A334074 a(n) is the numerator of the sum of reciprocals of primes not exceeding n and not dividing binomial(2*n, n).

0, 0, 1, 1, 1, 1, 12, 1, 10, 71, 16, 103, 215, 311, 311, 311, 431, 30, 791, 36, 575, 8586, 222349, 222349, 182169, 144961, 747338, 8630, 1343, 89513, 2904968, 520321, 45746, 1005129, 350073, 1890784, 72480703, 34997904, 257894479, 257894479, 1755387611, 1755387611

Offset: 1

Author

Amiram Eldar, Apr 13 2020

Keywords

• nonn
• frac

Comments

Erdős et al. (1975) could not decide if the fraction f(n) = a(n)/A334075(n) is bounded. They found its asymptotic mean (see formula).

Examples

			For n = 7, binomial(2*7, 7) = 3432 = 2^3 * 3 * 11 * 13, and there are 2 primes p <= 7 which are not divisors of 3432: 5 and 7. Therefore, a(7) = numerator(1/5 + 1/7) = numerator(12/35) = 12.
		

References

• R. K. Guy, Unsolved Problems in Number Theory, Springer, 1st edition, 1981. See section B33.

Links

• Chai Wah Wu, Table of n, a(n) for n = 1..3844
• Paul Erdős, Ronald L. Graham, Imre Z. Ruzsa and Ernst G. Straus, On the prime factors of C(2*n, n), Mathematics of Computation, Vol. 29, No. 129 (1975), pp. 83-92.

Crossrefs

Cf. A000984, A114124, A334075 (denominators).

Programs

Mathematica

a[n_] := Numerator[Plus @@ (1/Select[Range[n],PrimeQ[#] && !Divisible[Binomial[2n, n],#] &])]; Array[a, 50]

PARI

a(n) = {my(s=0, b=binomial(2*n,n)); forprime(p=2, n, if (b % p, s += 1/p)); numerator(s);} \\ Michel Marcus, Apr 14 2020

Python

from fractions import Fraction
from sympy import binomial, isprime
def A334074(n):
    b = binomial(2*n,n)
    return sum(Fraction(1,p) for p in range(2,n+1) if b % p != 0 and isprime(p)).numerator # Chai Wah Wu, Apr 14 2020

Formula

a(n) = numerator(Sum_{p prime <= n, binomial(2*n, n) (mod p) > 0} 1/p).

Limit_{k -> infinity} (1/k) Sum_{i=1..k} a(i)/A334075(i) = Sum_{k>=2} log(k)/2^k (A114124).

Limit_{k -> infinity} (1/k) Sum_{i=1..k} (a(i)/A334075(i))^2 = (Sum_{k>=2} log(k)/2^k)^2.