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A292480
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p-INVERT of the odd positive integers, where p(S) = 1 - S^2.
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17
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0, 1, 6, 20, 56, 160, 480, 1456, 4384, 13136, 39360, 118064, 354272, 1062928, 3188736, 9565936, 28697632, 86093264, 258280512, 774841520, 2324523104, 6973567888, 20920705152, 62762119792, 188286360736, 564859074896, 1694577214656, 5083731648560
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OFFSET
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0,3
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COMMENTS
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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).
In the following guide to p-INVERT sequences using s = (1,3,5,7,9,...) = A005408, in some cases t(1,3,5,7,9,...) is a shifted (or differently indexed) version of the cited sequence:
p(S) *********** t(1,3,5,7,9,...)
1 - S^3 (not yet in OEIS)
(1 - S)^2 (not yet in OEIS)
(1 - S)^3 (not yet in OEIS)
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LINKS
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FORMULA
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G.f.: x*(1 + x)^2/((1 - 3*x)*(1 - x + 2*x^2)).
a(n) = 4*a(n-1) - 5*a(n-2) + 6*a(n-3) for n >= 5.
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EXAMPLE
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s = (1,3,5,7,9,...), S(x) = x + 3 x^2 + 5 x^3 + 7 x^4 + ...,
p(S(x)) = 1 - ( x + 3 x^2 + 5 x^3 + 7 x^4 + ...)^2,
1/p(S(x)) = 1 + x^2 + 6 x^3 + 20 x^4 + 56 x^5 + ...
T(x) = (-1 + 1/p(S(x)))/x = x + 6 x^2 + 20 x^3 + 56 x^4 + ...
t(s) = (0,1,2,20,56,...).
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MATHEMATICA
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z = 60; s = x (x + 1)/(1 - x)^2; p = 1 - s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A005408 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A292480 *)
Join[{0}, LinearRecurrence[{4, -5, 6}, {1, 6, 20}, 30]] (* Vincenzo Librandi, Oct 03 2017 *)
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PROG
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(Magma) I:=[0, 1, 6, 20]; [n le 4 select I[n] else 4*Self(n-1)- 5*Self(n-2)+6*Self(n-3): n in [1..30]]; // Vincenzo Librandi, Oct 03 2017
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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