[seqfan] Re: A202286

Hans Havermann gladhobo at teksavvy.com
Wed Mar 14 18:42:54 CET 2012

> Isn't this sequence finite?

M. F. Hasler, Mar 13 2012, in his A202286 edit: "The sequence is  
finite, more specifically there cannot be more than 28 terms. 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 (because p and R(p) have the same number of  
digits). So we must have 1 <= k <= 19 in the p+q case, and 1 <= k <= 9  
in the p-q case."

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?

I think the separation into +q and -q cases isn't very helpful here:  
If p<R(p), k can only be 1 so, generally, p>R(p) which immediately  
suggests at most 9 +q's. 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.

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


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