[seqfan] Re: A093893 Subsequence
T. D. Noe
noe at sspectra.com
Fri May 29 20:52:41 CEST 2009
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
>
>
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>
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--
Tony Noe | voice: 503-690-2099
Software Spectra, Inc. | fax: 503-690-8159
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