A290930 p-INVERT of the positive integers, where p(S) = (1 - S^2)(1 - 2*S^2).

0, 3, 12, 37, 116, 372, 1188, 3763, 11860, 37261, 116760, 365056, 1139224, 3549635, 11045804, 34335421, 106633804, 330916268, 1026277180, 3181108619, 9855901108, 30524529485, 94506627952, 292521594048, 905220237168, 2800700318291, 8663793207244

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 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 (8, -25, 44, -54, 44, -25, 8, -1)

Formula

G.f.: (3 x - 12 x^2 + 16 x^3 - 12 x^4 + 3 x^5)/(1 - 8 x + 25 x^2 - 44 x^3 + 54 x^4 - 44 x^5 + 25 x^6 - 8 x^7 + x^8).

a(n) = 8*a(n-1) - 25*a(n-2) + 44*a(n-3) - 54*a(n-4) + 44*a(n-5) - 25*a(n-6) + 8*a(n-7) - a(n-8).

Mathematica

z = 60; s = x/(1 - x)^2; p = (1 - s^2)(1 - 2s^2);
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A290930 *)

Crossrefs

Cf. A000027, A290890.

Sequence in context: A211958 A255610 A022727 * A264423 A240193 A319077

Adjacent sequences: A290927 A290928 A290929 * A290931 A290932 A290933

Keyword

nonn,easy

Author

Clark Kimberling, Aug 19 2017

Status

approved