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A317111
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Number of permutations of [n] in which the length of every increasing run is 0 or 1 (mod 4).
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14
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1, 1, 1, 1, 2, 10, 50, 210, 840, 4200, 29400, 231000, 1755600, 13213200, 109309200, 1051050000, 11099088000, 120071952000, 1320791472000, 15317750448000, 192286654560000, 2577944809440000, 35885904294240000, 513695427204960000, 7641940962015360000
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OFFSET
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0,5
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COMMENTS
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Similarly, 1/(1 - x + x^2/2! - ... - x^(2m-1)/(2m-1)!) is the e.g.f. for permutations in which every increasing run has length 0 or 1 (mod 2m).
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LINKS
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FORMULA
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E.g.f.: 1/(1 - x + x^2/2! - x^3/3!).
a(0) = a(1) = a(2) = 1; a(n) = n * a(n-1) - n * (n-1) * a(n-2) / 2 + n * (n-1) * (n-2) * a(n-3) / 6 for n > 2. - Ilya Gutkovskiy, Jan 22 2024
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EXAMPLE
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For n=4 the a(4)=2 permutations are 4321 and 1234.
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MAPLE
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gser:=series(1/(1-x+x^2/2!-x^3/3!), x, 21): seq(n!*coeff(gser, x, n), n=0..20);
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MATHEMATICA
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With[{nmax = 25}, CoefficientList[Series[1/(1 -x +x^2/2! -x^3/3!), {x, 0, nmax}], x]*Range[0, nmax]!] (* G. C. Greubel, Nov 30 2018 *)
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PROG
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(PARI) my(x='x+O('x^25)); Vec(serlaplace(1/(1 -x +x^2/2 -x^3/6))) \\ G. C. Greubel, Nov 30 2018
(Magma) m:=25; R<x>:=PowerSeriesRing(Rationals(), m); b:=Coefficients(R!( 1/(1-x+x^2/2-x^3/6) )); [Factorial(n-1)*b[n]: n in [1..m]]; // G. C. Greubel, Nov 30 2018
(Sage)
f= 1/(1 -x +x^2/2 -x^3/6)
g=f.taylor(x, 0, 13)
L=g.coefficients()
coeffs={c[1]:c[0]*factorial(c[1]) for c in L}
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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STATUS
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approved
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