[seqfan] Re: A prime sequence that contains each sequential prime twice.

Charles Greathouse charles.greathouse at case.edu
Mon Jul 29 19:14:32 CEST 2013


Any infinite subset, that is.

Charles Greathouse
Analyst/Programmer
Case Western Reserve University


On Mon, Jul 29, 2013 at 12:57 PM, Andrew Weimholt <andrew.weimholt at gmail.com
> wrote:

> On Mon, Jul 29, 2013 at 5:52 AM, Hans Havermann <gladhobo at teksavvy.com>
> wrote:
> >
> > http://oeis.org/A026272 (and, adding zeros, http://oeis.org/A193564 ).
> Not *any* subset surely. Some will run into the issue of having to place a
> second copy in a position already occupied.
> >
>
> Any subset!!!
>
> Place terms and their copy in the natural order of their value, and the
> second position will always be free.
> When placing the first occurrence of X, we place it in the first open slot
> (say at position k)
> Suppose we find that the slot for the copy is already occupied. This is
> position k+1+X.
> But what value can occupy it? Say the value is N.
> N < X, because we placed N before X.
> The position of the first copy of N is somewhere to left of position k.
> Call it position m. Then m<k.
> The position of N's copy is then m+1+N, but we've already found N's copy
> blocking us at position k+1+X,
> so m+1+N = k+1+X, but m+1+N < k+1+X, because m<k and N<X.
> This is a contradiction, therefore position k+1+X cannot be already
> occupied.
>
> Andrew
>
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