OFFSET
1,4
COMMENTS
A simple insight shows that the prime factors decomposition of CB(n) = binomial(n, floor(n/2)) (i) does not contain any prime factor greater than n, (ii) contains exactly once all primes in the interval ((n+1)/2, n]. Hence, CB(n) is divisible by the product P2(n) of all primes in ((n+1)/2, n]. The relatively small elements of this sequence are a(n) = CB(n)/P2(n). For n > 6, they can be shown to be devoid of any prime factor exceeding n/3.
LINKS
Stanislav Sykora, Table of n, a(n) for n = 1..1000
FORMULA
a(n) = C(n, floor(n/2))/Product_{n/2 < prime p <= n} p.
EXAMPLE
CB(21) = binomial(21,10) = 21!/(10!11!) = 352716 is divisible by all primes in (11,21] to 1st power, i.e., by 13*17*19 = 4199. Hence a(21) = 352716/4199 = 84.
PROG
(PARI) {lambda1(n) = result=binomial(n, floor((n+1)\2)); forprime(p=1+floor((n+1)\2), n, result=result/p); }
CROSSREFS
KEYWORD
nonn
AUTHOR
Stanislav Sykora, May 27 2012
STATUS
approved