A231580 a(n) is the numerator of the probability that n segments of length 2, each placed randomly on a line segment of length 2n, will completely cover the line segment.

1, 2, 7, 34, 638, 4876, 220217, 6885458, 569311642, 7515775348, 197394815194, 78863079581996, 886395722771204, 848070074996694008, 222148423805582000341, 33494470531439170224754, 35665304857619152523926, 280147437461017444466304484

Offset: 1

Author

Philipp O. Tsvetkov, Nov 11 2013

Keywords

• nonn
• frac

Comments

For n=1 the length of the line to cover is equal to 2. There is only one way to cover it with 2-length segment and it will be the full cover. So, the probability is equal to 1. For n=2 the length of the line to cover is equal to 4. Let's start randomly and sequentially to cover it with 2-length segments. The first segment could be placed at 3 position with probability 1/3 in the following ways (xxoo, oxxo, ooxx). The second 2-length segment could be added only in the first and the last cases. So we have the following covers (xxxx, oxxo, xxxx). Thus the probability to find the full cover of 4-length line when it is randomly sequentially filled by 2-length segments is equal to 2/3.

Examples

			1, 2/3, 7/15, 34/105, 638/2835, 4876/31185, 220217/2027025, 6885458/91216125, 569311642/10854718875, 7515775348/206239658625, 197394815194/7795859096025, ...
		

Links

• Vincenzo Librandi, Table of n, a(n) for n = 1..260
• Philipp O. Tsvetkov, Stoichiometry of irreversible ligand binding to a one-dimensional lattice, Scientific Reports, Springer Nature (2020) Vol. 10, Article number: 21308.

Crossrefs

Cf. A231634 (denominators).

Programs

Maple

A231580f := proc(n)
    option remember;
    if n <= 0 then
        1;
    else
        add(procname(k)*procname(n-k-1),k=0..n-1)/(2*n-1) ;
    end if;
end proc:
A231580 := proc(n)
    numer(A231580f(n)) ;
end proc:
seq(A231580(n),n=1..30) ; # R. J. Mathar, Aug 28 2014

Mathematica

f[g_List, l_] := f[g, l] = Sum[f[g[[;; n]], l] f[g[[n + 1 ;;]], l], {n, Length[g] - 1}]/(Total[l + g] - 2 l + 1);
f[{}] = f[{}, _] = 1;
f[ConstantArray[0, #], 2] & /@ Range[2, 20] // Numerator

PARI

f=[1]; for(n=2, 25, f=concat(f, sum(k=1, n-1, (f[k]*f[n-k])) / (2*n-3))); f
vector(#f, k, numerator(f[k])) \\ Colin Barker, Jul 24 2014, for sequence shifted by 1 index

Formula

Numerator of f(n), where f(0)=1 and f(n) = Sum_{k=0..n-1} f(k)*f(n-k-1)/(2*n-1). - Michael Somos, Mar 01 2014

Extensions

Name edited by Jon E. Schoenfield, Nov 13 2018