A361077 a(n) = largest sqrt(2*n)-smooth divisor of binomial(2*n, n).

1, 1, 2, 4, 2, 36, 12, 24, 18, 4, 4, 24, 4, 200, 5400, 720, 90, 540, 300, 600, 180, 120, 120, 10800, 900, 3528, 10584, 784, 280, 1680, 112, 224, 882, 2940, 2940, 504, 28, 56, 4200, 19600, 980, 158760, 7560, 75600, 113400, 5040, 35280, 211680, 44100, 1800, 648, 432, 216, 45360, 1680

Offset: 0

Author

Matthieu Pluntz, Mar 01 2023

Keywords

• nonn

Comments

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

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

Binomial(2*n, n)/a(n) and A263931(n)/a(n) are coprime with a(n) and squarefree. By the Erdős squarefree conjecture, proved in 1996, no a(n) with n >= 5 is squarefree.

Examples

			binomial(10, 5) = 2^2*3^2*7. 2,3 <= sqrt(10), 7 > sqrt(10) so a(5) = 2^2*3^2 = 36.
		

Links

• Matthieu Pluntz, Table of n, a(n) for n = 0..10000
• Eric Weisstein's World of Mathematics, Erdős Squarefree Conjecture.

Crossrefs

Cf. A263922, A263931.

Programs

PARI

a(n) = my(m=sqrtint(2*n), f=factor(binomial(2*n, n), m+1)); for (k=1, #f~, if (f[k,1]>m, f[k,1]=1)); factorback(f); \\ Michel Marcus, Mar 03 2023

R

library(primes)
n = 2*10^4
pp = generate_primes(max = sqrt(n))
npinb = list()
for(p in pp){
  np = rep(0,n)
  for(t in 1:(log(n)/log(p))) np[p*(1:(n/p))] = np[1:(n/p)] + 1 #multiplicities of prime factor p in the integers
  npinb[[as.character(p)]] = cumsum(np)[2*(1:(n/2))] - 2*cumsum(np[1:(n/2)]) #multiplicities of prime factor p in the central binomial coefficients
}
res = rep(1,n/2)
for(p in pp){
  select = p^2 <= 2*(1:(n/2))
  res[select] = res[select]*p^npinb[[as.character(p)]][select]
}
#offset of res is 1

Formula

a(n) = binomial(2*n, n) / Product(p prime | p^2 > 2*n, floor(2*n/p) is odd).