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

[Andrew Weimholt]

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