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A184184 Triangle read by rows: T(n,k) is the number of permutations of [n] having k adjacent cycles (0 <= k <= n). An adjacent cycle is a cycle of the form (i, i+1, i+2, ...) (including 1-element cycles). 1
1, 0, 1, 0, 1, 1, 1, 2, 2, 1, 6, 8, 6, 3, 1, 34, 42, 27, 12, 4, 1, 216, 258, 156, 64, 20, 5, 1, 1566, 1824, 1068, 420, 125, 30, 6, 1, 12840, 14664, 8400, 3220, 930, 216, 42, 7, 1, 117696, 132360, 74580, 28080, 7950, 1806, 343, 56, 8, 1, 1193760, 1326120, 737640, 273960, 76440, 17094, 3192, 512, 72, 9, 1 (list; table; graph; refs; listen; history; text; internal format)
OFFSET
0,8
COMMENTS
Sum of entries in row n is n!.
T(n,0) = A184185(n).
T(n,1) = A013999(n-1).
Sum_{k>=0} k*T(n,k) = 1! + 2! + ... + n! = A007489(n).
LINKS
FindStat - Combinatorial Statistic Finder, The number of adjacent cycles of a permutation
FORMULA
G.f. of column k is (1/k!)*z^k*(1-z)*Sum_{i>=0} (k+i)!*(z-z^2)^i (private communication from Vladeta Jovovic, May 26 2009).
T(n,k) = (1/k!)*Sum_{i=ceiling((n-k-1)/2)..n-k} (-1)^(n-k-i)*(k+i)!*binomial(i+1, n-k-i).
The bivariate g.f. is G(t,z) = ((1-z)/(1-tz))*F((z-z^2)/(1-tz)), where F(z) = Sum_{j>=0} j!*z^j.
EXAMPLE
T(3,2) = 2 because we have (1)(23) and (12)(3).
T(4,2) = 6 because we have (1)(234), (1)(24)(3), (12)(34), (123)(4), (14)(2)(3), and (13)(2)(4).
Triangle starts:
1;
0, 1;
0, 1, 1;
1, 2, 2, 1;
6, 8, 6, 3, 1;
34, 42, 27, 12, 4, 1;
MAPLE
T := proc (n, k) options operator, arrow: add((-1)^(n-k-i)*factorial(k+i)*binomial(i+1, n-k-i), i = ceil((1/2)*n-(1/2)*k-1/2) .. n-k)/factorial(k) end proc: for n from 0 to 10 do seq(T(n, k), k = 0 .. n) end do; # yields sequence in triangular form
CROSSREFS
Sequence in context: A077873 A123305 A118024 * A074297 A339604 A020824
KEYWORD
nonn,tabl
AUTHOR
Emeric Deutsch, Feb 16 2011 (based on communication from Vladeta Jovovic)
STATUS
approved

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Last modified March 28 14:21 EDT 2024. Contains 371254 sequences. (Running on oeis4.)