A395192 a(n) is the least k with the largest abundancy where k is a proper divisor of n.

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

Offset

2,3

Comments

The abundancy of k is sigma(k)/k, where sigma is the sum of divisors function, A000203.

a(n) is of the form n/p where p is a prime divisor of n. Proof: sigma(k)/k is multiplicative with sigma(p^m)/p^m = (p^(m+1)-1)/((p-1)*p^m) which increases as m increases. To ensure sigma(k)/k is maximal, divide by as few primes as possible. To ensure k is a proper divisor of n, divide by one prime.

When we have an abundancy-defined set, S, such as A005101, A005231, A023196 or A023197, we can determine whether a member, k, is a primitive member of S using only the abundancy of a(k).

For all positive integer m, a(p*m) = m for all except finitely many primes p. Thus all positive integers appear infinitely often. These values p*m are all values k such that a(k) = m.

The integers m for which a(p*m) = m for all primes p are given in A137825.

For all k in A362619 we have a(k) = k/A006530(k) but this property is not exclusive to A362619. For example 200 is not in A362619 but a(200) = 200/A006530(200) = 200/5 = 40.

Links

Paolo Xausa, Table of n, a(n) for n = 2..10000

Formula

a(n) = n/A137826(m), where m = min({k : n|A137825(k)}).

a(p) = 1 for prime p.

a(p*q) = p for primes p <= q.

Example

Usually a number n has a unique proper divisor with the largest abundancy. The first exception is 400, where this property is shared by its divisors 80 and 200, whose abundancy is 93/40, so a(400) = 80.

Mathematica

A395192[n_] := Min[MaximalBy[n/FactorInteger[n][[All, 1]], DivisorSigma[1, #]/# &]];
Array[A395192, 100, 2] (* Paolo Xausa, Apr 21 2026 *)

Prog

(PARI) a(n) = {my(f = factor(n), r = 0, res = 0, c); forstep(i = #f~, 1, -1, c = sigma(n/f[i, 1])/(n/f[i, 1]); if(c > r, r = c; res = n/f[i, 1])); res}

Crossrefs

Cf. A004394.

See the comments for the relationships with A000203, A005101, A005231, A006530, A023196, A023197, A362619.

See the formula section for the relationships with A137825, A137826.

Sequence in context: A094521 A321757 A159272 * A098372 A385999 A396079

Adjacent sequences: A395189 A395190 A395191 * A395193 A395194 A395195

Keyword

nonn

Author

David A. Corneth and Peter Munn, Apr 15 2026

Status

approved