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p-INVERT of (0,0,1,2,3,4,5,...), the nonnegative integers A000027 preceded by one zero, where p(S) = 1 - S - S^2.
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%I #9 Aug 24 2017 08:23:44

%S 0,0,1,2,3,6,13,26,50,96,184,351,669,1278,2447,4692,9004,17285,33182,

%T 63687,122208,234461,449774,862776,1655010,3174766,6090231,11683285,

%U 22413104,42997349,82486280,158241688,303570021,582365698,1117202719,2143225358

%N p-INVERT of (0,0,1,2,3,4,5,...), the nonnegative integers A000027 preceded by one zero, 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 A290890 for a guide to related sequences.

%H Clark Kimberling, <a href="/A290991/b290991.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 (4, -6, 5, -3, 1, 1)

%F a(n) = 4*a(n-1) - 6*a(n-2) + 5*a(n-3) - 3*a(n-4) + a(n-5) + a(n-6).

%F G.f.: x^2*(1 - 2*x + x^2 + x^3) / (1 - 4*x + 6*x^2 - 5*x^3 + 3*x^4 - x^5 - x^6). - _Colin Barker_, Aug 24 2017

%t z = 60; s = x^3/(1 - x)^2; p = 1 - s - s^2;

%t Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* 0,0,1,2,3,4,5,... *)

%t Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A290991 *)

%o (PARI) concat(vector(2), Vec(x^2*(1 - 2*x + x^2 + x^3) / (1 - 4*x + 6*x^2 - 5*x^3 + 3*x^4 - x^5 - x^6) + O(x^40))) \\ _Colin Barker_, Aug 24 2017

%Y Cf. A000027, A289780, A290990.

%K nonn,easy

%O 0,4

%A _Clark Kimberling_, Aug 21 2017