A396897 a(n) = n divided by the largest ratio of consecutive divisors of n.

1, 1, 2, 1, 3, 1, 4, 3, 4, 1, 6, 1, 4, 5, 8, 1, 9, 1, 10, 7, 4, 1, 12, 5, 4, 9, 14, 1, 15, 1, 16, 9, 4, 7, 18, 1, 4, 9, 20, 1, 21, 1, 16, 15, 4, 1, 24, 7, 20, 9, 16, 1, 27, 11, 28, 9, 4, 1, 30, 1, 4, 21, 32, 13, 33, 1, 16, 9, 28, 1, 36, 1, 4, 25, 16, 11, 36, 1, 40, 27

Offset

2,3

Comments

a(n) is always a positive integer: if d_{i+1}/d_i is the maximal ratio of consecutive divisors, then a(n) = (n/d_{i+1})*d_i, which is an integer since d_{i+1} divides n.

Links

Table of n, a(n) for n=2..81.

Formula

a(n) = n / max_i (d_{i+1}/d_i), where 1 = d_1 < ... < d_t = n are the divisors of n.

a(p) = 1 for prime p.

For squarefree n = p_1*p_2*...*p_k with p_1 < ... < p_k, a(n) = min_{0<=j<=k-1} (n/p_{j+1}) * Product_{i=1..j} p_i.

Example

For n = 10 the divisors are 1, 2, 5, 10, with consecutive ratios 2, 5/2, 2; the largest is 5/2, so a(10) = 10/(5/2) = 4.
For n = 12 the divisors are 1, 2, 3, 4, 6, 12, with largest consecutive ratio 2, so a(12) = 12/2 = 6.

Mathematica

a[n_]:=With[{d=Divisors[n]}, n/Max[Rest[d]/Most[d]]]; Array[a, 80, 2]

Prog

(PARI) a(n)=my(d=divisors(n)); n/vecmax(vector(#d-1, i, d[i+1]/d[i]));

Crossrefs

Cf. A033676, A163925, A163926, A396205, A396715, A396716.

Sequence in context: A385927 A323906 A020952 * A332294 A079554 A247892

Adjacent sequences: A396894 A396895 A396896 * A396898 A396899 A396901

Keyword

nonn,easy

Author

Eric Fodge, Jun 09 2026

Status

approved