A111004 Number of permutations avoiding a consecutive 132 pattern.

1, 1, 2, 5, 16, 63, 296, 1623, 10176, 71793, 562848, 4853949, 45664896, 465403791, 5108121216, 60069714207, 753492215808, 10042248398625, 141712039383552, 2110880441637045, 33097631526180864, 544903371859138335, 9398216812334008320, 169463659008217238055

Offset: 0

Author

David Callan, Oct 01 2005

Keywords

• nonn

Comments

a(n) is the number of permutations on [n] that avoid the consecutive pattern 132 (pattern entries must occur consecutively in the permutation).

In the Mathematica code below, a[n, k] is the number of such permutations with first entry k and they are counted recursively by the length, say ell, of the longest increasing left factor L. (For ell >= 2 the first entry following L must be < the penultimate entry of L or else a consecutive 132 would occur.) The first sum counts ell = 1, the second ell = 2, the third ell >= 3; m is the penultimate entry of L and j is the first entry in the (reduced) subpermutation following L. Note that j is indexed from 0 to cover the case when L is the entire permutation.

Asymptotically, a(n)/n! ~ c/r^n where r = 1.2755477364172... is the unique positive root of Integrate[exp(-t^2/2), {t,0,r}] = 1 and c = exp(r^2/2)/r = 1.7685063678958....

Examples

			The first 3 entries of 2431 form a consecutive 132 pattern.
The 4!-a(4) = 8 permutations on [4] that DO contain a consecutive 132 pattern are 1243, 1324, 1423, 1432, 2143, 2431, 3142, 4132. Also, for example, 1342 contains a scattered 1-3-2 pattern but not a consecutive 132.
		

Links

• Ray Chandler, Table of n, a(n) for n = 0..200
• A. Baxter, B. Nakamura, and D. Zeilberger. Automatic generation of theorems and proofs on enumerating consecutive Wilf-classes
• Sergi Elizalde, Asymptotic enumeration of permutations avoiding generalized patterns, arXiv:math/0505254 [math.CO], 2005.
• S. Elizalde and M. Noy, Consecutive patterns in permutations, Adv. Appl. Math. 30 (2003), 110-125.
• M. E. Jones and J. B. Remmel, Pattern matching in the cycle structures of permutations, Pure Math. Appl. (PU.M.A.) 22 (2011), 173-208.

Crossrefs

Cf. A049774, A197365, A324130, A324134.

Row m = 0 of A327722.

Programs

Mathematica

Clear[a]; a[0, 0] = a[0] = 1; a[n_, 0]/;n>=1 := 0; a[n_, k_]/;k>n := 0; a[n_, k_]/;1<=k<=n<=2 := 1; a[n_, k_]/;n>=3 := a[n, k] = Sum[a[n-1, j], {j, k-1}] + (n-k)Sum[a[n-2, j], {j, k-1}] + Sum[(n-m) Binomial[m-k-1, ell-3]a[n-ell, j], {ell, 3, n-k+1}, {m, k+ell-2, n-1}, {j, 0, m-ell+1}]; a[n_]/;n>=1 := a[n] = Sum[a[n, k], {k, n}]; Table[a[n], {n, 0, 15}]
(* or, faster *) ExpGfToList[f_, n_, x_] := CoefficientList[Normal[Series[f, {x, 0, n}]] /. x^(pwr_) -> pwr!*x^pwr, x]; ExpGfToList[1/( 1-(Pi/2)^(1/2)*Erf[z/2^(1/2)] ), 25, z]

Formula

E.g.f.: Sum_{n >= 0} a(n) x^n/n! = 1/( 1 - (Pi/2)^(1/2)*Erf(x/2^(1/2)) ).

a(n) = A197365(n,0). - Peter Bala, Oct 14 2011

From Sergei N. Gladkovskii, Nov 28 2011: (Start)

E.g.f.: A(x) = 1/( 1 - (Pi/2)^(1/2)*erf(x/2^(1/2)) ) = (1 + (x^3)/(2*(x-1)*W(0) -(x^2)))/(1 - x) with

W(k) = 2*(k^2) + (5 - 4*(x^2))*k + 3 - 2*(x^2) + 2*(x^2)*(k+1)*((2*k + 3)^2)/W(k+1) (continued fraction). (End)