A115112 Number of different ways to select n elements from two sets of n elements under the precondition of choosing at least one element from each set.

0, 4, 18, 68, 250, 922, 3430, 12868, 48618, 184754, 705430, 2704154, 10400598, 40116598, 155117518, 601080388, 2333606218, 9075135298, 35345263798, 137846528818, 538257874438, 2104098963718, 8233430727598, 32247603683098, 126410606437750, 495918532948102

Offset: 1

Author

Hieronymus Fischer, Jan 22 2006

Keywords

• nonn
• easy

Comments

Also number of lattice paths from (0,0) to (n,n) that use steps (1,0) and (0,1) and do not include (n,0) or (0,n). - Ran Pan, Apr 10 2015

Examples

			a(5) = binomial(10,5) - 2 = 250.
		

Links

• Vincenzo Librandi, Table of n, a(n) for n = 1..300
• Guo-Niu Han, Enumeration of Standard Puzzles, 2011. [Cached copy]
• Guo-Niu Han, Enumeration of Standard Puzzles, arXiv:2006.14070 [math.CO], 2020.
• Gejza Jenca and Peter Sarkoci, Linear extensions and order-preserving poset partitions, arXiv:1112.5782 [math.CO], 2011-2015. - From _N. J. A. Sloane_, Apr 08 2012
• Ran Pan, Exercise K, Project P.

Crossrefs

Cf. A000984, A115111, A115246.

Programs

Magma

[Binomial(2*n, n)-2: n in [1..25]]; // Vincenzo Librandi, Apr 10 2015

Maple

seq(sum((binomial(n,m))^2,m=1..n-1),n=1..24); # Zerinvary Lajos, Jun 19 2008

Mathematica

Table[Sum[Binomial[n, i] Binomial[n, n - i], {i, 1, n - 1}], {n, 1, 10}]

Formula

a(n) = binomial(2*n, n) - 2 = A000984(n) - 2.

a(n) = Sum_{i=1..n-1} binomial(n,i)^2.

Recurrence: n*(3*n - 5)*a(n) = (15*n^2 - 31*n + 12)*a(n-1) - 2*(2*n - 3)*(3*n - 2)*a(n-2). - Vaclav Kotesovec, Oct 19 2012

a(n) ~ 4^n/sqrt(Pi*n). - Vaclav Kotesovec, Oct 19 2012

E.g.f.: exp(2*x) * BesselI(0,2*x) - 2*exp(x) + 1. - Ilya Gutkovskiy, Mar 04 2021