[seqfan] proof of finiteness of A073396

[Max Alekseyev]

Let m = \prod_{i=1}^n p_i^{k_i}, where p_1 < ... < p_n are primes and
k_i >= 1, be an element of A073396, i.e.,

m = (\sum_{i=1}^n p_i) * (\sum_{i=1}^n p_i*k_i).

First notice that if n=1, then m = p^k = k*p^2, implying that m = 2^4 or 3^3.
For the rest assume that n >= 2.

Denote
K = \sumi=1^n k_i = Omega(m) = A001222(m)
s_t = \sum_{i=1}^t p_i
S_t = \sum_{i=1}^t p_i*k_i

Then m = s_n * S_n and thus
0 == m == s_{n-1} * S_{n-1} (mod p_n),
implying that p_n <= S_{n-1}.

Consider two cases, depending on whether k_n = 1 or k_n > 1.

1) if k_n > 1, then s_n <= S_n < K*p_n (the last inequality is strict
since n>=2)

m = s_n * S_n < K^2 * p_n^2

Let m' = m / p_n^2, which is an integer with Omega(m') = K-2.
Then
2^(K-2) <= m' < K^2
implying that K <= 7 and m' <= 49.

Now, we go over all possible values of m' and solve the equation
A008472(m') * (A001414(m') + p) = m' * p^2 (this corresponds to the
case k_n >= 3)
or the equation
(A008472(m') + p) * (A001414(m') + p) = m' * p^2 (this corresponds to
the case k_n = 2)
with respect to prime p = p_n.
In either case, p must divide A008472(m')*A001414(m') and we can
simply let p go over prime divisors of A008472(m')*A001414(m') and try
m'*p as m.

2) If k_n = 1, then s_n <= S_n = S_{n-1} + p_n <= 2 * S_{n-1} <= 2 *
(K-1) * p_{n-1}, implying that

m = s_n * S_n <= 4 * (K-1)^2 * p_{n-1}^2

Let m' = m / (p_n * p_{n-1}), which is an integer with Omega(m') = K-2.
Then
2^(K-2) <= m' = m / (p_n * p_{n-1}) <= 4 * (K-1)^2 * p_{n-1}/p_n < 4 * (K-1)^2,
implying that K <= 10 and m' < 324.

Now, we go over all possible values of m' and look for m of the form m
= m' * p * q where p < q are primes (in fact, p = p_{n-1} and q =
p_n).
Consider two cases:

2.1) if p divides m' (this corresponds to k_{n-1} >= 2), we let p go
over all prime divisors of m'. Similarly to the case 1), we further
have that q must divide A008472(m'*p)*A001414(m'*p), which is a finite
number of candidates to consider.

2.2) if p does not divide m' (this corresponds to the case k_{n-1} =
1), we solve the equation
(A008472(m') + p + q) * (A001414(m') + p + q) = m' * p * q
with respect to primes p < q.
We notice that q divides (A008472(m') + p) * (A001414(m') + p), which
leads to two cases:

2.2.1) q = A008472(m') + p  or  q = A001414(m') + p. In either case, p
must divide
2*A008472(m')*A001414(m')*(A008472(m')+A001414(m'))
and we let p go over the prime divisors of this number.

2.2.2) q is a proper divisor of A008472(m') + p or A001414(m') + p. Then
p < q <= (A001414(m') + p)/2,
implying that p < A001414(m').
So we let p go over the primes below A001414(m'), and let q go over
the prime divisors of (A008472(m') + p) * (A001414(m') + p).

In summary, we have a finite number of cases to consider. I carefully
went over them all and found that there are no other elements of
A073396 besides currently known 16, 27, and 150.

Regards,
Max