A129178 Triangle read by rows: T(n,k) is the number of permutations p of {1,2,...,n} such that invc(p)=k (n >= 0; 0 <= k <= (n-1)(n-2)/2) (see comment for invc definition).

1, 1, 2, 4, 2, 8, 8, 6, 2, 16, 24, 28, 26, 16, 8, 2, 32, 64, 96, 120, 126, 110, 82, 52, 26, 10, 2, 64, 160, 288, 432, 564, 658, 680, 638, 542, 416, 284, 172, 90, 38, 12, 2, 128, 384, 800, 1376, 2072, 2824, 3526, 4058, 4344, 4346, 4066, 3562, 2912, 2218, 1566, 1016, 598

Offset: 0

Author

Emeric Deutsch, Apr 11 2007

Keywords

• nonn
• tabf

Comments

invc(p) is defined (by Carlitz) in the following way: express p in standard cycle form (i.e., cycles ordered by increasing smallest elements with each cycle written with its smallest element in the first position), then remove the parentheses and count the inversions in the obtained word.

Row n has 1+(n-1)*(n-2)/2 - delta_{0,n} terms. Row sums are the factorials (A000142). T(n,0) = 2^(n-1) = A011782(n) = A000079(n-1). T(n,1) = (n-2)*2^(n-2) = A036289(n-2) for n>=2. T(n,k) = A121552(n,n+k).

It appears that Sum_{k>=0} k*T(n,k) = A126673(n).

Examples

			T(3,0)=4, T(3,1)=2 because we have 123=(1)(2)(3), 132=(1)(23), 213=(12)(3), 231=(123) with the resulting word (namely 123) having 0 inversions and 312=(132) and (321)=(13)(2) with the resulting word (namely 132) having 1 inversion.
Triangle starts:
   1;
   1;
   2;
   4,   2;
   8,   8,   6,   2;
  16,  24,  28,  26,  16,   8,   2;
  32,  64,  96, 120, 126, 110,  82,  52,  26,  10,  2;
  ...
		

References

• L. Carlitz, Generalized Stirling numbers, Combinatorial Analysis Notes, Duke University, 1968, 1-7.

Links

• Alois P. Heinz, Rows n = 0..50, flattened
• Toufik Mansour, Mark Shattuck, A q-analog of the hyperharmonic numbers, Afrika Matematika 25.1 (2014): 147-160.
• M. Shattuck, Parity theorems for statistics on permutations and Catalan words, INTEGERS, Electronic J. of Combinatorial Number Theory, Vol. 5, Paper A07, 2005.

Crossrefs

Cf. A000142, A011782, A000079, A036289, A121552, A126673.

Programs

Maple

s:=j->2+sum(t^i, i=1..j): for n from 0 to 9 do P[n]:=sort(expand(simplify(product(s(j), j=0..n-2)))) od: for n from 0 to 9 do seq(coeff(P[n], t, j), j=0..degree(P[n])) od;  # yields sequence in triangular form

Mathematica

nMax = 9; s[j_] := 2 + Sum[t^i, {i, 1, j}]; P[0] = P[1] = 1; P[2] = 2; For[ n = 3, n <= nMax, n++, P[n] = Sort[Expand[Simplify[Product[s[j], {j, 0, n-2}]]]]]; Table[Coefficient[P[n], t, j], {n, 0, nMax}, {j, 0, Exponent[ P[n], t]}] // Flatten (* Jean-François Alcover, Jan 24 2017, adapted from Maple *)

Formula

Generating polynomial of row n is P[n](t) = 2*(2+t)*(2+t+t^2)*...*(2 + t + t^2 + ... + t^(n-2)) for n >= 3, P[1](t)=1, P[2](t)=2.

Extensions

One term for row n=0 prepended by Alois P. Heinz, Dec 16 2016