A263931 a(n) = binomial(2*n, n) / Product(p prime | n < p <= 2*n).

1, 1, 2, 4, 2, 36, 12, 24, 90, 20, 4, 168, 28, 1400, 5400, 720, 90, 5940, 23100, 46200, 180180, 17160, 1560, 140400, 11700, 45864, 179928, 13328, 52360, 5969040, 397936, 795872, 3133746, 12345060, 726180, 2863224, 159068, 318136, 1255800, 4958800, 247940

Offset

0,3

Comments

The highest exponent in the prime factorization of a(n) is A263922(n), n>=2.

a(n) is even for n>=2.

By the Erdős squarefree conjecture, proved in 1996, no a(n) with n >= 5 is squarefree. - Robert FERREOL, Sep 06 2022

Links

David A. Corneth, Table of n, a(n) for n = 0..5806 (terms < 10^1000)

Eric Weisstein's World of Mathematics, Erdős Squarefree Conjecture.

Formula

a(n) = A000984(n)/A261130(n).

Maple

a := n -> binomial(2*n, n)/convert(select(isprime, {$n+1..2*n}), `*`):
seq(a(n), n=0..40);

Prog

(PARI) a(n) = { my(res = 1); forprime(p = 2, n, res*= p^(val(2*n, p) - 2*val(n, p))); forprime(p = n + 1, 2*n, res*= p^(val(2*n, p) - 2*val(n, p) - 1)); res } val(n, p) = my(r=0); while(n, r+=n\=p); r \\ David A. Corneth, Apr 03 2021
(Python)
from math import comb
from sympy import primorial
def A263931(n): return comb(m:=n<<1, n)*primorial(n, nth=False)//primorial(m, nth=False) if n else 1 # Chai Wah Wu, Sep 07 2022

Crossrefs

Cf. A000984, A261130, A263922, A356637.

Sequence in context: A006496 A352642 A361077 * A259685 A301460 A130172

Adjacent sequences: A263928 A263929 A263930 * A263932 A263933 A263934

Keyword

nonn

Author

Peter Luschny, Oct 31 2015

Status

approved