[seqfan] Re: No 3 terms in Geom Progression
David Applegate
david at research.att.com
Thu Mar 1 19:32:45 CET 2012
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*
> --
> Dear Friends, I will soon be retiring from AT&T. New coordinates:
> Neil J. A. Sloane, President, OEIS Foundation
> 11 South Adelaide Avenue, Highland Park, NJ 08904, USA
> Phone: 732 828 6098; home page: http://NeilSloane.com
> Email: njasloane at gmail.com
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