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A292535 p-INVERT of the squares (A000290), where p(S) = 1 + S - 2 S^2. 1
1, 7, 38, 189, 909, 4368, 21093, 102051, 493702, 2387689, 11546425, 55837024, 270025769, 1305841103, 6315023830, 30539305893, 147687325509, 714212301776, 3453913488845, 16703042620715, 80775512179990, 390628431683601, 1889069687788593, 9135495517900480 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,2

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 A292479 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 (7, -15, 24, -11, 5, -1)

FORMULA

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

a(n) = 7*a(n-1) - 15*a(n-2) + 24*a(n-3) - 11*a(n-4) + 5*a(n-5) - a(n-6) for n >= 7.

MATHEMATICA

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

Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A005408 *)

Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A292535 *)

CROSSREFS

Cf. A000290, A292479.

Sequence in context: A241524 A291822 A099453 * A026763 A217340 A037696

Adjacent sequences:  A292532 A292533 A292534 * A292536 A292537 A292538

KEYWORD

nonn,easy

AUTHOR

Clark Kimberling, Oct 04 2017

STATUS

approved

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Last modified April 3 17:03 EDT 2020. Contains 333197 sequences. (Running on oeis4.)