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%I #6 Aug 15 2017 19:42:20
%S 2,11,53,253,1205,5740,27336,130200,620129,2953634,14067934,67004505,
%T 319137367,1520027050,7239773429,34482491204,164237487721,
%U 782250685197,3725800625523,17745705518523,84521448139914,402569240665810,1917406730442806,9132462688572345
%N p-INVERT of the primes (A000040), 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 = Sum[Prime[k] x^k, {k, 1, z}]; p = 1 - s - s^2;
%t Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000040 *)
%t Drop[CoefficientList[Series[1/p, {x, 0, z}], x] , 1](* A289847 *)
%Y Cf. A000040, A030017 ("INVERT" applied to the primes), A289928.
%K nonn,easy
%O 0,1
%A _Clark Kimberling_, Aug 14 2017
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