cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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

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Author

Peter Bala, Oct 14 2011

Keywords

Comments

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)

Examples

			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.
		

Crossrefs

T(2n+1,n) gives A001147.
T(2n+2,n) gives 2*A076729.
Cf. A162975, A264781 (pattern 12354).

Programs

  • Maple
    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
  • Mathematica
    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 *)

Formula

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.