|
%I #11 Aug 13 2017 17:34:35
%S 1,3,9,26,74,211,600,1708,4860,13832,39364,112029,318827,907366,
%T 2582312,7349121,20915193,59523497,169400608,482104856,1372044007,
%U 3904762096,11112739032,31626246588,90006565434,256153755080,728999555983,2074692805003,5904462080604
%N p-INVERT of (1,1,2,2,3,3,...) (A008619), where p(S) = 1 - S - 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 A289780 for a guide to related sequences.
%H Clark Kimberling, <a href="/A289806/b289806.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (3, 1, -5, 2, 2, -1)
%F G.f.: (1 - x^2 + x^3)/(1 - 3 x - x^2 + 5 x^3 - 2 x^4 - 2 x^5 + x^6).
%F a(n) = 3*a(n-1) + a(n-2) - 5*a(n-3) + 2*a(n-4) + 2*a(n-5) - a(n-6).
%t z = 60; s = x/((1 - x) (1 - x^2)); p = 1 - s - s^2;
%t Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A008619 shifted *)
%t Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A289806 *)
%Y Cf. A008619, A289780.
%K nonn,easy
%O 0,2
%A _Clark Kimberling_, Aug 12 2017
|
