# [seqfan] Re: A093893 Subsequence

Hagen von EItzen math at von-eitzen.de
Thu May 28 22:58:31 CEST 2009

```mathar wrote:
> In http://list.seqfan.eu/pipermail/seqfan/2009-May/001521.html Leroy spake
>
> lq> Consider sequence A093893,
> lq> This is the list of positive integers n such that the partial sum of any 2 or more divisors of n is composite.
> lq>
> lq> What I wonder about is the subsequence, which doesn't seem to be in the EIS, where the nth term is the smallest term of A093893 with exactly n divisors.
> lq>
> lq> (Starts at a(2).)
> lq>
> lq> 3, 49, 87, etc.
> lq>
> lq> It seems that it is very unlikely that this sequence is infinite, or even that it is not short.
> lq>
> lq> Can it be proved that this sequence is finite or infinite?
>
> I think this starts 3, 49, 87, 130321, 4753, >1000000, 285541  (n=2 to 8)
> The values for n=7 and n=9 to n=31  are all larger than 1 million (if they exist).
>
If n has exactly 7 divisors, then n = p^6 for some prime p.
We need that none of the following is prime (even number of summands are
trivially composite as well
as those avoiding 1):
Three summands:
1+p+p^2
1+p+p^3
...
1+p+p^6
1+p^2+p^3
1+p^2+p^4
...
1+p^5+p^6
Five summands:
1+p+p^2+p^3+p^4 = (p^7-1)/(p-1) - p^5 -p^6
...
(p^7-1)/(p-1) - p^2 -p^3
Seven summands:: (p^7-1)/(p-1)

In PARI,
good(p)=local(v=vector(6,k,p^k);s7=(p^7-1)/(p-1);ok=!isprime(s7));if(ok,for(i=2,6,if(ok,for(j=1,i-1,if(isprime(v[i]+v[j]+1)||isprime(s7-v[i]-v[j]),ok=0)))));ok
p=3;while(!good(p), p=nextprime(p+1)); p^6
quickly produces
7212549413161
(i.e. 139^6)

Hence the sequence is now known to start

3, 49, 87, 130321, 4753, 7212549413161, 285541
(and if I'm not wrong the n=11 term is 31^10 = 819628286980801)

I suspect that the sequence *is* infinite:
For given n and indeterminate p, consider all partial sums obtained by taking 1 plus an even number of
elements of {p,p^2,...,p^(n-1)} (there are 2^(n-1) such sums).
Overly heuristically, each is prime with probability <1/(2*ln(p)), hence for big primes p, the probability
of all being composite tends to 1. In fact, simply using "< 1/2" instead of "<1/(2*ln(p))" one would expect
a success among the first approximately 2^2^(n-1) primes.

Hagen

```