A292489 p-INVERT of the odd positive integers, where p(S) = 1 - S - 6 S^2.

1, 10, 60, 312, 1656, 8928, 48024, 257904, 1385352, 7442784, 39985272, 214811280, 1154025000, 6199749504, 33306803352, 178933509936, 961281138888, 5164272731808, 27743925989304, 149048175357648, 800728728609384, 4301739993919680, 23110157427289560

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 A292480 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 (5, -1, 15, 6)

Formula

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

a(n) = 5*a(n-1) - a(n-2) + 16*a(n-3) + 6*a(n-4) for n >= 5.

Mathematica

z = 60; s = x (x + 1)/(1 - x)^2; p = 1 - s - 6 s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A005408 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A292489 *)

Prog

(PARI) x='x+O('x^99); Vec(((1+x)*(1+4*x+7*x^2))/((1-5*x-2*x^2)*(1+3*x^2))) \\ Altug Alkan, Oct 03 2017

Crossrefs

Cf. A005408, A292480.

Sequence in context: A003472 A112502 A293081 * A083585 A250575 A155633

Adjacent sequences: A292486 A292487 A292488 * A292490 A292491 A292492

Keyword

nonn,easy

Author

Clark Kimberling, Oct 03 2017

Status

approved