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 A292483 p-INVERT of the odd positive integers, where p(S) = (1 - S)^3. 1
 3, 15, 61, 240, 912, 3376, 12240, 43632, 153360, 532656, 1831248, 6240240, 21100176, 70858800, 236510928, 785115504, 2593432080, 8528565168, 27932538960, 91144257264, 296391022992, 960802812720, 3105562639824, 10010945435760, 32189993590032, 103264606820016 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,1 COMMENTS 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). See A292480 for a guide to related sequences. LINKS Clark Kimberling, Table of n, a(n) for n = 0..1000 Index entries for linear recurrences with constant coefficients, signature (9, -27, 27) FORMULA G.f.: -(((1 + x) (3 - 15 x + 22 x^2 - 7 x^3 + x^4))/(-1 + 3 x)^3). a(n) = 9*a(n-1) - 27*a(n-2) + 27*a(n-3)  for n >= 6. a(n) = 16*3^(n-5)*(51 + 22*n + 2*n^2) for n>2. - Colin Barker, Oct 03 2017 MATHEMATICA z = 60; s = x (x + 1)/(1 - x)^2; p = (1 - s)^3; Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A005408 *) Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A292483 *) CROSSREFS Cf. A005408, A292480. Sequence in context: A295505 A001655 A058749 * A218227 A072336 A122671 Adjacent sequences:  A292480 A292481 A292482 * A292484 A292485 A292486 KEYWORD nonn,easy AUTHOR Clark Kimberling, Oct 02 2017 STATUS approved

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Last modified December 10 23:29 EST 2019. Contains 329910 sequences. (Running on oeis4.)