A238977 Number of ballot sequences of length n with exactly 2 fixed points.

0, 0, 1, 1, 3, 8, 24, 74, 246, 848, 3088, 11644, 45844, 186336, 784928, 3403128, 15212744, 69802944, 328988096, 1587831568, 7848954928, 39651793024, 204691645824, 1078028406176, 5790745961568, 31687186373888, 176575788105984, 1001061518465984, 5771865641210176

Offset: 0

Author

Joerg Arndt and Alois P. Heinz, Mar 07 2014

Keywords

• nonn
• easy

Comments

The fixed points are in the first 2 positions.

Also the number of standard Young tableaux with n cells such that the first column contains 1 and 2, but not 3. An alternate definition uses the first row.

Examples

			a(2) = 1: [1,2].
a(3) = 1: [1,2,1].
a(4) = 3: [1,2,1,1], [1,2,1,2], [1,2,1,3].
a(5) = 8: [1,2,1,1,1], [1,2,1,1,2], [1,2,1,1,3], [1,2,1,2,1], [1,2,1,2,3], [1,2,1,3,1], [1,2,1,3,2], [1,2,1,3,4].
		

Links

• Joerg Arndt and Alois P. Heinz, Table of n, a(n) for n = 0..800
• Wikipedia, Young tableau

Crossrefs

Column k=2 of A238802.

Cf. A000085, A007310.

Programs

Maple

a:= proc(n) option remember; `if`(n<3, n*(n-1)/2,
      ((3*n^2+3*n-33)*a(n-1) +(n-4)*(10*n^2-27*n-6)*a(n-2)
      +(n-4)*(n-5)*(7*n-18)*a(n-3)) / (10*n^2-64*n+105))
    end:
seq(a(n), n=0..40);

Mathematica

b[n_, l_List] := b[n, l] = If[n <= 0, 1, b[n - 1, Append[l, 1]] + Sum[If[i == 1 || l[[i - 1]] > l[[i]], b[n - 1, ReplacePart[l, i -> l[[i]] + 1]], 0], {i, 1, Length[l]}]]; a[n_] := b[n - 3, {2, 1}]; a[0] = a[1] = 0; Table[Print["a(", n, ") = ", an = a[n]]; an, {n, 0, 40}] (* Jean-François Alcover, Feb 06 2015, after Maple *)

Formula

See Maple program.

a(n) ~ sqrt(2)/6 * exp(sqrt(n)-n/2-1/4) * n^(n/2) * (1 + 7/(24*sqrt(n))). - Vaclav Kotesovec, Mar 07 2014

Recurrence (for n>=4): (n-3)*(n^2 - 6*n + 11)*a(n) = (n^3 - 9*n^2 + 32*n - 39)*a(n-1) + (n-4)*(n-2)*(n^2 - 4*n + 6)*a(n-2). - Vaclav Kotesovec, Mar 08 2014

From Peter Bala, Oct 05 2021: (Start)

a(n) = (1/3)*( A000085(n) - A000085(n-3) ) for n >= 3.

a(n) = (1/3)*Sum_{k = 0..floor(n/2)} (1 - binomial(n-2*k,3)/binomial(n,3))* binomial(n,2*k) * (2*k)!/(2^k*k!) for n >= 3.

Conjecture: a(n+3) == 1 (mod n) iff n is coprime to 2 and 3, that is, iff n is a term of A007310. (End)