CROSSREFS

Cf. A000027, A290890.

Sequence in context: A213140 A245751 A033876 * A009174 A178345 A183547

Adjacent sequences:  A291028 A291029 A291030 * A291032 A291033 A291034

KEYWORD

nonn,easy

AUTHOR

Clark Kimberling, Aug 19 2017

STATUS

approved

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Last modified October 23 19:37 EDT 2019. Contains 328373 sequences. (Running on oeis4.)