%I #9 Jun 02 2023 21:50:47

%S 0,7,14,70,224,868,3080,11368,41216,150640,548576,2000992,7293440,

%T 26592832,96946304,353449600,1288577024,4697851648,17127165440,

%U 62441440768,227645874176,829940392960,3025756030976,11031154419712,40216845025280,146620616568832

%N p-INVERT of (1,1,1,1,1,...), 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 A291000 for a guide to related sequences.

%H Clark Kimberling, <a href="/A291008/b291008.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (2,6).

%F G.f.: 7*x/(1 - 2*x - 6*x^2).

%F a(n) = 2*a(n-1) + 6*a(n-2) for n >= 3.

%F a(n) = 7*A083099(n).

%F a(n) = (sqrt(7)*((1+sqrt(7))^n - (1-sqrt(7))^n)) / 2. - _Colin Barker_, Aug 23 2017

%F a(n) = 7*i^(n-1)*6^((n-1)/2)*ChebyshevU(n-1, -i/sqrt(6)). - _G. C. Greubel_, Jun 01 2023

%t z = 60; s = x/(1 - x); p = 1 - s^7;

%t Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000012 *)

%t Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A291008 *)

%t LinearRecurrence[{2,6}, {0,7}, 40] (* _G. C. Greubel_, Jun 01 2023 *)

%o (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 (SageMath)

%o A291008=BinaryRecurrenceSequence(2,6,0,7)

%o [A291008(n) for n in range(41)] # _G. C. Greubel_, Jun 01 2023

%Y Cf. A000012, A083099, A289780, A291000.

%K nonn,easy

%O 0,2

%A _Clark Kimberling_, Aug 22 2017