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A292398
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p-INVERT of A010892, where p(S) = 1 - S - S^2 - S^3.
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2
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1, 3, 10, 32, 102, 323, 1021, 3224, 10177, 32121, 101378, 319960, 1009830, 3187145, 10059029, 31747584, 100199485, 316242607, 998102878, 3150142840, 9942261690, 31379074783, 99036453193, 312571964808, 986517893269, 3113579153493, 9826861945870
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OFFSET
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0,2
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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 - x^2 + x^3 + x^4)/(-1 + 4 x - x^2 - 6 x^3 + x^4 + 4 x^5 + x^6)).
a(n) = 4*a(n-1) - a(n-2) - 6*a(n-3) + a(n-4) + 4*a(n-5) + a(n-6) for n >= 7.
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MAPLE
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if n=0 then 1 elif n=1 then 3 elif n=2 then 10 elif n=3 then 32 elif n=4 then 102 elif n=5 then 323 elif n>=6 then 4*procname(n-1)-procname(n-2)-6*procname(n-3)+procname(n-4)+4*procname(n-5)+procname(n-6) fi; end:
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MATHEMATICA
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z = 60; s = x/(1 - x + x^2); p = 1 - s - s^2 - s^3;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A010892 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A292398 *)
LinearRecurrence[{4, -1, -6, 1, 4, 1}, {1, 3, 10, 32, 102, 323}, 30] (* Harvey P. Dale, Feb 24 2022 *)
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PROG
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(GAP)
a:=[1, 3, 10, 32, 102, 323];; for n in [7..10^2] do a[n]:=4*a[n-1]-a[n-2]-6*a[n-3]+a[n-4]+4*a[n-5]+a[n-6]; od; A292398:=a; # Muniru A Asiru, Oct 02 2017
(PARI) Vec(-(1 - x - x^2 + x^3 + x^4)/(-1 + 4*x - x^2 - 6*x^3 + x^4 + 4*x^5 + x^6) + O(x^20)) \\ Felix Fröhlich, Oct 02 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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