A193639 Triangle T(n,k) of ways n couples can sit in a row with exactly k of them together

1, 0, 2, 8, 8, 8, 240, 288, 144, 48, 13824, 15744, 8064, 2304, 384, 1263360, 1401600, 710400, 211200, 38400, 3840, 168422400, 183582720, 92620800, 28108800, 5529600, 691200, 46080, 30865121280, 33223034880, 16717639680, 5148057600, 1061222400, 149022720, 13547520, 645120

Offset

0,3

Comments

Row n sums to (2n)!

Dot product of row n and (0,1,2,3,...n) is equal to (2n)!

Dot product of row n and (0,0,1,2,...n-1) is equal to T(n,0)

Links

Andrew Woods, Rows n = 0..50 of triangle, flattened

Formula

T(n, k) = 2*(2*n-k)*T(n-1, k-1) + ((2*n-1-k)*(2*n-2-k)+2*k)*(T(n-1, k) + 2*(k+1)*(2*n-2-k)*T(n-1, k+1) + (k+2)*(k+1)*T(n-1, k+2)

T(n, n) = 2^n * n! = (2n)!!

T(n, k) = sum(i=k..n, (-1)^(i-k) * 2^i * (2n-i)! * binomial(n, i) * binomial(i, k))

T(n, 0) = A007060(n).

T(n, n) = A000165(n).

Example

Triangle begins:
1
0 2
8 8 8
240 288 144 48
13824 15744 8064 2304 384
There are T(3, 2) = 144 ways to arrange three couples in a row so that exactly two of them are together.

Mathematica

Table[Table[Sum[(-1)^k Binomial[n-i, k](2n-i-k)! 2^(k+i), {k, 0, n-i}]*Binomial[n, i], {i, 0, n}], {n, 0, 10}]//Grid (* Geoffrey Critzer, Apr 21 2014 *)

Crossrefs

Sequence in context: A289841 A143812 A245508 * A154481 A092280 A070987

Adjacent sequences: A193636 A193637 A193638 * A193640 A193641 A193642

Keyword

nonn,tabl

Author

Andrew Woods, Aug 01 2011

Status

approved