|
%I #12 Aug 13 2017 10:46:50
%S 1,2,3,7,16,31,64,134,274,567,1168,2405,4967,10232,21094,43505,89672,
%T 184892,381203,785886,1620327,3340606,6887304,14199737,29275538,
%U 60357622,124439898,256558196,528948160,1090536002,2248364880,4635470266,9556979689,19703689739
%N p-INVERT of A175676 (starting at n=3), 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) + c(1)*x + c(2)*x^2 + ... 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 <a href="/index/Rec#order_12">Index entries for linear recurrences with constant coefficients</a>, signature (1, 1, 4, -2, 0, -6, 1, 0, 4, 0, 0, -1)
%F a(n) = a(n-1) + a(n-2) + 4*a(n-3) - 2*a(n-4) - 6*a(n-7) + a(n-8) + 4*a(n-10) - a(n-13).
%F G.f.: (1 + x - 2*x^3 + x^6) / (1 - x - x^2 - 4*x^3 + 2*x^4 + 6*x^6 - x^7 - 4*x^9 + x^12). - _Colin Barker_, Aug 13 2017
%t z = 60; s = x/((x - 1)^2*(1 + x + x^2)^2); p = 1 - s - s^2;
%t Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A175676, shifted *)
%t Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A289844 *)
%o (PARI) Vec((1 + x - 2*x^3 + x^6) / (1 - x - x^2 - 4*x^3 + 2*x^4 + 6*x^6 - x^7 - 4*x^9 + x^12) + O(x^60)) \\ _Colin Barker_, Aug 13 2017
%Y Cf. A175686, A289780.
%K nonn,easy
%O 0,2
%A _Clark Kimberling_, Aug 12 2017
|
