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A177250 Triangle read by rows: T(n,k) is the number of permutations of [n] having k adjacent 3-cycles (0 <= k <= floor(n/3)), i.e., no cycles of the form (i, i+1, i+2). 5
1, 1, 2, 5, 1, 22, 2, 114, 6, 697, 22, 1, 4923, 114, 3, 39612, 696, 12, 357899, 4923, 57, 1, 3588836, 39612, 348, 4, 39556420, 357900, 2460, 20, 475392841, 3588836, 19806, 116, 1, 6187284605, 39556420, 178950, 820, 5, 86701097310, 475392840, 1794420 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,3

COMMENTS

Row n contains 1 + floor(n/3) entries.

Sum of entries in row n = n! (A000142).

T(n,0) = A177251(n).

Sum_{k>=0} k*a(n,k) = (n-2)! (n>=3).

LINKS

Table of n, a(n) for n=0..42.

R. A. Brualdi and E. Deutsch, Adjacent q-cycles in permutations, arXiv:1005.0781 [math.CO], 2010.

FORMULA

T(n,k) = Sum_{j=0..floor(n/3)} (-1)^(k+j)*binomial(j,k)*(n-2j)!/j!.

EXAMPLE

T(7,2)=3 because we have (123)(456)(7), (123)(4)(567), and (1)(234)(567).

Triangle starts:

    1;

    1;

    2;

    5,  1;

   22,  2;

  114,  6;

  697, 22,  1;

MAPLE

T := proc (n, k) options operator, arrow: sum((-1)^(k+j)*binomial(j, k)*factorial(n-2*j)/factorial(j), j = 0 .. floor((1/3)*n)) end proc: for n from 0 to 14 do seq(T(n, k), k = 0 .. floor((1/3)*n)) end do; # yields sequence in triangular form

MATHEMATICA

T[n_, k_] := Sum[(-1)^(k + j)*Binomial[j, k]*(n - 2 j)!/j!, {j, 0, n/3}];

Table[T[n, k], {n, 0, 14}, {k, 0, n/3}] // Flatten (* Jean-Fran├žois Alcover, Nov 20 2017 *)

CROSSREFS

Cf. A000166, A008290, A177248, A177249, A177251, A177252, A177253.

Sequence in context: A186766 A047921 A242783 * A102786 A222637 A190950

Adjacent sequences:  A177247 A177248 A177249 * A177251 A177252 A177253

KEYWORD

nonn,tabf

AUTHOR

Emeric Deutsch, May 07 2010

EXTENSIONS

Crossreferences corrected by Emeric Deutsch, May 09 2010

STATUS

approved

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Last modified November 19 05:29 EST 2018. Contains 317333 sequences. (Running on oeis4.)