[seqfan] Re: A093893 Subsequence
franktaw at netscape.net
franktaw at netscape.net
Fri May 29 18:47:28 CEST 2009
Well, a(10) will be a number of the form p^4 * q (conceivably p^9, but
it shouldn't be hard to eliminate those once you find the answer). It
shouldn't be too hard to search those.
Franklin T. Adams-Watters
-----Original Message-----
From: Robert G. Wilson, v <rgwv at rgwv.com>
Dear Hagen,
No, a(11) does not equal to 31^10. The divisors of 31^10 =
819628286980801
are {1, 31, 961, 29791, 923521, 28629151, 887503681, 27512614111,
852891037441,
26439622160671, 819628286980801}. The 1486th subset {1, 31, 961, 29791,
923521,
28629151, 887503681} sums to 917087137, a prime. a(11)= 211^10.
What I need is a value for a(10).
Bob.
Hagen von EItzen wrote:
> 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,f
or(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
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