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%I #4 Aug 15 2017 19:43:22
%S 1,4,14,48,162,547,1842,6206,20906,70438,237326,799629,2694199,
%T 9077599,30585239,103051135,347211149,1169861760,3941626163,
%U 13280557904,44746308037,150764154490,507971076799,1711511703373,5766612400708,19429501132982,65464000013233
%N p-INVERT of (1,2,3,5,7,11,13,...); i.e., 1 and the primes (A008578), 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.
%t z = 60; s = x + Sum[Prime[k] x^(k + 1), {k, 1, z}]; p = 1 - s - s^2;
%t Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A008578 *)
%t Drop[CoefficientList[Series[1/p, {x, 0, z}], x] , 1] (* A289928 *)
%Y Cf. A008578, A289847.
%K nonn,easy
%O 0,2
%A _Clark Kimberling_, Aug 15 2017
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