256360

Appears in sequences

• Numbers m such that DivisorSigma(4*k-2, m) mod m = 0 holds presumably for all k; that is, (4k-2)-power-sums of divisors of m are divisible by m for all k.at n=26A066290
• For every positive integer m, let u(m) = (d(1),d(2),...,d(k)) be the unitary divisors of m. The sequence (a(n)) consists of successive numbers m which d(k)/d(1) + d(k-1)/d(2) + ... + d(k)/d(1) is an integer.at n=27A229996