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A349979
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Irregular triangle read by rows: T(n,k) is the number of n-permutations whose second-longest cycle has length exactly k; n>=0, 0<=k<=floor(n/2).
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7
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1, 1, 1, 1, 2, 4, 6, 15, 3, 24, 61, 35, 120, 290, 270, 40, 720, 1646, 1974, 700, 5040, 11025, 14707, 8288, 1260, 40320, 85345, 117459, 90272, 29484, 362880, 749194, 1023390, 974720, 446040, 72576, 3628800, 7347374, 9813210, 10666480, 6332040, 2128896
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OFFSET
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0,5
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COMMENTS
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If the permutation has no second cycle, then its second-longest cycle is defined to have length 0.
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LINKS
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FORMULA
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EXAMPLE
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Triangle begins:
[0] 1;
[1] 1;
[2] 1, 1;
[3] 2, 4;
[4] 6, 15, 3;
[5] 24, 61, 35;
[6] 120, 290, 270, 40;
[7] 720, 1646, 1974, 700;
[8] 5040, 11025, 14707, 8288, 1260;
[9] 40320, 85345, 117459, 90272, 29484;
...
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MAPLE
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b:= proc(n, l) option remember; `if`(n=0, x^l[1], add((j-1)!*
b(n-j, sort([l[], j])[2..3])*binomial(n-1, j-1), j=1..n))
end:
T:= n-> (p-> seq(coeff(p, x, i), i=0..n/2))(b(n, [0$2])):
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MATHEMATICA
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b[n_, l_] := b[n, l] = If[n == 0, x^l[[1]], Sum[(j - 1)!*b[n - j, Sort[ Append[l, j]][[2 ;; 3]]]*Binomial[n - 1, j - 1], {j, 1, n}]];
T[n_] := With[{p = b[n, {0, 0}]}, Table[Coefficient[p, x, i], {i, 0, n/2}]];
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CROSSREFS
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Column 0 gives 1 together with A000142.
Column 1 gives 1 - (n-1)! + A006231(n).
T(2n,n) gives A110468(n-1) for n>=1.
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KEYWORD
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nonn,tabf
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AUTHOR
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STATUS
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approved
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