A212791 Central binomial coefficient CB(n) purged of all primes exceeding (n+1)/2.

1, 1, 1, 2, 2, 4, 1, 2, 18, 36, 6, 12, 12, 24, 45, 90, 10, 20, 2, 4, 84, 168, 14, 28, 700, 1400, 2700, 5400, 360, 720, 45, 90, 2970, 5940, 11550, 23100, 23100, 46200, 90090, 180180, 8580, 17160, 780, 1560, 70200, 140400

Offset: 1

Author

Stanislav Sykora, May 27 2012

Keywords

• nonn

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.

Examples

			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.
		

Links

• Stanislav Sykora, Table of n, a(n) for n = 1..1000

Crossrefs

Cf. A001405, A212792, A000040 (primes).

Programs

PARI

{lambda1(n) = result=binomial(n,floor((n+1)\2));forprime(p=1+floor((n+1)\2),n,result=result/p);}

Formula

a(n) = C(n, floor(n/2))/Product_{n/2 < prime p <= n} p.