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A291225
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p-INVERT of (0,1,0,1,0,1,...), where p(S) = (1 - S)^5.
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2
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5, 15, 40, 100, 236, 535, 1175, 2515, 5270, 10846, 21980, 43950, 86850, 169840, 329042, 632135, 1205205, 2281925, 4293270, 8030558, 14940700, 27659095, 50968455, 93518940, 170905555, 311159365, 564521620, 1020800470, 1840124050, 3307314163, 5927828905
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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 A291219 for a guide to related sequences.
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LINKS
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Index entries for linear recurrences with constant coefficients, signature (5, -5, -10, 15, 11, -15, -10, 5, 5, 1)
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FORMULA
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a(n) = 5*a(n-1) - 5*a(n-2) - 10*a(n-3) + 15*a(n-4) + 11*a(n-5) - 15*a(n-6) - 10*a(n-7) + 5*a(n-8) + 5*a(n-9) + a(n-10) for n >= 11.
G.f.: (5 - 10*x - 10*x^2 + 25*x^3 + 11*x^4 - 25*x^5 - 10*x^6 + 10*x^7 + 5*x^8) / (1 - x - x^2)^5. - Colin Barker, Aug 28 2017
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MATHEMATICA
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z = 60; s = x/(1 - x^2); p = (1 - s)^5;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000035 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A291225 *)
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PROG
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(PARI) Vec((5 - 10*x - 10*x^2 + 25*x^3 + 11*x^4 - 25*x^5 - 10*x^6 + 10*x^7 + 5*x^8) / (1 - x - x^2)^5 + O(x^40)) \\ Colin Barker, Aug 28 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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