%I #10 Dec 26 2018 16:00:37
%S 0,7,28,119,532,2352,10388,45913,202916,896777,3963288,17515680,
%T 77410200,342112855,1511961052,6682082183,29531331004,130513137552,
%U 576800248892,2549157374953,11265950967908,49789649104601,220044376637232,972481802150208,4297864230688560
%N p-INVERT of the positive integers, where p(S) = 1 - 7*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 See A290890 for a guide to related sequences.
%H Clark Kimberling, <a href="/A290913/b290913.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, 1, 4, -1)
%F G.f.: (7 x)/(1 - 4 x - x^2 - 4 x^3 + x^4).
%F a(n) = 4*a(n-1) + a(n-2) + 4*a(n-3) - a(n-4).
%F a(n) = 7*A290914(n) for n >= 0.
%t z = 60; s = x/(1 - x)^2; p = 1 - 7 s^2;
%t Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
%t u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A290913 *)
%t u/7 (* A290914 *)
%t LinearRecurrence[{4,1,4,-1},{0,7,28,119},30] (* _Harvey P. Dale_, Dec 26 2018 *)
%Y Cf. A000027, A290890, A290914.
%K nonn,easy
%O 0,2
%A _Clark Kimberling_, Aug 18 2017