[seqfan] Re: Pairwise sums are all semiprimes.

[zak seidov]

Maximilian, you are right!

As Douglas McNeil rightly wrote
in the message to me,

1) a(4) = 54 not 90
2) sequence is finite, as
   there is no a(5):
   any a(5) + (1,3,32,54) makes
   one multiple of 4.

Thanks to you two,
Zak

P.S. Q about n-tuples with square pair-wise sums
should be non-trivial (I hope ;)) 


--- On Sat, 6/12/10, Maximilian Hasler <maximilian.hasler at gmail.com> wrote:

> From: Maximilian Hasler <maximilian.hasler at gmail.com>
> Subject: Re: [seqfan] Pairwise sums are all semiprimes.
> To: "Zakir Seidov" <zakseidov at yahoo.com>
> Cc: "Sequence Fanatics Discussion list" <seqfan at list.seqfan.eu>
> Date: Saturday, June 12, 2010, 8:40 AM
> I get
> a=[1,3,32,54]
> then the sequence stops, because:
> 
> * if the next term is a(5)=2k+1, then
> a(5)+a(1)=2(k+1)
> a(5)+a(2)=2(k+2)
> which cannot both be semiprime (since either k+1 or k+2
> have a second factor 2)
> 
> * if the next term is a(5)=2k, then
> a(5)+a(3)=2(k+16)
> a(5)+a(4)=2(k+27)
> which cannot both be semiprime (since either k+16 or k+27
> have a
> second factor 2).
> 
> So you might object that 54 is not the correct term.
> But in view of what precedes, whatever you choose as
> preceding terms,
> among a(1)..a(4) there are at least and at most 2 even and
> 2 odd terms
> (using the same reasoning as above), and there is no
> possible 5th term.
> 
> Maximilian
> 
> 
> 
> On Sat, Jun 12, 2010 at 5:51 AM, zak seidov <zakseidov at yahoo.com>
> wrote:
> > a(1) = 1; for n>1, a(n) = smallest number >
> a(n-1) such that the pairwise sums of elements are all
> semiprimes.
> > 1,3,32,90
> > more terms?
> > Zak
> >
> >
> >
> >
> >
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