[seqfan] Re: A202286

Maximilian Hasler maximilian.hasler at gmail.com
Thu Mar 15 00:48:26 CET 2012

On Wed, Mar 14, 2012 at 1:42 PM, Hans Havermann wrote:
> M. F. Hasler, Mar 13 2012, in his A202286 edit: "The sequence is finite,
> (...)
> Proof: To have p = k*R(p) -/+ q, with q < p,
> we must have 0 < k = (p +/- q) / R(p) < 2p / (p/10) = 20,
> since the prime p cannot end in 0 and therefore R(p) > p/10.
> (....)

> Isn't maximum k the sole determinant of the number of terms in this
> sequence? If there was a 20-ways term wouldn't there have to be a k>19 in
> order for this to occur?

AFAICS there is no flaw in the above inequality, so only 0 < k < 20 is possible.

> I think the separation into +q and -q cases isn't very helpful here:

maybe -- my main goal in mentioning this was to draw attention to
the fact that the choice of the sign multiplies ("a priori") the number
of possibilities [in terms of k-values] by 2.

Indeed I have not "optimized" the maximum number of possibilities in
my statement
(see below).

> If p<R(p), k can only be 1

I agree

> so, generally, p>R(p) which immediately suggests at most 9 +q's.

This ("at most 9") may be true, but maybe not "so" immediate.

Note that the limited number of (19) possible k-values, multiplied by
the 2 possible choices of the sign, already completes the proof of

There cannot exist any prime p such that this equation holds in more
than 19 x 2 ways.
Basically, that's where I stopped my reasoning, because the goal was reached.
But it was so obvious that the maximal k is only possible for one
choice of sign,
that's why I added the half-phrase about "- q  implies  k <= 9".
[Note the way in which I used the +/- in my comment].

But it is not true that the other case ("+ q") is possible only for k >= 10.

> If that were ever to happen, it would have to happen for k=1..9
> with an additional of at most 10 -q's (for k=10..19).
> Of course k can be <10
> for some -q but that reduces the number of available +q's.

Well,... it's not immediate to me what this means for the number of solutions.

> Here are my eight {k,q} for p=845534401:

I suggest you add them as a second example !


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