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A292301
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p-INVERT of A010892, where p(S) = 1 + S - S^2.
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5
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-1, 1, 1, -1, 0, 1, -1, -1, 1, 0, -1, 1, 1, -1, 0, 1, -1, -1, 1, 0, -1, 1, 1, -1, 0, 1, -1, -1, 1, 0, -1, 1, 1, -1, 0, 1, -1, -1, 1, 0, -1, 1, 1, -1, 0, 1, -1, -1, 1, 0, -1, 1, 1, -1, 0, 1, -1, -1, 1
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OFFSET
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0
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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).
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LINKS
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FORMULA
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G.f.: -((-1 + x)^2/(1 - x + x^2 - x^3 + x^4)).
a(n) = a(n-1) - a(n-2) + a(n-3) - a(n-4) for n >= 5.
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MAPLE
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A292301 := proc(n) option remember: if n = 0 then -1 elif n = 1 then 1 elif n = 2 then 1 elif n = 3 then -1 elif n >= 4 then procname(n-1) - procname(n-2) + procname(n-3) - procname(n-4) fi; end:
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MATHEMATICA
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z = 60; s = x/(1 - x + x^2); p = 1 + s - s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A010892 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A292301 *)
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PROG
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(GAP) a:=[-1, 1, 1, -1];; for n in [5..10^3] do a[n] := a[n-1] - a[n-2] + a[n-3] -a [n-4]; od; A292301 := a; # Muniru A Asiru, Oct 16 2017
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CROSSREFS
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KEYWORD
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easy,sign
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AUTHOR
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STATUS
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approved
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