A Triangle Based On Coprimality

[Isabel Lugo]

What seems to happen is that after not too long, the triangle settles
down into alternating rows of prime and composite numbers, starting
with:

row #9: 59 61 67 71 73 79 83 89 97
row #10: 32 33 34 35 36 38 39 40 42 44

and basically, row 2n has enough small prime factors among its numbers
that row 2n+1 is all primes because any composites have been filtered
out.  Then since row 2n+1 is all primes and the primes are less dense
than the composites, there are unused composites that are smaller than
any of the primes in row 2n+1, and these become row 2n+2.

That's hardly a proof, but it seems like the beginning of one.  If
that pattern is indeed ture, then determining the row in which an
integer appears would be straightforward.

Isabel Lugo

On Tue, 31 Aug 04 13:51:45 -0600, Leroy Quet <qq-quet at mindspring.com> wrote:
> [posted too to sci.math.]
> 
> Consider the following triangle of integers:
> 
> 1
> 2  3
> 5  7  11
> 4  6  8  9
> 13 17 19 23 25
> 12 14 16 18 21 22
> ....
> 
> The top row is a 1.
> The m_th row of the triangle contains the lowest m as-yet-unused positive
> integers, in order from left to right, which are each coprime with every
> integer in the (m-1)th row.
> 
> (Triangle not in the EIS, it seems.)
> 
> Is there a relatively easy way to determine in which row the integer k
> appears?
> 
> thanks,
> Leroy Quet
> 
>


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