I'm including Michel Marcus because (a) I don't know if he's on this mailing list, and (b) he has helped me a lot in this research. The search for k-perfect numbers is well-known. We consider the function sigma(n), the sum of the divisors of n. Because sigma(n) >= n (with equality only when n = 1), it makes sense to look for numbers n such that sigma(n) = kn for some k; such a number is called k-perfect. (The "classical" perfect numbers are 2-perfect in this terminology.) The k-perfect numbers, for all indices k, are presented at OEIS sequence A007691. The following variation is due to Greg Martin, who presented it at the Western Number Theory Conference at Asilomar in 1999; it is "Western Number Theory problem 99:08". (See Myerson's compendium of WNT problems at https://www.math.colostate.edu/~achter/wntc/problems/problems2001.pdf .) Douglas Iannucci made some progress on the problem in his 2006 paper -- see http://math.colgate.edu/~integers/g41/g41.pdf . I'm using Iannucci's notation and nomenclature. Suppose d is a divisor of n. Consider the number of prime factors of the codivisor n/d, counting multiplicity. Call this the index of d in n. For example, the index of 2 in 12 is 2, because 12 = 2*(2*3). The index of 10 in 1000 is 4, because 1000 = 10*2*2*5*5. Now compute a _weighted_ sum of the divisors of n, where the weight of a divisor is 1 if the divisor has even index, and -1 if its index is odd. Call this weighted sum rho(n). For example, rho(12) = 12 - 6 - 4 + 3 + 2 - 1 = 6. Note that rho(n) <= n (with equality only when n = 1), so it makes sense to look for numbers n such that k*rho(n) = n for some k; Iannucci calls such a number k-imperfect. (I personally would have preferred "k-contraperfect", but that ship has sailed.) Because 12 = 2*rho(12), 12 is 2-imperfect. The function rho(n) is multiplicative, with rho(p^e) = p^e - p^(e-1) + ... +/- 1; the sign of the trailing unit depends on the parity of e. Like k-perfect numbers, there are lots of known examples of k-imperfect numbers. Many of these are listed at https://oeis.org/A127724 . Because they get big so quickly, we soon lose track of whether they are consecutive; we are pretty sure we know the smallest 50, but after that, the gaps have not been searched exhaustively. About 1800 examples are known; part of my recent research has added around 200 new ones. While k-perfect numbers are known for all 1 <= k <= 11, for k-imperfect numbers, we only had examples of k = 1, 2, 3, or 4 ... until earlier today, when I found one with k = 5. Martin and Iannucci had only found values of k up to 3, but many examples with k = 4 were subsequently found by Corneth, Johnson, Lelechenko, and Marcus (Michel Marcus found more than a thousand). We didn't know for sure if there were any 5-imperfect numbers until this afternoon. The smallest known 4-imperfect number is 993803899780063855042560 (24 digits). My discovery today is so far the only known 5-imperfect number; it has 208 digits, and the exact value is 1947793410288108579327587698415272737289992039373107522449638016140636142596276017072442826236838486130285072853813458640948347868163450516845165654669897619318450994951647517899051394662400000000000000000000. Michel Marcus and I have both checked it (he with a package he wrote in Pari, I with a complete cludge in Emacs Lisp). I would welcome additional confirmation. I have just emerged from a "dry valley" in my search space, and expect to find at least a few more 5-imperfect numbers in the next few days, but I thought I shouldn't let the occasion of the discovery of the first example go unmentioned.