Further define Q(N), the quotient-factor of N, Q(N) = F(N)/N.

In common usage:

Q(N) = 1, N is perfect

Q(N) < 1, N is deficient

Q(N) > 1, N is abundant.

Previously existing result:

N is perfect if N = 2^n x (2^(n+1) - 1), provided (2^(n+1) - 1) is prime.

It is not known whether all perfect numbers are of this form.

Preliminary results:

If P is prime, F(P) = 1 and Q(P) = 1/P. Clearly, for e, there exists N s.t. that Q(N) < e.

Primes are

*maximally deficient*.

If N = P^n, where P is prime, F(N) = (P^n - 1)/(P - 1), Q(N) = (N-1)/N(P-1)

For particular P, Q(P^n) tends to 1/(P-1) as n tends to infinity.

In the case P = 2, F(2^n) = 2^n - 1 and Q(2^n) tends to 1.

Integers of the form 2^n are

*minimally deficient*.

Question: Are there other N such that F(N) = N - 1?

If N is abundant, any multiple of N is abundant with greater factor-quotient. (See following lemma; more fractions are added when a multiple is derived.) Formally, if M divides N, then Q(N) > Q(M).

**Lemma**

For all n, there exists N s.t. Q(N) > n.

*(Abundancy increases without limit.)*

**Proof**

Q(N) is the sum of fractions of the form 1/n. Since Σ1/n diverges (well-known result), Q(N) is unbounded.

Therefore it is possible to define N is abundant-S if Q(N) >= S and assert that there exists an infinity of abundant-S numbers, whatever the value of S. For instance, 120 is abundant-2 (the smallest such).

Question: Find for each integer N, find the smallest M such that Q(M) ³ N

Further Result:

Since the sum of the odd fractions can also be shown to diverge[1], there is also an infinity of odd abundant-S numbers. The first odd abundant number is 945 and the smallest odd abundant-2 number is believed to be 1 018 976 683 725, which has Q-value 2.0107.

Define N is super-abundant if Q(N) > 1 and Q(N) > Q(n) if N > n. (Informally, a new record is set.)

The first super-abundant number is the first abundant number, 12. Q(12) = 1.33. (The record-holders that are not abundant are 2, 4 and 6.) The set of super-abundant numbers has infinitely many members. (A record-holder will be beaten by any multiple, from result above.)

Question: Find the super-abundant numbers.

[1] The proof is similar to that of the full series. Consider the series 1/3 + 1/5 + 1/7 +1/9 ... If we successively group 1, 3, 9, 27 ... terms ie 1/3 + (1/5 + 1/7 + 1/9) + (1/11 + ... + 1/27) + ... it is clear that the series is greater than 1/3 + 1/3 + 1/3 +1/3 ... and therefore diverges (slowly).

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