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A291677
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Number of permutations p of [2n] such that 0p has exactly n alternating runs.
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4
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1, 1, 7, 148, 6171, 425976, 43979902, 6346283560, 1219725741715, 301190499710320, 92921064554444490, 35025128774218944648, 15838288022236083603486, 8462453158197423495502224, 5274234568391796228927038748, 3792391176672742840187796835728
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OFFSET
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0,3
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LINKS
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FORMULA
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a(n) ~ c * d^n * n! * (n-1)!, where d = 3.4210546206711870249402157940795853513... and c = 0.32723781013647536133280275922604008889245... - Vaclav Kotesovec, Apr 29 2018
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EXAMPLE
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a(2) = 7: 1243, 1342, 1432, 2341, 2431, 3421, 4321.
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MAPLE
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b:= proc(n, k) option remember; `if`(k=0,
`if`(n=0, 1, 0), `if`(k<0 or k>n, 0,
k*b(n-1, k)+b(n-1, k-1)+(n-k+1)*b(n-1, k-2)))
end:
a:= n-> b(2*n, n):
seq(a(n), n=0..20);
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MATHEMATICA
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b[n_, k_] := b[n, k] = If[k == 0, If[n == 0, 1, 0], If[k < 0 || k > n, 0, k*b[n - 1, k] + b[n - 1, k - 1] + (n - k + 1)*b[n - 1, k - 2]]];
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PROG
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(Python)
from sympy.core.cache import cacheit
@cacheit
def b(n, k): return (1 if n==0 else 0) if k==0 else 0 if k<0 or k>n else k*b(n - 1, k) + b(n - 1, k - 1) + (n - k + 1)*b(n - 1, k - 2)
def a(n): return b(2*n, n)
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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