%I #17 May 29 2022 18:07:05

%S 0,1,4,11,28,72,188,493,1292,3383,8856,23184,60696,158905,416020,

%T 1089155,2851444,7465176,19544084,51167077,133957148,350704367,

%U 918155952,2403763488,6293134512,16475640049,43133785636,112925716859,295643364940,774004377960

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

%C 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).

%C Note that in A290890, s = (1,2,3,4,...); i.e., A000027(n+1) for n>=0, whereas in A290990, s = (0,1,2,3,4,...); i.e., A000027(n) for n>=0.

%C Guide to p-INVERT sequences using s = (1,2,3,4,5,...) = A000027:

%C p(S) t(1,2,3,4,5,...)

%C 1 - S A001906

%C 1 - S^2 A290890; see A113067 for signed version

%C 1 - S^3 A290891

%C 1 - S^4 A290892

%C 1 - S^5 A290893

%C 1 - S^6 A290894

%C 1 - S^7 A290895

%C 1 - S^8 A290896

%C 1 - S - S^2 A289780

%C 1 - S - S^3 A290897

%C 1 - S - S^4 A290898

%C 1 - S^2 - S^4 A290899

%C 1 - S^2 - S^3 A290900

%C 1 - S^3 - S^4 A290901

%C 1 - 2S A052530; (1/2)*A052530 = A001353

%C 1 - 3S A290902; (1/3)*A290902 = A004254

%C 1 - 4S A003319; (1/4)*A003319 = A001109

%C 1 - 5S A290903; (1/5)*A290903 = A004187

%C 1 - 2*S^2 A290904; (1/2)*A290904 = A290905

%C 1 - 3*S^2 A290906; (1/3)*A290906 = A290907

%C 1 - 4*S^2 A290908; (1/4)*A290908 = A099486

%C 1 - 5*S^2 A290909; (1/5)*A290909 = A290910

%C 1 - 6*S^2 A290911; (1/6)*A290911 = A290912

%C 1 - 7*S^2 A290913; (1/7)*A290913 = A290914

%C 1 - 8*S^2 A290915; (1/8)*A290915 = A290916

%C (1 - S)^2 A290917

%C (1 - S)^3 A290918

%C (1 - S)^4 A290919

%C (1 - S)^5 A290920

%C (1 - S)^6 A290921

%C 1 - S - 2*S^2 A290922

%C 1 - 2*S - 2*S^2 A290923; (1/2)*A290923 = A290924

%C 1 - 3*S - 2*S^2 A290925

%C (1 - S^2)^2 A290926

%C (1 - S^2)^3 A290927

%C (1 - S^3)^2 A290928

%C (1 - S)(1 - S^2) A290929

%C (1 - S^2)(1 - S^4) A290930

%C 1 - 3 S + S^2 A291025

%C 1 - 4 S + S^2 A291026

%C 1 - 5 S + S^2 A291027

%C 1 - 6 S + S^2 A291028

%C 1 - S - S^2 - S^3 A291029

%C 1 - S - S^2 - S^3 - S^4 A201030

%C 1 - 3 S + 2 S^3 A291031

%C 1 - S - S^2 - S^3 + S^4 A291032

%C 1 - 6 S A291033

%C 1 - 7 S A291034

%C 1 - 8 S A291181

%C 1 - 3 S + 2 S^3 A291031

%C 1 - 3 S + 2 S^2 A291182

%C 1 - 4 S + 2 S^3 A291183

%C 1 - 4 S + 3 S^3 A291184

%H Clark Kimberling, <a href="/A290890/b290890.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4, -5, 4, -1)

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

%F a(n) = 4*a(n-1) - 5*a(n-2) + 4*a(n-3) - a(n-4).

%e (See the examples at A289780.)

%t z = 60; s = x/(1 - x)^2; p = 1 - s^2;

%t Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)

%t Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A290890 *)

%Y Cf. A000027, A113067, A289780, A113067 (signed version of same sequence).

%K nonn,easy

%O 0,3

%A _Clark Kimberling_, Aug 15 2017