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A291002
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p-INVERT of (1,1,1,1,1,...), where p(S) = (1 - S)*(1 - 2*S)*(1 - 3*S).
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2
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6, 31, 146, 652, 2816, 11896, 49496, 203752, 832376, 3381736, 13683896, 55206952, 222242936, 893219176, 3585623096, 14380739752, 57637717496, 230895178216, 924613703096, 3701553914152, 14815513224056, 59289946122856, 237243465219896, 949224905162152
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OFFSET
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0,1
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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).
See A291000 for a guide to related sequences.
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LINKS
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FORMULA
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G.f.: (6 - 23*x + 23*x^2)/(1 - 9*x + 26*x^2 - 24*x^3).
a(n) = 9*a(n-1) - 26*a(n-2) + 24*a(n-3) for n >= 4.
a(n) = (2^n - 16*3^n + 27*4^n) / 2. - Colin Barker, Aug 23 2017
E.g.f.: (1/2)*(exp(2*x) - 16*exp(3*x) + 27*exp(4*x)). - G. C. Greubel, Apr 27 2023
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MATHEMATICA
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z = 60; s = x/(1-x); p = (1-s)*(1-2*s)*(1-3*s);
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000012 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A291002 *)
LinearRecurrence[{9, -26, 24}, {6, 31, 146}, 41] (* G. C. Greubel, Apr 27 2023 *)
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PROG
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(Magma) [(2^n-16*3^n+27*4^n)/2: n in [0..40]]; // G. C. Greubel, Apr 27 2023
(SageMath) [(2^n-16*3^n+27*4^n)/2 for n in range(41)] # G. C. Greubel, Apr 27 2023
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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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