A353691 a(n) is the least number k > n such that h(k)/h(n) is an integer, where h(n) is the harmonic mean of the divisors of n, or -1 if no such k exists.

6, 120, 28, 234, 30, 270, 42, 29792, 252, 1120, 66, 234, 78, 840, 140, 200, 102, 2016, 114, 1170, 945, 1320, 138, 1080, 150, 1560, 756, 270, 174, 3360, 186, 1272960, 308, 2040, 210, 9720, 222, 2280, 364, 148960, 246, 1890, 258, 2574, 1260, 2760, 282, 600, 294

Offset

1,1

Comments

Does a(n) exist for all n? If m is a harmonic number (A001599) and gcd(n, m) = 1, then a(n) exists and a(n) <= m*n, since h(m*n) = h(m)*h(n) and h(m) is an integer.

Links

Amiram Eldar, Table of n, a(n) for n = 1..639

Formula

a(p) = 6*p for a prime p > 3.

Example

a(2) = 120 since 120 is the least number > 2 such that h(120)/h(2) = (16/3)/(4/3) = 4 is an integer.

Mathematica

h[n_] := DivisorSigma[0, n]/DivisorSigma[-1, n]; a[n_] := Module[{k = n + 1, hn = h[n]}, While[!IntegerQ[h[k]/hn], k++]; k]; Array[a, 30]

Prog

(Python)
from math import prod, gcd
from sympy import factorint
def A353691_helper(n):
    f = factorint(n).items()
    return prod(p**e*(p-1)*(e+1) for p, e in f), prod(p**(e+1)-1 for p, e in f)
def A353691(n):
    Hnp, Hnq = A353691_helper(n)
    g = gcd(Hnp, Hnq)
    Hnp //= g
    Hnq //= g
    k = n+1
    Hkp, Hkq = A353691_helper(k)
    while (Hkp*Hnq) % (Hkq*Hnp):
        k += 1
        Hkp, Hkq = A353691_helper(k)
    return k # Chai Wah Wu, May 07 2022

Crossrefs

Cf. A001599, A099377, A099378.

Similar sequences: A069789, A069797, A069805, A353692.

Sequence in context: A054957 A373239 A271648 * A290341 A246827 A127726

Adjacent sequences: A353688 A353689 A353690 * A353692 A353693 A353694

Keyword

nonn

Author

Amiram Eldar, May 04 2022

Status

approved