A390834 The maximum number of proper divisors of a semiperfect number k that sum to k.

3, 4, 4, 4, 6, 5, 5, 6, 6, 4, 7, 5, 6, 8, 5, 8, 3, 7, 9, 6, 9, 9, 5, 3, 6, 8, 7, 3, 12, 9, 9, 3, 8, 10, 7, 9, 8, 6, 12, 3, 7, 14, 3, 10, 7, 9, 9, 6, 7, 12, 12, 9, 3, 10, 7, 10, 15, 3, 14, 3, 8, 11, 13, 6, 7, 13, 3, 14, 7, 13, 7, 7, 8, 12, 3, 10, 11, 12, 15, 7

Offset

1,1

Links

Robert Israel, Table of n, a(n) for n = 1..1000

Formula

If A005835(n) is a perfect number then a(n) = A032741(A005835(n)).

Example

A005835(17) = 78 = 39 + 26 + 13, and no other subset of {1, 2, 3, 6, 13, 26, 39} sums to 78, so a(17) = 3.
A005835(34) = 144 = 1 + 2 + 3 + 4 + 6 + 8 + 12 + 24 + 36 + 48, and no other subset of {1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48, 72} that sums to 144 has more terms, so a(34) = 10.

Maple

# after obtaining a list of semiperfect numbers SP
f:= proc(n) local k, D, d, x; uses Optimization, NumberTheory;
    k:= SP[n];
    D:= Divisors(k) minus {k};
    LPSolve(add(x[d], d=D), {add(d*x[d], d=D) = k}, maximize, assume=binary, depthlimit=1000)[1];
end proc:
map(f, [$1..nops(SP)]); # Robert Israel, Jan 07 2026

Prog

(Python)
from sympy import proper_divisors
def max_divs(n: int) -> int:
    if n <= 1: return -1
    dp = {0: 0}
    for d in proper_divisors(n):
        n_dp = dict(dp)
        for s, count in dp.items():
            n_sum = s + d
            if n_sum <= n:
                n_dp[n_sum] = max(n_dp.get(n_sum, -1), count + 1)
        dp = n_dp
    return dp.get(n, -1)
print([a for n in range(1, 400) if (a := max_divs(n)) > 0]) # Peter Luschny, Jan 08 2026

Crossrefs

Cf. A005835, A210455, A000396, A032741, A390728, A391005, A392361 (minimum).

Sequence in context: A179842 A259052 A268395 * A304032 A022484 A361497

Adjacent sequences: A390831 A390832 A390833 * A390835 A390836 A390837

Keyword

nonn

Author

Peter Luschny, Nov 25 2025

Extensions

Corrected by Robert Israel, Jan 07 2026

Status

approved