1,4

T(n,1)=A000255(n-1); T(n,2)=A000166 (the derangement numbers); T(n,3)=A000274(n); T(n,4)=A000313(n); T(n,5)=A001260(n);

This triangle is essentially A010027 (ascending pairs in permutations of [n]) with a different offset. The same triangle gives the number of permutations of [n] having k unit ascents (n>=1; 0<=k<=n-1). For permutations sorted by number of non-unitary (i.e. >1) descents (also called "big" descents), see A120434. For permutations sorted by number of unitary moves (i.e. ascent or descent), see A001100. - Olivier GERARD (olivier.gerard(AT)gmail.com), Oct 09 2007

With offsets n=0 (k=0) this is a binomial convolution triangle, a Sheffer triangle of the Appell type: ((exp(-x))/(1-x)^2),x). See the e.g.f. given below.

F. N. David, M. G. Kendall and D. E. Barton, Symmetric Function and Allied Tables, Cambridge, 1966, p. 263 (Table 7.5.1).

Ch. A. Charalambides, Enumerative Combinatorics, Chapman & Hall/CRC, Boca Raton, Florida, 2002, p. 179, Table 5.4 for S_{n,k} (without row n=1 and column k=0).

G.f.=exp(-x+tx)/(1-x)^2 (if offset is 0), i.e. T(n,k)=(n-1)!*[x^(n-1) t^k]exp(-x+tx)/(1-x)^2.

T(n,k)=binomial(n-1,k)*A000255(n-1), n-1>=k>=0, else 0.

T(4,2)=3 because we have 14/3/2, 2/14/3 and 3/2/14 (the small descents are shown by a /).

Triangle starts:

1;

1,1;

3,2,1;

11,9,3,1;

53,44,18,4,1;

T(4,2)=3 because we have 14/3/2, 2/14/3 and 3/2/14 (the unit descents are shown by a /).

G:=exp(-x+t*x)/(1-x)^2: Gser:=simplify(series(G, x=0, 15)): for n from 0 to 10 do P[n+1]:=sort(n!*coeff(Gser, x, n)) od: for n from 1 to 11 do seq(coeff(P[n], t, k), k=0..n-1) od; # yields sequence in triangular form

Cf. A000255, A000166, A000274, A000313, A001260.

Cf. A010027, A120434, A001100.

Sequence in context: A077756 A115080 A104219 this_sequence A117442 A118435 A115085

Adjacent sequences: A123510 A123511 A123512 this_sequence A123514 A123515 A123516

nonn,tabl

Emeric Deutsch (deutsch(AT)duke.poly.edu), Oct 02 2006