A162975 -id:A162975 - cp's OEIS frontend

A049774 Number of permutations of n elements not containing the consecutive pattern 123.

1, 1, 2, 5, 17, 70, 349, 2017, 13358, 99377, 822041, 7477162, 74207209, 797771521, 9236662346, 114579019469, 1516103040833, 21314681315998, 317288088082405, 4985505271920097, 82459612672301846, 1432064398910663705, 26054771465540507273, 495583804405888997218

Offset: 0

Tuwani A. Tshifhumulo (tat(AT)caddy.univen.ac.za)

Permutations on n letters without double falls. A permutation w has a double fall at k if w(k) > w(k+1) > w(k+2) and has an initial fall if w(1) > w(2).
Hankel transform is A055209. - Paul Barry, Jan 12 2009
Increasing colored 1-2 trees of order n with choice of two colors for the right branches of the vertices of outdegree 2 except those vertices on the path from the root to the leftmost leaf. - Wenjin Woan, May 21 2011

			Permutations without double increase and without pattern 123:
a(3) = 5: 132, 213, 231, 312, 321.
a(4) = 17: 1324, 1423, 1432, 2143, 2314, 2413, 2431, 3142, 3214, 3241, 3412, 3421, 4132, 4213, 4231, 4312, 4321.

F. N. David and D. E. Barton, Combinatorial Chance, Hafner, New York, 1962, pp. 156-157.
I. P. Goulden and D. M. Jackson, Combinatorial Enumeration, Wiley, N.Y., 1983, (5.2.17).

Alois P. Heinz, Table of n, a(n) for n = 0..464 (first 201 terms from Ray Chandler)
Martin Aigner, Catalan and other numbers: a recurrent theme, in Algebraic Combinatorics and Computer Science, a Tribute to Gian-Carlo Rota, pp.347-390, Springer, 2001.
Juan S. Auli, Pattern Avoidance in Inversion Sequences, Ph. D. thesis, Dartmouth College, ProQuest Dissertations Publishing (2020), 27964164.
Juan S. Auli, Sergi Elizalde, Consecutive Patterns in Inversion Sequences, arXiv:1904.02694 [math.CO], 2019. See Table 1.
Paul Barry, Constructing Exponential Riordan Arrays from Their A and Z Sequences, Journal of Integer Sequences, 17 (2014), #14.2.6.
Paul Barry, On a transformation of Riordan moment sequences, arXiv:1802.03443 [math.CO], 2018.
Nicolas Basset, Counting and generating permutations using timed languages, HAL Id: hal-00820373, 2013.
A. Baxter, B. Nakamura, and D. Zeilberger. Automatic generation of theorems and proofs on enumerating consecutive Wilf-classes; Local copy [Pdf file only, no active links].
S. Elizalde, Asymptotic enumeration of permutations avoiding generalized patterns arXiv:math/0505254 [math.CO], 2015.
S. Elizalde and M. Noy, Consecutive patterns in permutations, Adv. Appl. Math. 30 (2003), 110-123.
Steven Finch, Pattern-Avoiding Permutations [Archived version]
Steven Finch, Pattern-Avoiding Permutations [Cached copy, with permission]
Ira M. Gessel and Yan Zhuang, Counting permutations by alternating descents , arXiv:1408.1886 [math.CO], 2014. See Eq. (3). - _N. J. A. Sloane_, Aug 11 2014
Kaarel Hänni, Asymptotics of descent functions, arXiv:2011.14360 [math.CO], Nov 29 2020, p. 14.
Mingjia Yang and Doron Zeilberger, Increasing Consecutive Patterns in Words, arXiv:1805.06077 [math.CO], 2018.
Christopher Zhu, Enumerating Permutations and Rim Hooks Characterized by Double Descent Sets, arXiv:1910.12818 [math.CO], 2019.

Cf. A065429, A080635, A111004, A117158, A177523, A177533.
Column k=0 of A162975.
Column k=3 of A242784.
Equals 1 + A000303. - Greg Dresden, Feb 22 2020

b:= proc(u, o, t) option remember;
     `if`(u+o=0, 1, add(b(u-j, o+j-1, 0), j=1..u)+
     `if`(t=1, 0,   add(b(u+j-1, o-j, 1), j=1..o)))
    end:
a:= n-> b(n, 0$2):
seq(a(n), n=0..23);  # Alois P. Heinz, Nov 04 2021

Table[Simplify[ n! SeriesCoefficient[ Series[ Sqrt[3] Exp[x/2]/(Sqrt[3] Cos[Sqrt[3]/2 x] - Sin[Sqrt[3]/2 x]), {x, 0, n}], n] ], {n, 0, 40}]
(* Second program: *)
b[u_, o_, t_, k_] := b[u, o, t, k] = If[t == k, (u + o)!, If[Max[t, u] + o < k, 0, Sum[b[u + j - 1, o - j, t + 1, k], {j, 1, o}] + Sum[b[u - j, o + j - 1, 1, k], {j, 1, u}]]];
a[n_] := b[0, n, 0, 2] - b[0, n, 0, 3] + 1;
a /@ Range[0, 40] (* Jean-François Alcover, Nov 09 2020, after Alois P. Heinz in A000303 *)

E.g.f.: 1/Sum_{i>=0} (x^(3*i)/(3*i)! - x^(3*i+1)/(3*i+1)!). [Corrected g.f. --> e.g.f. by Vaclav Kotesovec, Feb 15 2015]
Equivalently, e.g.f.: exp(x/2) * r / sin(r*x + (2/3)*Pi) where r = sqrt(3)/2. This has simple poles at (3*m+1)*x0 where x0 = Pi/sqrt(6.75) = 1.2092 approximately and m is an arbitrary integer. This yields the asymptotic expansion a(n)/n! ~ x0^(-n-1) * Sum((-1)^m * E^(3*m+1) / (3*m+1)^(n+1)) where E = exp(x0/2) = 1.8305+ and m ranges over all integers. - Noam D. Elkies, Nov 15 2001
E.g.f.: sqrt(3)*exp(x/2)/(sqrt(3)*cos(x*sqrt(3)/2) - sin(x*sqrt(3)/2) ); a(n+1) = Sum_{k=0..n} binomial(n, k)*a(k)*b(n-k) where b(n) = number of n-permutations without double falls and without initial falls. - Emanuele Munarini, Feb 28 2003
O.g.f.: A(x) = 1/(1 - x - x^2/(1 - 2*x - 4*x^2/(1 - 3*x - 9*x^2/(1 - ... - n*x - n^2*x^2/(1 - ...))))) (continued fraction). - Paul D. Hanna, Jan 17 2006
a(n) = leftmost column term of M^n*V, where M = an infinite tridiagonal matrix with (1,2,3,...) in the super, sub, and main diagonals and the rest zeros. V = the vector [1,0,0,0,...]. - Gary W. Adamson, Jun 16 2011
E.g.f.: A(x)=1/Q(0); Q(k)=1-x/((3*k+1)-(x^2)*(3*k+1)/((x^2)-3*(3*k+2)*(k+1)/Q(k+1))); (continued fraction). - Sergei N. Gladkovskii, Nov 25 2011
a(n) ~ n! * exp(Pi/(3*sqrt(3))) * (3*sqrt(3)/(2*Pi))^(n+1). - Vaclav Kotesovec, Jul 28 2013
E.g.f.: T(0)/(1-x), where T(k) = 1 - x^2*(k+1)^2/( x^2*(k+1)^2 - (1-x-x*k)*(1-2*x-x*k)/T(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Oct 17 2013

Corrected and extended by Vladeta Jovovic, Apr 14 2001

A242783 Number T(n,k) of permutations of [n] with exactly k (possibly overlapping) occurrences of the consecutive step pattern given by the binary expansion of n, where 1=up and 0=down; triangle T(n,k), n>=0, read by rows.

Original entry on oeis.org

1, 1, 2, 5, 1, 21, 3, 70, 50, 450, 270, 4326, 602, 99, 12, 1, 34944, 5376, 209863, 139714, 13303, 1573632, 1366016, 530432, 158720, 21824925, 15302031, 2715243, 74601, 302273664, 161855232, 14872704, 2854894485, 2600075865, 712988175, 59062275

Offset: 0

Alois P. Heinz, May 22 2014

Sum_{k>0} k*T(n,k) = A249249(n).

			T(7,3) = 12 because 12 permutations of {1,2,3,4,5,6,7} have exactly 3 (possibly overlapping) occurrences of the consecutive step pattern up, up, up given by the binary expansion of 7 = 111_2: (1,2,3,4,5,7,6), (1,2,3,4,6,7,5), (1,2,3,5,6,7,4), (1,2,4,5,6,7,3), (1,3,4,5,6,7,2), (2,1,3,4,5,6,7), (2,3,4,5,6,7,1), (3,1,2,4,5,6,7), (4,1,2,3,5,6,7), (5,1,2,3,4,6,7), (6,1,2,3,4,5,7), (7,1,2,3,4,5,6).
Triangle T(n,k) begins:
: n\k :       0        1       2       3  4  ...
+-----+------------------------------------
:  0  :       1;
:  1  :       1;                             [row  1 of A008292]
:  2  :       2;                             [row  2 of A008303]
:  3  :       5,       1;                    [row  3 of A162975]
:  4  :      21,       3;                    [row  4 of A242819]
:  5  :      70,      50;                    [row  5 of A227884]
:  6  :     450,     270;                    [row  6 of A242819]
:  7  :    4326,     602,     99,     12, 1; [row  7 of A220183]
:  8  :   34944,    5376;                    [row  8 of A242820]
:  9  :  209863,  139714,  13303;            [row  9 of A230695]
: 10  : 1573632, 1366016, 530432, 158720;    [row 10 of A230797]

Alois P. Heinz, Rows n = 0..130, flattened

Column k=0-10 give: A242785, A246221, A246222, A246223, A246224, A246225, A246226, A246227, A246228, A246229, A243105.
Row sums give A000142.
Cf. A242784, A249249, A295987, A335308.

T:= proc(n) option remember; local b, k, r, h;
      k:= iquo(n,2,'r'); h:= 2^ilog2(n);
      b:= proc(u, o, t) option remember; `if`(u+o=0, 1, expand(
      add(b(u-j, o+j-1, irem(2*t,   h))*`if`(r=0 and t=k, x, 1), j=1..u)+
      add(b(u+j-1, o-j, irem(2*t+1, h))*`if`(r=1 and t=k, x, 1), j=1..o)))
      end: forget(b);
      (p-> seq(coeff(p, x, i), i=0..degree(p)))(b(n, 0, 0))
    end:
seq(T(n), n=0..15);

T[n_] := T[n] = Module[{b, k, r, h}, {k, r} = QuotientRemainder[n, 2]; h = 2^Floor[Log[2, n]]; b[u_, o_, t_] := b[u, o, t] = If[u + o == 0, 1, Expand[ Sum[b[u - j, o + j - 1, Mod[2*t, h]]*If[r == 0 && t == k, x, 1], {j, 1, u}] + Sum[b[u + j - 1, o - j, Mod[2*t + 1, h]]*If[r == 1 && t == k, x, 1], {j, 1, o}]]]; Function[p, Table[Coefficient[p, x, i], {i, 0, Exponent[p, x]}]][b[n, 0, 0]]]; Table[T[n], {n, 0, 15}] // Flatten (* Jean-François Alcover, Feb 20 2016, after Alois P. Heinz *)

A197365 T(n,k) gives the number of permutations of the set [n] that contain k occurrences of the subword (132); irregular array read by rows (n >= 0 and 0 <= k <= max(0, floor((n-1)/2))).

Original entry on oeis.org

1, 1, 2, 5, 1, 16, 8, 63, 54, 3, 296, 368, 56, 1623, 2649, 753, 15, 10176, 20544, 9024, 576, 71793, 172596, 104814, 13572, 105, 562848, 1569408, 1228608, 259968, 7968, 4853949, 15398829, 14824314, 4532034, 306729, 945, 45664896, 162412416, 185991936, 75929856

Offset: 0

Peter Bala, Oct 14 2011

A permutation p(1)p(2)...p(n) in the symmetric group S_n contains the subword (132) if there are 3 consecutive elements p(i)p(j)p(k) that have the same order relations as (132), that is, p(i) < p(j) > p(k) and p(i) < p(k). For the enumeration of permutations containing the subword (123) see A162975.
From Petros Hadjicostas, Nov 05 2019: (Start)
The attached Maple program gives a recurrence for the o.g.f. of each row in terms of u for T(n,k), the number of permutations of [n] containing exactly k occurrences of the consecutive pattern 123...(r+1)(r+3)(r+2) for r >= 0. In the program, t = r + 2. Here, n >= 0 and 0 <= k <= max(0, (n-1)/t).
Using that recurrence we may get any row or column from the irregular triangular array T(n, k) for any r >= 0. (Here r = 0, while in array A264781 we have r = 2.)
The recurrence follows from manipulation of the bivariate o.g.f/e.g.f. 1/W(u,z) = Sum_{n, k >= 0} T(n, k)*u^k*z^n/n!, whose reciprocal W(u,z) is the solution of the o.d.e. in Theorem 3.2 in Elizalde and Noy (2003) (with m = a = r + 1). The number t = r + 2 is the order of the o.d.e. in terms of the variable z.
(End)

			Table begins
.n\k.|......0......1.....2......3
= = = = = = = = = = = = = = = = =
..0..|......1
..1..|......1
..2..|......2
..3..|......5......1
..4..|.....16......8
..5..|.....63.....54.....3
..6..|....296....368....56
..7..|...1623...2649...753....15
..8..|..10176..20544..9024...576
...
T(4,0) = 16: The 16 permutations of S_4 not containing the subword (132) are (1234), (2134), (2314), (3124), (3214), (1342), (2341), (3241), (2413), (3412), (3421), (4123), (4213), (4231), (4312), (4321).
T(4,1) = 8: The 8 permutations of S_4 with 1 occurrence of the subword (132) are 1243, 1324, 1423, 1432, 2143, 2431, 3142, 4132.

Alois P. Heinz, Rows n = 0..100, flattened
A. Baxter, B. Nakamura, and D. Zeilberger, Automatic generation of theorems and proofs on enumerating consecutive Wilf-classes, 2011.
S. Elizalde and M. Noy, Consecutive patterns in permutations, Adv. Appl. Math. 30 (2003), 110-125.
Petros Hadjicostas, General Maple program for a recurrence for the number of permutations of [n] containing exactly k times the consecutive pattern 123...(r+1)(r+3)(r+2) for r >= 0. [Here r = 0 and in the program t = r + 2 = 2.]

Columns k=0-10 give: A111004, A263885, A263886, A263887, A263888, A263889, A263890, A263891, A263892, A263893, A263894.
T(2n+1,n) gives A001147.
T(2n+2,n) gives 2*A076729.
Cf. A162975, A264781 (pattern 12354).

b:= proc(u, o, t) option remember; `if`(u+o=0, 1, expand(
      add(b(u-j, o+j-1, 0)*`if`(j (p-> seq(coeff(p, x, i), i=0..degree(p)))(b(n, 0$2)):
seq(T(n), n=0..14);  # Alois P. Heinz, Oct 30 2013

b[u_, o_, t_] := b[u, o, t] = If[u+o == 0, 1, Expand[Sum[b[u-j, o+j-1, 0]*If[jJean-François Alcover, Mar 05 2015, after Alois P. Heinz *)

E.g.f.: 1/(1 - int_{t = 0..z} exp((u-1)*t^2/2!)) = sqrt(1 - u)/(sqrt(1 - u) -sqrt(Pi/2) * erf(z/2*sqrt(1 - u))) = 1 + z + 2*z^2/2! + (5 + u)*z^3/3! + (16 + 8*u)*z^4/4! + ....

n-th row sum = n!. First column is A111004.

A162976 Triangle read by rows: T(n,k) is the number of permutations of {1,2,...,n} having k double descents and initial descents (n>=0; 0<=k<=max(0,n-1)) [we say that i is a doubledescent of a permutation p if p(i) > p(i+1) > p(i+2); we say that a permutation p has an initial descent if p(1) > p(2)].

Original entry on oeis.org

1, 1, 1, 1, 3, 2, 1, 9, 11, 3, 1, 39, 48, 28, 4, 1, 189, 297, 166, 62, 5, 1, 1107, 1902, 1419, 476, 129, 6, 1, 7281, 14391, 11637, 5507, 1235, 261, 7, 1, 54351, 118044, 111438, 56400, 19096, 3020, 522, 8, 1, 448821, 1078245, 1119312, 673128, 239146, 61986

Offset: 0

Emeric Deutsch, Jul 26 2009

Sum of entries in row n is n! = A000142(n).

T(n,0) = A080635(n).

			T(4,2) = 3 because each of the permutations 4312, 4213, and 3214 has one doubledescent and one initial descent.
Triangle starts:
:   1;
:   1;
:   1,   1;
:   3,   2,   1;
:   9,  11,   3,  1;
:  39,  48,  28,  4, 1;
: 189, 297, 166, 62, 5, 1;

I. P. Goulden and D. M. Jackson, Combinatorial Enumeration, John Wiley and Sons, N.Y., 1983 (p. 195).

Alois P. Heinz, Rows n = 0..141, flattened
Yan Zhuang, Monoid networks and counting permutations by runs, arXiv preprint, 2015.
Y. Zhuang, Counting permutations by runs, J. Comb. Theory Ser. A 142 (2016), pp. 147-176.

Cf. A000142, A080635, A162975.

eq := s^2-(t+1)*s+1 = 0: sol := solve(eq, s): a := sol[1]: b := sol[2]: G := (exp(b*z)-exp(a*z))/(b*exp(a*z)-a*exp(b*z)): Gser := simplify(series(G, z = 0, 15)): P[0]:=1: for n to 11 do P[n] := sort(expand(factorial(n)*coeff(Gser, z, n))) end do: for n from 0 to 11 do seq(coeff(P[n], t, j), j = 0 .. max(0,n-1)) end do;
# second Maple program:
b:= proc(u, o, t) option remember; `if`(u+o=0, 1, expand(
      add(b(u-j, o+j-1, 1), j=1..u)+
      add(b(u+j-1, o-j, 2)*`if`(t=2, x, 1), j=1..o)))
    end:
T:= n-> (p-> seq(coeff(p, x, i), i=0..degree(p)))(b(0, n, 1)):
seq(T(n), n=0..15);  # Alois P. Heinz, Dec 09 2016

b[u_, o_, t_] := b[u, o, t] = If[u+o == 0, 1, Expand[Sum[b[u-j, o+j-1, 1], {j, 1, u}] + Sum[b[u + j - 1, o - j, 2]*If[t == 2, x, 1], {j, 1, o}]]]; T[n_] := Function[p, Table[Coefficient[p, x, i], {i, 0, Exponent[p, x]}] ][b[0, n, 1]]; Table[T[n], {n, 0, 15}] // Flatten (* Jean-François Alcover, Dec 22 2016, after Alois P. Heinz *)

E.g.f.: G(t,z) = [exp(bz)-exp(az)]/[b*exp(az)-a*exp(bz)], where a+b=1+t and ab=1.

One term for row n=0 prepended by Alois P. Heinz, Dec 09 2016

A220183 Triangle read by rows: T(n,k) is the number of permutations of {1,2,...,n} having k triple descents (n>=0,0<=k<=n-3). We say that i is a triple descent of a permutation p if p(i) > p(i+1) > p(i+2) > p(i+3).

Original entry on oeis.org

1, 1, 2, 6, 23, 1, 111, 8, 1, 642, 67, 10, 1, 4326, 602, 99, 12, 1, 33333, 5742, 1093, 137, 14, 1, 288901, 59504, 12425, 1852, 181, 16, 1, 2782082, 666834, 151635, 24970, 3029, 231, 18, 1, 29471046, 8054684, 1981499, 355906, 48455, 4902, 287, 20, 1

Offset: 0

Geoffrey Critzer, Dec 12 2012

Row sums = n!.

T(n,0) = A117158.

			:     1;
:     1;
:     2;
:     6;
:    23,    1;
:   111,    8,    1;
:   642,   67,   10,   1;
:  4326,  602,   99,  12,  1;
: 33333, 5742, 1093, 137, 14, 1;
T(5,1) = 8 because we have: (4,5,3,2,1), (3,5,4,2,1), (2,5,4,3,1), (5,4,3,1,2), (1,5,4,3,2), (5,4,2,1,3), (5,3,2,1,4), (4,3,2,1,5).

Alois P. Heinz, Rows n = 0..143, flattened
P. Flajolet and R. Sedgewick, Analytic Combinatorics, 2009; see page 210

Cf. A162975, A242783, A242784.

b:= proc(u, o, t) option remember; `if`(u+o=0, 1, expand(
      add(b(u-j, o+j-1, 1), j=1..u)+
      add(b(u+j-1, o-j, [2, 3, 3][t])*`if`(t=3, x, 1), j=1..o)))
    end:
T:= n-> (p-> seq(coeff(p, x, i), i=0..degree(p)))(b(n, 0, 1)):
seq(T(n), n=0..14);  # Alois P. Heinz, Oct 29 2013

nn=10; u=y-1; a=Apply[Plus, Table[Normal[Series[y x^4/(1-y x - y x^2-y x^3), {x,0,nn}]][[n]]/(n+3)!, {n,1,nn-3}]]/.y->u; Range[0,nn]! CoefficientList[Series[1/(1-x-a), {x,0,nn}], {x,y}]//Grid

E.g.f.: 1/(1 - x - Sum_{k,n} I(n,k)(y - 1)^k*x^n/n!) where I(n,k) is the coefficient of y^k*x^n in the ordinary generating function expansion of y x^4/(1 - y*x - y*x^2 - y*x^3) See Flajolet and Sedgewick reference in Links section.

A230621 Number of permutations of [n] with exactly two (possibly overlapping) occurrences of the consecutive step pattern {up}^2.

Original entry on oeis.org

0, 0, 0, 0, 1, 8, 86, 803, 8221, 86214, 966114, 11405511, 142934124, 1892755874, 26487024478, 390658292572, 6063383527327, 98824236282650, 1688354110698402, 30179347977813309, 563462569163994435, 10970288500929001986, 222384832378410907480

Offset: 0

Alois P. Heinz, Oct 25 2013

nonn

			a(4) = 1: 1234.
a(5) = 8: 12354, 12453, 13452, 21345, 23451, 31245, 41235, 51234.
a(6) = 86: 123546, 123645, 123654, ..., 631245, 641235, 651234.
a(7) = 803: 1235476, 1236475, 1236547, ..., 7631245, 7641235, 7651234.

Alois P. Heinz, Table of n, a(n) for n = 0..462

Column k=2 of A162975.

Cf. A230620.

b:= proc(u, o, t) option remember;
      `if`(t=7, 0, `if`(u+o=0, `if`(t in [4, 6], 1, 0),
      add(b(u-j, o+j-1, [1, 1, 5, 6, 5, 6][t]), j=1..u)+
      add(b(u+j-1, o-j, [2, 3, 4, 7, 3, 4][t]), j=1..o)))
    end:
a:= n-> b(n, 0, 1):
seq(a(n), n=0..25);

b[u_, o_, t_] := b[u, o, t] = If[u + o == 0, 1,
    Sum[b[u - j, o + j - 1, 1], {j, 1, u}] +
    Sum[b[u + j - 1, o - j, 2]*If[t == 2, x, 1], {j, 1, o}] // Expand];
a[n_] := Coefficient[b[n, 0, 1], x, 2];
a /@ Range[0, 25] (* Jean-François Alcover, Dec 21 2020, after Alois P. Heinz in A162975 *)

a(n) ~ c * (3*sqrt(3)/(2*Pi))^n * n! * n^2, where c = 0.0359701024355206... . - Vaclav Kotesovec, Sep 06 2014

A230620 Number of permutations of [n] with at least two (possibly overlapping) occurrences of the consecutive step pattern {up}^2.

Original entry on oeis.org

0, 0, 0, 0, 1, 9, 97, 983, 10616, 119932, 1441405, 18383351, 249155667, 3581896559, 54540748818, 877824410030, 14904605652001, 266431586957773, 5004557444810885, 98594548150006583, 2033673324306909868, 43845407809459639440, 986496730691143433269

Offset: 0

Alois P. Heinz, Oct 25 2013

nonn

			a(4) = 1: 1234.
a(5) = 9: 12345, 12354, 12453, 13452, 21345, 23451, 31245, 41235, 51234.
a(6) = 97: 123456, 123465, 123546, ..., 631245, 641235, 651234.
a(7) = 983: 1234567, 1234576, 1234657, ..., 7631245, 7641235, 7651234.

Alois P. Heinz, Table of n, a(n) for n = 0..200

Cf. A230621.

b:= proc(u, o, t) option remember;
      `if`(t=5, (u+o)!, `if`(u+o+t<4, 0,
      add(b(u-j, o+j-1, [1, 1, 4, 4][t]), j=1..u)+
      add(b(u+j-1, o-j, [2, 3, 5, 3][t]), j=1..o)))
    end:
a:= n-> b(n, 0, 1):
seq(a(n), n=0..25);

b[u_, o_, t_] := b[u, o, t] =
    If[t == 5, (u + o)!, If[u + o + t < 4, 0,
    Sum[b[u - j, o + j - 1, {1, 1, 4, 4}[[t]]], {j, 1, u}] +
    Sum[b[u + j - 1, o - j, {2, 3, 5, 3}[[t]]], {j, 1, o}]]];
a[n_] := b[n, 0, 1];
a /@ Range[0, 25] (* Jean-François Alcover, Dec 22 2020, after Alois P. Heinz *)

a(n) = Sum_{i=2..n-2} A162975(n,i).

a(n) ~ n!. - Vaclav Kotesovec, Sep 06 2014

cp's OEIS Frontend

A049774 Number of permutations of n elements not containing the consecutive pattern 123.

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A242783 Number T(n,k) of permutations of [n] with exactly k (possibly overlapping) occurrences of the consecutive step pattern given by the binary expansion of n, where 1=up and 0=down; triangle T(n,k), n>=0, read by rows.

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A197365 T(n,k) gives the number of permutations of the set [n] that contain k occurrences of the subword (132); irregular array read by rows (n >= 0 and 0 <= k <= max(0, floor((n-1)/2))).

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A220183 Triangle read by rows: T(n,k) is the number of permutations of {1,2,...,n} having k triple descents (n>=0,0<=k<=n-3). We say that i is a triple descent of a permutation p if p(i) > p(i+1) > p(i+2) > p(i+3).

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A230621 Number of permutations of [n] with exactly two (possibly overlapping) occurrences of the consecutive step pattern {up}^2.

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A230620 Number of permutations of [n] with at least two (possibly overlapping) occurrences of the consecutive step pattern {up}^2.

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