1,4

This triangle is essentiallyA010027(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), seeA120434.For permutations sorted by number of unitary moves (i.e., ascent or descent), seeA001100.-Olivier Gérard,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.

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).

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

Alois P. Heinz,Rows n = 1..150, flattened

Bhadrachalam Chitturi and Krishnaveni K S,Adjacencies in Permutations,arXiv preprint arXiv:1601.04469 [cs.DM], 2016. See Table 0.

FindStat - Combinatorial Statistic Finder,The number of adjacencies of a permutation

Sergey Kitaev, Philip B. Zhang,Distributions of mesh patterns of short lengths,arXiv:1811.07679 [math.CO], 2018.

J. Liese, J. Remmel,Q-analogues of the number of permutations with k-excedances,PU. M. A. Vol. 21 (2010), No. 2, pp. 285-320 (see E_{n,1}(x) in Table 1 p. 291).

F. Poussin,Énumération des permutations par nombre de marches,RAIRO, Informatique théorique, 13 no. 3, 1979, p. 251-255.

T(n,1) =A000255(n-1).

T(n,2) =A000166(n-1) (the derangement numbers).

T(n,3) =A000274(n).

T(n,4) =A000313(n).

T(n,5) =A001260(n);

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.

Triangle starts:

1;

1, 1;

3, 2, 1;

11, 9, 3, 1;

53, 44, 18, 4, 1;

309, 265, 110, 30, 5, 1;

2119, 1854, 795, 220, 45, 6, 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 /).

T(4,2)=3 because we have 14/3/2, 2/14/3 and 3/2/14 (the small 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

Needs[ "Combinatorica`" ];

Table[Map[Count[#, 1]&, Map[Differences, Permutations[n]]]//Distribution, {n, 1, 10}]//Grid

(*Geoffrey Critzer,Dec 15 2012 *)

T[n_, k_]:= (n-1)! SeriesCoefficient[Exp[-x + t x]/(1-x)^2, {x, 0, n-1}, {t, 0, k}];

Table[T[n, k], {n, 1, 10}, {k, 0, n-1}] // Flatten (*Jean-François Alcover,Sep 25 2019 *)

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

Cf.A010027(mirror image),A120434,A001100.

Sequence in context:A115080A222730A104219*A117442A118435A115085

Adjacent sequences:A123510A123511A123512*A123514A123515A123516

nonn,tabl

Emeric Deutsch,Oct 02 2006

approved