A333072 Least k such that Sum_{i=1..n} k^i / i is a positive integer.

1, 2, 6, 6, 30, 10, 70, 70, 210, 168, 1848, 1848, 18018, 8580, 2574, 2574, 102102, 102102, 831402, 2771340, 3233230, 587860, 43266496, 117630786, 162249360, 145088370, 145088370, 2897310, 672175920, 672175920, 18232771830, 18232771830, 44279588730, 8886561060

Offset: 1

Author

Jinyuan Wang, Mar 10 2020

Keywords

• nonn

Comments

Note that the denominator of (Sum_{i=1..n} k^i/i) - k^p/p can never be divisible by p, where n/2 < p <= n. Therefore, for the expression to be an integer, such p must divide k. Thus, a(n) = k is divisible by A055773(n).

Links

• Chai Wah Wu, Table of n, a(n) for n = 1..61

Crossrefs

Cf. A034386, A055773, A332734, A333196.

Programs

PARI

a(n) = {my(m = prod(i=primepi(n/2)+1, primepi(n), prime(i)), k = m); while (denominator(sum(i=2, n, k^i/i)) != 1, k += m); k; }

Python

from sympy import primorial, lcm
def A333072(n):
    f = 1
    for i in range(1,n+1):
        f = lcm(f,i)
    f, glist = int(f), []
    for i in range(1,n+1):
        glist.append(f//i)
    m = 1 if n < 2 else primorial(n,nth=False)//primorial(n//2,nth=False)
    k = m
    while True:
        p,ki = 0, k
        for i in range(1,n+1):
            p = (p+ki*glist[i-1]) % f
            ki = (k*ki) % f
        if p == 0:
            return k
        k += m # Chai Wah Wu, Apr 04 2020

Formula

a(n) <= A034386(n).