[seqfan] Re: No 3 terms in Geom Progression

[Neil Sloane]

David, Thanks, very nice! It is now A208746. Neil

On Thu, Mar 1, 2012 at 1:32 PM, David Applegate <david at research.att.com>wrote:

> 1..16 \ {1,4} still contains 3,6,12 and 9,12,16.
>
> I believe the corrected and extended sequence is:
>
>
> 1,2,3,3,4,5,6,7,7,8,9,10,11,12,13,13,14,14,15,15,16,17,18,19,19,20,21,21,22,23,
>
> 24,24,25,26,27,27,28,29,30,31,32,33,34,34,35,36,37,38,38,38,39,39,40,41,42,43,
>
> 44,45,46,46,47,48,49,49,50,51,52,52,53,54,55,55,56,57,57,57,58,59,60,61,61,62
>
> However, my computation used a floating-point IP solver for the
> packing subproblems, so although it's almost certainly correct I
> wouldn't bet my life on it.
>
> My approach was to enumerate geometric progressions using
>
>  for (i=1;i<=N;i++) {
>    for (j=2; j*j<=i; j++) {
>      if (i % (j*j) != 0) continue;
>      for (k=1; k<j; k++) {
>        print i*k*k/(j*j), i*k/j, i;
>      }
>    }
>  }
>
> and then solve the integer program of maximizing the subset of {1..N}
> subject to not taking all 3 of any progression.
>
> -Dave
>
> > From seqfan-bounces at list.seqfan.eu Thu Mar  1 12:01:22 2012
> > Date: Thu, 1 Mar 2012 12:01:05 -0500
> > From: Neil Sloane <njasloane at gmail.com>
> > To: seqfans <seqfan at seqfan.eu>
> > Subject: [seqfan] No 3 terms in Geom Progression
>
> > *Dear Sequence Fans,*
> > *A003002 gives the size of the largest subset of [1,2,...,n] which
> contains
> > no*
> > *3-term arithmetic progression.*
> > *But what is **
> > he largest subset of [1,2,...,n] which contains no
> > *
> > *3-term geometric progression?*
> > *E.g. if n=16, it looks like omitting 1 and 4 works, so a(16) = 14*
> > *A quick hand calculation gives (for n>=1):*
> > *1,2,3,3,4,5,6,7,7,8,9,10,11,12,13,14*
> > *
> > *
> > *Could someone correct/extend this?*
> > *Neil*
>
> > --
>
>