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A138771 Triangle read by rows: T(n,k) is the number of permutations of {1,2,...,n} whose 2nd cycle has k entries; each cycle is written with the smallest element first and cycles are arranged in increasing order of their first elements (n>=1; 0<=k<=n-1). For example, 1432=(1)(24)(3) has 2 entries in the 2nd cycle; 3421=(1324) has 0 entries in the 2nd cycle. 3
1, 1, 1, 2, 3, 1, 6, 11, 5, 2, 24, 50, 26, 14, 6, 120, 274, 154, 94, 54, 24, 720, 1764, 1044, 684, 444, 264, 120, 5040, 13068, 8028, 5508, 3828, 2568, 1560, 720, 40320, 109584, 69264, 49104, 35664, 25584, 17520, 10800, 5040 (list; table; graph; refs; listen; history; text; internal format)
OFFSET
1,4
COMMENTS
T(n,0)=(n-1)!=A000142(n-1).
T(n,1)=A000254(n-1).
T(n,2)=A001705(n-2).
T(n,3)=2*A001711(n-4).
T(n,4)=6*A001716(n-5).
T(n,n-1)=(n-2)! (n>=2).
Sum(kT(n,k),k=0..n-1)=(n-1)!(n-1)(n+2)/4=A138772(n).
LINKS
FORMULA
T(n,k)=(n-1)T(n-1,k)+(n-2)! (1<=k<=n-1). The row generating polynomials P[n](t) satisfy: P[n+1](t)=nP[n](t)+(n-1)!(t+t^2+...+t^n).
EXAMPLE
T(4,2)=5 because we have (1)(23)(4), (1)(24)(3), (13)(24), (12)(34) and (14)(23).
Triangle starts;
1;
1,1;
2,3,1;
6,11,5,2;
24,50,26,14,6;
120,274,154,94,54,24;
MAPLE
T:=proc (n, k) if k = 0 then factorial(n-1) elif n <= k then 0 else (n-1)*T(n-1, k)+factorial(n-2) end if end proc: for n to 9 do seq(T(n, k), k=0..n-1) end do;
CROSSREFS
From Johannes W. Meijer, Oct 16 2009: (Start)
A000142 equals for n=>1 the row sums.
a(n) = A165680(n) * A165675(n-1).
(End)
Sequence in context: A155856 A086960 A165675 * A121748 A174893 A008275
KEYWORD
nonn,tabl
AUTHOR
Emeric Deutsch, Apr 10 2008
STATUS
approved

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Last modified April 20 05:25 EDT 2024. Contains 371798 sequences. (Running on oeis4.)