A082971 Number of permutations of {1,2,...,n} containing exactly 3 occurrences of the 132 pattern.
1, 14, 82, 410, 1918, 8657, 38225, 166322, 716170, 3059864, 12994936, 54924212, 231235054, 970347575, 4060697955, 16952812170, 70629116910, 293720506860, 1219498444500, 5055891511980, 20933654593020, 86571545598642, 357628915621698, 1475896409177780
Offset: 4
Examples
a(4)=1 because we have 1432 (the 132 occurrences are 143, 142 and 132).
Links
- Vincenzo Librandi, Table of n, a(n) for n = 4..1000
- Miklós Bóna, The Number of Permutations with Exactly r 132-Subsequences Is P-Recursive in the Size!, Advances in Applied Mathematics, Volume 18, Issue 4, May 1997, Pages 510-522.
- Miklós Bóna, Permutations with one or two 132-subsequences, Discrete Math., 181 (1998) 267-274.
- T. Mansour and A. Vainshtein, Counting occurrences of 132 in a permutation, arXiv:math/0105073 [math.CO], 2001.
Programs
-
Magma
[1] cat [(n^6+51*n^5-407*n^4-99*n^3+7750*n^2-22416*n+20160)* Factorial(2*n-9)/(6*Factorial(n)*Factorial(n-5)): n in [5..30]]; // Vincenzo Librandi, Oct 30 2018
-
Maple
P:=2*x^3-5*x^2+7*x-2: Q:=-22*x^6-106*x^5+292*x^4-302*x^3+135*x^2-27*x+2: g:= (P+Q/(1-4*x)^(5/2))*1/2: gser:=series(g,x=0,30): seq(coeff(gser,x,n),n=4..25); # Emeric Deutsch, Mar 27 2008
-
Mathematica
a[4] = 1; a[n_] := (n^6 + 51 n^5 - 407 n^4 - 99 n^3 + 7750 n^2 - 22416 n + 20160) (2 n - 9)!/(6 n! (n - 5)!); Table[a[n], {n, 4, 25}] (* Jean-François Alcover, Oct 30 2018 *) -
PARI
a(n)=(2*n-9)!/n!/6/(n-5)!*(n^6+51*n^5-407*n^4-99*n^3 +7750*n^2 -22416*n+20160)
Formula
a(n) = (2*n-9)!/n!/6/(n-5)! *(n^6+51*n^5-407*n^4-99*n^3 +7750*n^2 -22416*n +20160).
a(n) = (n^6 + 51*n^5 - 407*n^4 - 99*n^3 + 7750*n^2 - 22416*n + 20160)*(2*n-9)!/(6*n!*(n-5)!) for n>=5; a(4)=1. G.f.: (1/2)*(P(x) + Q(x)/(1-4*x)^(5/2)), where P(x) = 2*x^3 - 5*x^2 + 7*x - 2, Q(x) = -22*x^6 - 106*x^5 + 292*x^4 - 302*x^3 + 135*x^2 - 27*x + 2. - Emeric Deutsch, Mar 27 2008
Extensions
Edited by N. J. A. Sloane, May 21 2008 at the suggestion of R. J. Mathar