[seqfan] Re: proof of finiteness of A073396

Max Alekseyev maxale at gmail.com
Tue May 7 16:12:14 CEST 2013

No. Prime factorization is not defined for 0 (and neither are A008472
and A001414).
Max

On Tue, May 7, 2013 at 7:02 AM, Andrew Weimholt
<andrew.weimholt at gmail.com> wrote:
> Should 0 be in the sequence?
>
>
>
> On Sat, May 4, 2013 at 3:25 PM, Max Alekseyev <maxale at gmail.com> wrote:
>
>> 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
>>
>> 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
>>
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