OFFSET
1,2
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.
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.
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
EXAMPLE
1, 2/3, 7/15, 34/105, 638/2835, 4876/31185, 220217/2027025, 6885458/91216125, 569311642/10854718875, 7515775348/206239658625, 197394815194/7795859096025, ...
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
PROG
(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
CROSSREFS
KEYWORD
nonn,frac
AUTHOR
Philipp O. Tsvetkov, Nov 11 2013
EXTENSIONS
Name edited by Jon E. Schoenfield, Nov 13 2018
STATUS
approved