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A291008 p-INVERT of (1,1,1,1,1,...), where p(S) = 1 - 7*S^2.

%I A291008 #9 Jun 02 2023 21:50:47
%S A291008 0,7,14,70,224,868,3080,11368,41216,150640,548576,2000992,7293440,
%T A291008 26592832,96946304,353449600,1288577024,4697851648,17127165440,
%U A291008 62441440768,227645874176,829940392960,3025756030976,11031154419712,40216845025280,146620616568832
%N A291008 p-INVERT of (1,1,1,1,1,...), where p(S) = 1 - 7*S^2.
%C A291008 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 A291008 See A291000 for a guide to related sequences.
%H A291008 Clark Kimberling, <a href="/A291008/b291008.txt">Table of n, a(n) for n = 0..1000</a>
%H A291008 <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (2,6).
%F A291008 G.f.: 7*x/(1 - 2*x - 6*x^2).
%F A291008 a(n) = 2*a(n-1) + 6*a(n-2) for n >= 3.
%F A291008 a(n) = 7*A083099(n).
%F A291008 a(n) = (sqrt(7)*((1+sqrt(7))^n - (1-sqrt(7))^n)) / 2. - _Colin Barker_, Aug 23 2017
%F A291008 a(n) = 7*i^(n-1)*6^((n-1)/2)*ChebyshevU(n-1, -i/sqrt(6)). - _G. C. Greubel_, Jun 01 2023
%t A291008 z = 60; s = x/(1 - x); p = 1 - s^7;
%t A291008 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000012 *)
%t A291008 Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291008 *)
%t A291008 LinearRecurrence[{2,6}, {0,7}, 40] (* _G. C. Greubel_, Jun 01 2023 *)
%o A291008 (Magma) [n le 2 select 7*(n-1) else 2*Self(n-1) + 6*Self(n-2): n in [1..30]]; // _G. C. Greubel_, Jun 01 2023
%o A291008 (SageMath)
%o A291008 A291008=BinaryRecurrenceSequence(2,6,0,7)
%o A291008 [A291008(n) for n in range(41)] # _G. C. Greubel_, Jun 01 2023
%Y A291008 Cf. A000012, A083099, A289780, A291000.
%K A291008 nonn,easy
%O A291008 0,2
%A A291008 _Clark Kimberling_, Aug 22 2017