A289802 p-INVERT of the quarter-squares (A002620), where p(S) = 1 - S - S^2.

1, 4, 15, 53, 185, 643, 2234, 7764, 26988, 93819, 326149, 1133811, 3941521, 13702079, 47633109, 165588965, 575643853, 2001134880, 6956629199, 24183622175, 84070541130, 292257951771, 1015988587832, 3531923782817, 12278174929397, 42683134990390

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 A289780 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, -5, -4, 12, -5, -4, 4, -1)

Formula

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

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

Mathematica

z = 60; s = x/((1 - x)^2*(1 - x^2)); p = 1 - s - s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A002620 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A289802 *)

Crossrefs

Cf. A002620, A289780.

Sequence in context: A210781 A367818 A303271 * A071719 A370034 A289927

Adjacent sequences: A289799 A289800 A289801 * A289803 A289804 A289805

Keyword

nonn,easy

Author

Clark Kimberling, Aug 12 2017

Status

approved