cp's OEIS Frontend

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

%I A290930 #6 Aug 19 2017 13:24:40
%S A290930 0,3,12,37,116,372,1188,3763,11860,37261,116760,365056,1139224,
%T A290930 3549635,11045804,34335421,106633804,330916268,1026277180,3181108619,
%U A290930 9855901108,30524529485,94506627952,292521594048,905220237168,2800700318291,8663793207244
%N A290930 p-INVERT of the positive integers, where p(S) = (1 - S^2)(1 - 2*S^2).
%C A290930 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 A290930 See A290890 for a guide to related sequences.
%H A290930 Clark Kimberling, <a href="/A290930/b290930.txt">Table of n, a(n) for n = 0..1000</a>
%H A290930 <a href="/index/Rec#order_08">Index entries for linear recurrences with constant coefficients</a>, signature (8, -25, 44, -54, 44, -25, 8, -1)
%F A290930 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).
%F A290930 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).
%t A290930 z = 60; s = x/(1 - x)^2; p = (1 - s^2)(1 - 2s^2);
%t A290930 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
%t A290930 Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A290930 *)
%Y A290930 Cf. A000027, A290890.
%K A290930 nonn,easy
%O A290930 0,2
%A A290930 _Clark Kimberling_, Aug 19 2017