[seqfan] Re: A093893 Subsequence

[Max Alekseyev]

a(10) = 211^4 * 421 = 834472284661

Max

On Fri, May 29, 2009 at 2:52 PM, T. D. Noe <noe at sspectra.com> wrote:
> An upper bound for a(10) is 211^9.
> a(11) is 211^10.
> An upper bound for a(12) is 421^11.
> a(13) is 2311^12.
> An upper bound for a(14) is 2311^13.
>
> Tony
>
>>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
>>
>>
>>_______________________________________________
>>
>>Seqfan Mailing list - http://list.seqfan.eu/
>
> --
>
> Tony Noe                  | voice:     503-690-2099
> Software Spectra, Inc.    | fax:       503-690-8159
> 14025 N.W. Harvest Lane   | e-mail:    noe at sspectra.com
> Portland, OR  97229, USA  | Web site:  http://www.sspectra.com
>
>
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