# [seqfan] Re: Biprimes of K

Eric Angelini Eric.Angelini at kntv.be
Wed Sep 17 15:40:32 CEST 2014

```> Unless I misunderstood the procedure,

... my fault.
I'll explain more:
_Undelined_ is NOT "eliminated", sorry!
An integer might be underlined several times.
We are interested in integer that have never been underlined (once or more).
Best,
É.
----------------------------------------
Hello SeqFans,
what are the "biprimes" of K?
K = http://oeis.org/A026242

K = 1, 1, 2, 3, 2, 4, 3, 5, 6, 4, 7, 8, 5, 9, 6, 10, 11, 7, 12, 8, 13, 14, 9, 15, 16, 10, 17, 11, 18, 19, 12, 20, 21, 13, 22, 14, 23, 24, 15, 25, 16, 26, 27, 17, 28, 29, 18, 30, 19, 31, 32, 20, 33, 21, 34, 35, 22, 36, 37, 23, 38, 24, 39, 40, 25, 41, ...

and underline one on two terms
(this is 2-3-6-7-5-...)
We then go back to the first "3"
and underline one on three terms
(this is 3-4-5-10-12-...)
We then go back to the first "4"
and underline one on four terms
(this is 4-9-7-14-...)
We then go back to the first "5"
and underline one on five terms
(this is 5-7-9-11-...)
etc.
At the end, we will see that some
terms are not underlined: the
"biprimes" of K.

BIP = 2,3,4,5,8,15,...

What are the next ones (if there are)?

Best,
É.

-----Message d'origine-----
De : SeqFan [mailto:seqfan-bounces at list.seqfan.eu] De la part de M. F. Hasler
Envoyé : mercredi 17 septembre 2014 15:35
À : Sequence Fanatics Discussion list
Objet : [seqfan] Re: Biprimes of K

On Wed, Sep 17, 2014 at 3:16 AM, L. Edson Jeffery <lejeffery2 at gmail.com> wrote:
> Please check your code. Applying the process to 3 should mark
> (underline) the first occurrence of 20 and applying it to 7 should
> mark the second occurrence of 20. So 20 is not in the BIP sequence,
> meaning that your

I don't agree, since the the first 7 is eliminated in the first step and the second 7 is eliminated when the procedure is applied to 4 so it is never used.
(Also, 7 is not in Eric's list and I think as is the case for Eratothenes' sieve, only the "(bi)primes" of this list are to be used in that way.) Unless I misunderstood the procedure, of course...

Maximilian

(PARI) (ugly, only for my records...)
M=Mat([vector(90,n,n)~,A026242[1..90]~])~
do(k)=forstep(i=k+M[2,k],#M,M[2,k],M[2,i]=0);M
comp(A,B)=setminus(Set(A[2,1..#A]),Set(B[2,1..#B]))
do(3) \\ "uses" 2 => %1

do(4) \\ "uses" 3 => %2
[1 1 2 3 0 4 0 5 0 0 0 8 0 9 0 0 0 7 0 8 0 0 0 15 0 10 0 0 0 19 0 20 0
0 0 14 0 24 0 0 0 26 0 17 0 0 0 30 0 31 0 0 0 21 0 35 0 0 0 23 0 24 0
0 0 41 0 26 0 0 0 45 0 46 0 0 0 30 0 50 0 0 0 52 0 33 0 0 0 56]

comp(%1,%2) \\ numbers eliminated in this step:
% = [11, 13, 25, 27, 29, 36, 40, 47, 51, 55]

do(6) \\ "uses" 4 => %6
[1 1 2 3 0 4 0 5 0 0 0 8 0 0 0 0 0 0 0 8 0 0 0 15 0 0 0 0 0 0 0 20 0 0
0 14 0 0 0 0 0 0 0 17 0 0 0 30 0 0 0 0 0 0 0 35 0 0 0 23 0 0 0 0 0 0 0
26 0 0 0 45 0 0 0 0 0 0 0 50 0 0 0 52 0 0 0 0 0 0]

comp(%2,%6) \\ numbers eliminated in this step:
% = [7, 9, 10, 19, 21, 24, 31, 33, 41, 46, 56]

do(8) \\ "uses" 5 => %8
[1 1 2 3 0 4 0 5 0 0 0 8 0 0 0 0 0 0 0 8 0 0 0 15 0 0 0 0 0 0 0 20 0 0
0 14 0 0 0 0 0 0 0 17 0 0 0 0 0 0 0 0 0 0 0 35 0 0 0 23 0 0 0 0 0 0 0
0 0 0 0 45 0 0 0 0 0 0 0 50 0 0 0 52 0 0 0 0 0 0]

comp(%6,%8) \\ numbers eliminated in this step:
% = [26, 30]

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