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A290890
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p-INVERT of the positive integers, where p(S) = 1 - S^2.
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65
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0, 1, 4, 11, 28, 72, 188, 493, 1292, 3383, 8856, 23184, 60696, 158905, 416020, 1089155, 2851444, 7465176, 19544084, 51167077, 133957148, 350704367, 918155952, 2403763488, 6293134512, 16475640049, 43133785636, 112925716859, 295643364940, 774004377960
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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).
Note that in A290890, s = (1,2,3,4,...); i.e., A000027(n+1) for n>=0, whereas in A290990, s = (0,1,2,3,4,...); i.e., A000027(n) for n>=0.
Guide to p-INVERT sequences using s = (1,2,3,4,5,...) = A000027:
p(S) t(1,2,3,4,5,...)
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LINKS
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FORMULA
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G.f.: x/(1 - 4 x + 5 x^2 - 4 x^3 + x^4).
a(n) = 4*a(n-1) - 5*a(n-2) + 4*a(n-3) - a(n-4).
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EXAMPLE
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MATHEMATICA
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z = 60; s = x/(1 - x)^2; p = 1 - s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A290890 *)
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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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