Let's define:

sopf(n) = "sum of prime factors of the value n, without repetition"

Examples:

sopf(714)=sopf(2x3x7x17) = 2+3+7+17

sopf(81) = sopf(3x3x3x3) = 3 

Jason Earls sent to these pages the following conjecture

For any k value there is always at least one n value such that: sopf(n)=sopf(n+k)

 

"I was very interested in Jud McCranie's experimental data for the "sopf(n)=sopf(n+k)" conjecture, and I thought it was about time I presented my theoretical results to go with them. So far I have managed to prove:

For *a large proportion* of k values there is always at least one n value such that sopf(n)=sopf(n+k).

Of course, I need to say more about *a large proportion*. I expect, that with the data Jud has generated, we may very easily be able to construct a proof. Here is the key part of the proof.

Suppose there are two integers M,N>1 such that sopf(M)=sopf(N). Then evidently that demonstrates a solution for k=N-M. Choose any integer Q> 1 such that any of Q's prime factors are factors either of *both* M and N, or *neither* M and N. Then Q.M, Q.N, gives a solution for Q.k.

Now, to construct the proof, we need to construct a table of sopf(n).

We will find some interesting results on the way. We may write a little program that generates each sopf(n), looks through all previous values of n to find an equal value, and applies the lemma above to generate an interesting 'rule'.

Here are some examples of rules we can generate.

sopf(2)=sopf(4).

Let Q be *any* integer. Then sopf(2Q)=sopf(4Q), and so we have a solution for k=2Q with n=2Q (In other words, in Jud's table, n*<=k for even k at least). This covers all even values of k.

sopf(5)=sopf(6).

Let Q be an integer with no factors 2, 3, or 5. Then sopf(5Q)=sopf(6Q) and so we have a solution for k=Q.

At this point then the only k values that remain are odd multiples of 3 or 5. (I think this covers over 76% of the possible k).

We may continue for other rules. For example sopf(7)=sopf(10). *But* often we can improve the rule we find. For instance sopf(7)=sopf(10)=sopf(49)=sopf(50). Let Q be an integer with no factors 2, 5, or 7. Then sopf(49Q)=sopf(50Q) gives a solution for k=Q, and sopf(7Q)=sopf(10Q) gives a solution for k=3Q.

Now the only k remaining unsolved have factors of 5 *or* factors of * both* 3 and 7. Now it is interesting to look at Jud's tables - 5 and 21 are in fact the next *difficult* values of k. This is not a coincidence. Let us look at k a multiple of 5. Jud's table tells us sopf(114Q)=sopf(119Q) is a solution for k=5Q, *provided* Q has no common factors with 114 or 119. In other words, we have solved for multiples of 5 * provided* they have no factors 2, 3, 7, 17, 19. Even more k values are covered.

By identifying more pairs such that sopf(M)=sopf(N), we can cover more and more values of k. Perhaps if Jud has already generated the table of sopf's, he might just be able to finish the proof and find a set of rules that 'cover' all the k... I am sure *almost all* k will fall to this method! Perhaps even there are better rules to support Jud's new sub-question, that n*<=k....

As a quick note, the "rule generator" I mention above works in question 2) as well, the same lemma holds even if sopf() is defined *with* repetition.

His answer was:

"This I do not know yet... but I think the answer is yes. There seem to be two different kinds of rules. For some, M and N have no common factors and so sopf(QM)=sopf(QN) *only* for certain values of Q. But there are other rules (such as 2,4) for which *any* value of Q will work. So perhaps what happens is, with many of the first rule all that remains is "k is solved except for multiples of ????". Then we find the second kind of rule for ????...."

On your web page at Conjecture 25 you discuss Jason Earls's conjecture that for any k there is at least one n such that sopf(n)=sopf(n+k). You claim that Chris Nash's method will yield a complete solution. Here's a proof that it won't:

Suppose we have a finite set of pairs (Mi, Ni) which 'covers' all odd numbers. I.e. for each index i, sopf(Mi) = sopf(Ni) and for every positive odd number k there's an index i and a positive integer Q such that Q (Ni - Mi) = k and every prime factor of Q is a factor of both Mi and Ni or of neither Mi nor Ni.

Let k be the product of all odd primes which divide some Mi or Ni, and suppose that k is covered by the pair (Mi, Ni); thus Q (Ni - Mi) = k for some Q, all of whose prime factors divide either both or neither of Mi and Ni. Since k is odd, one of Mi and Ni is even and the other is odd. Also, there's some odd prime p which divides exactly one of Mi or Ni; otherwise sopf of the even one would be 2 more than sopf of the odd one.

By definition of k, p divides k. Since p doesn't divide Ni - Mi, p must divide Q. But every prime factor of Q is a factor of either both or neither of Mi and Ni, so we have a contradiction. Hence our finite set of pairs doesn't cover all odd numbers.

It doesn't mean that Jason's conjecture is false, just that Nash's original method isn't sufficient to prove it.

***

Mr. Dean Hickerson wrote (10/10/01) the following about the Nash's second approach:

"The new proposal doesn't work either. Essentially the same proof as before shows that no finite set of pairs will work:

As before, let k be the product of all odd primes which divide some M_i or N_i, and suppose that k is covered by the pair (M_i, N_i); i.e. we have k=a1*a4*(a2*M_i - a3*N_i) where a1, a2, a3, and a4 are as described in Nash's proposal. As before, there's an odd prime p which divides exactly one of M_i or N_i. If p divides M_i, then p doesn't divide either a3 or N_i, so p doesn't divide a2*M_i - a3*N_i. Similarly, if p divides N_i, then p doesn't divide either a2 or M_i, so p doesn't divide a2*M_i - a3*N_i.

And p obviously doesn't divide either a1 or a4, so p doesn't divide k, contradicting the definition of k."

***