[seqfan] Re: proof of finiteness of A073396
Andrew Weimholt
andrew.weimholt at gmail.com
Tue May 7 13:02:23 CEST 2013
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
> 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
>
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