%I #42 Dec 27 2021 21:46:55

%S 1,1,1,1,2,3,1,3,9,11,1,4,18,44,53,1,5,30,110,265,309,1,6,45,220,795,

%T 1854,2119,1,7,63,385,1855,6489,14833,16687,1,8,84,616,3710,17304,

%U 59332,133496,148329,1,9,108,924,6678,38934,177996,600732,1334961,1468457,1

%N Triangle read by rows: T(n,k) is the number of permutations of [n] having k consecutive ascending pairs (0 <= k <= n-1).

%C A "consecutive ascending pair" in a permutation p_1, p_2, ..., p_n is a pair p_i, p_{i+1} = p_i + 1.

%C From _Emeric Deutsch_, May 15 2010: (Start)

%C The same triangle, but with rows indexed differently, also arises as follows: U(n,k) = number of permutations of [n] having k blocks (1 <= k <= n), where a block of a permutation is a maximal sequence of consecutive integers which appear in consecutive positions. For example, the permutation 5412367 has 4 blocks: 5, 4, 123, and 67.

%C When seen as coefficients of polynomials with decreasing exponents: evaluations are A001339 (x=2), A081923 (x=3), A081924 (x=4), A087981 (x=-1).

%C The sum of the entries in row n is n!.

%C U(n,n) = A000255(n-1) = d(n-1) + d(n), U(n,n-1)=d(n), where d(j)=A000166(j) (derangement numbers). (End)

%C This is essentially the reversal of the exponential Riordan array [exp(-x)/(1-x)^2,x] (cf. A123513). - _Paul Barry_, Jun 17 2010

%C U(n-1, k-2) * n*(n-1)/k = number of permutations of [n] with k elements not fixed by the permutation. - _Michael Somos_, Aug 19 2018

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

%H Alois P. Heinz, <a href="/A010027/b010027.txt">Rows n = 1..150, flattened</a>

%H A. N. Myers, <a href="http://dx.doi.org/10.1006/jcta.2002.3279">Counting permutations by their rigid patterns</a>, J. Combin. Theory, A 99 (2002), 345-357. [_Emeric Deutsch_, May 15 2010]

%F E.g.f.: exp(x*(y-1))/(1-x)^2. - _Vladeta Jovovic_, Jan 03 2003

%F From _Emeric Deutsch_, May 15 2010: (Start)

%F U(n,k) = binomial(n-1,k-1)*(k-1)!*Sum_{j=0..k-1} (-1)^(k-j-1)*(j+1)/(k-j-1)! (1 <= k <= n).

%F U(n,k) = (k+1)!*binomial(n,k)*(1/n)*Sum_{i=0..k+1} (-1)^i/i!.

%F U(n,k) = (1/n)*binomial(n,k)*d(k+1), where d(j)=A000166(j) (derangement numbers). (End)

%e Triangle starts:

%e 1;

%e 1, 1;

%e 1, 2, 3;

%e 1, 3, 9, 11;

%e 1, 4, 18, 44, 53;

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

%e 1, 6, 45, 220, 795, 1854, 2119;

%e 1, 7, 63, 385, 1855, 6489, 14833, 16687;

%e 1, 8, 84, 616, 3710, 17304, 59332, 133496, 148329;

%e 1, 9, 108, 924, 6678, 38934, 177996, 600732, 1334961, 1468457;

%e ...

%e For n=3, the permutations 123, 132, 213, 231, 312, 321 have respectively 2,0,0,1,1,0 consecutive ascending pairs, so row 3 of the triangle is 3,2,1. - _N. J. A. Sloane_, Apr 12 2014

%e In the alternative definition, T(4,2)=3 because we have 234.1, 4.123, and 34.12 (the blocks are separated by dots). - _Emeric Deutsch_, May 16 2010

%p U := proc (n, k) options operator, arrow: factorial(k+1)*binomial(n, k)*(sum((-1)^i/factorial(i), i = 0 .. k+1))/n end proc: for n to 10 do seq(U(n, k), k = 1 .. n) end do; # yields sequence in triangular form; # _Emeric Deutsch_, May 15 2010

%t t[n_, k_] := Binomial[n, k]*Subfactorial[k+1]/n; Table[t[n, k], {n, 1, 12}, {k, 1, n}] // Flatten (* _Jean-François Alcover_, Jan 07 2014, after _Emeric Deutsch_ *)

%Y Diagonals, reading from the right-hand edge: A000255, A000166, A000274, A000313, A001260, A001261. A045943 is another diagonal.

%Y Cf. A123513 (mirror image).

%Y A289632 is the analogous triangle with the additional restriction that all consecutive pairs must be isolated, i.e., must not be back-to-back to form longer consecutive sequences.

%K tabl,nonn

%O 1,5

%A _N. J. A. Sloane_

%E More terms from _Vladeta Jovovic_, Jan 03 2003

%E Original definition from David, Kendall and Barton restored by _N. J. A. Sloane_, Apr 12 2014