A292533 p-INVERT of the squares (A000290), where p(S) = (1 - S)^2.

2, 11, 46, 187, 748, 2948, 11480, 44273, 169374, 643601, 2431526, 9140616, 34212350, 127563959, 474022478, 1756118055, 6488228880, 23912815820, 87935847700, 322713694333, 1182114988606, 4322734288413, 15782353895178, 57537481431056, 209479529802682

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 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 (8, -20, 18, -12, 4, -1)

Formula

G.f.: -(((1 + x) (-2 + 7 x - 5 x^2 + 2 x^3))/(-1 + 4 x - 2 x^2 + x^3)^2).

a(n) = 8*a(n-1) - 20*a(n-2) + 18*a(n-3) - 12*a(n-4) + 4*a(n-5) - a(n-6) for n >= 7.

Mathematica

z = 60; s = x (x + 1)/(1 - x)^3; p = (1 - s)^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A005408 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A292533 *)

Crossrefs

Cf. A000290, A292479.

Sequence in context: A319428 A000176 A042927 * A140305 A229019 A142346

Adjacent sequences: A292530 A292531 A292532 * A292534 A292535 A292536

Keyword

nonn,easy

Author

Clark Kimberling, Oct 04 2017

Status

approved