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A291028 p-INVERT of the positive integers, where p(S) = 1 - 6*S + S^2. 2

%I #9 Sep 08 2022 08:46:19

%S 6,47,362,2787,21456,165180,1271644,9789793,75367038,580215573,

%T 4466808294,34387867640,264736107506,2038079457267,15690220398162,

%U 120791667500967,929918545909756,7159007901103540,55113853093361544,424295774604244773,3266454697733704038

%N p-INVERT of the positive integers, where p(S) = 1 - 6*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="/A291028/b291028.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (10,-19,10,-1).

%F G.f.: (6 - 13 x + 6 x^2)/(1 - 10 x + 19 x^2 - 10 x^3 + x^4).

%F a(n) = 10*a(n-1) - 19*a(n-2) + 10*a(n-3) - a(n-4).

%t z = 60; s = x/(1 - x)^2; p = 1 - 6 s + s^2;

%t Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)

%t Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A291028 *)

%t LinearRecurrence[{10, -19, 10, -1}, {6, 47, 362, 2787}, 40] (* _Vincenzo Librandi_, Aug 20 2017 *)

%o (Magma) I:=[6,47,362,2787]; [n le 4 select I[n] else 10*Self(n-1)-19*Self(n-2)+10*Self(n-3)-Self(n-4): n in [1..40]]; // _Vincenzo Librandi_, Aug 20 2017

%Y Cf. A000027, A290890.

%K nonn,easy

%O 0,1

%A _Clark Kimberling_, Aug 19 2017

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Last modified September 19 08:32 EDT 2024. Contains 376007 sequences. (Running on oeis4.)