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 A289844 p-INVERT of A175676 (starting at n=3), where p(S) = 1 - S - S^2. 2
 1, 2, 3, 7, 16, 31, 64, 134, 274, 567, 1168, 2405, 4967, 10232, 21094, 43505, 89672, 184892, 381203, 785886, 1620327, 3340606, 6887304, 14199737, 29275538, 60357622, 124439898, 256558196, 528948160, 1090536002, 2248364880, 4635470266, 9556979689, 19703689739 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,2 COMMENTS 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). See A289780 for a guide to related sequences. LINKS Index entries for linear recurrences with constant coefficients, signature (1, 1, 4, -2, 0, -6, 1, 0, 4, 0, 0, -1) FORMULA 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). 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 MATHEMATICA z = 60; s = x/((x - 1)^2*(1 + x + x^2)^2); p = 1 - s - s^2; Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A175676, shifted *) Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A289844 *) PROG (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 CROSSREFS Cf. A175686, A289780. Sequence in context: A250193 A004782 A049956 * A153056 A235112 A081207 Adjacent sequences:  A289841 A289842 A289843 * A289845 A289846 A289847 KEYWORD nonn,easy AUTHOR Clark Kimberling, Aug 12 2017 STATUS approved

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Last modified March 18 16:00 EDT 2019. Contains 321292 sequences. (Running on oeis4.)