[seqfan] Re: An old sequence from van der Poorten

[Vladimir Shevelev]

Dear Seq Fans,

I submitted A261728:

a(1)=1; a(2*n) = 3*n; for odd n>1, a(n) is the smallest number not already present which is entailed by the rules (i) k present => 3*k+1 present; (ii) 2*k present => k present. 
 
 DATA  1, 3, 4, 6, 2, 9, 7, 12, 10, 15, 5, 18, 13, 21, 16, 24, 8, 27, 19, 30, 22, 33, 11, 36, 25, 39, 28, 42, 14, 45, 31, 48, 34, 51, 17, 54, 37, 57, 40, 60, 20, 63, 43, 66, 46, 69, 23, 72, 49, 75, 52, 78, 26, 81, 55, 84, 58, 87, 29, 90, 61, 93, 64, 96, 32, 99, 67, 102, 70, 105, 35, 108, 73, 111, 76, 114, 38, 117, 79 
 
Conjecture. The only fixed points are 6*k+1, k>=0; if n==3 (mod 6), then a(n) = n+1; if n==5 (mod 6), then a(n) = (n-1)/2. 

If the conjecture is true, then the sequence 
is a permutation of the positive integers. 

On the other hand, the statement 'the sequence is a permutation of the positive integers' is equivalent to the (3*n+1)-conjecture.

Proof. Indeed, if (3*n+1)-conjecture is true 

then, according to the rule, the sequence is a permutaion of the positive numbers. Conversely, we use induction. Let for all m<=N, the (3*n+1)-conjecture is true. Consider N+1. 

1) If N+1 is even = 2^k*t, where t is odd. 

Since t<N, then, beginning with N+1=>t etc., by the rule of (3*n+1)-problem, by the inductional supposition (IS), we reach 1. 

2) If N+1 = 6*t+1, then 2*t<n and, by IS, 

2*t=>N+1=>...=>1; 

3)Finally, let N+1 be odd of the form 6*t+5. Then 4*t+3<N. Therefore, by IS, we have 4*t+3=>12*t+10=>N+1=>...=>1. QED 

Thus the above Conjecture  implies the (3*n+1)-conjecture. 

Best regards,
Vladimir

________________________________________
From: SeqFan [seqfan-bounces at list.seqfan.eu] on behalf of Vladimir Shevelev [shevelev at exchange.bgu.ac.il]
Sent: 29 August 2015 20:45
To: Sequence Fanatics Discussion list
Subject: [seqfan] Re: An old sequence from van der Poorten

Dear Max,

You are right, indeed there are difficulties to prove that. So, in A261690
I did the following comment.

"29 Aug 2015 Max Alekseyev noted that, while

(3*n+1)-cojecture indeed implies that the sequence is a permutation of the positive integers not divisible by 3, but the opposite statement is far from obvious

at all. The author cannot yet prove this, so his previous comment is only a conjecture. In connection with this,

consider the following conjecture which naturally could be called (n-1)/3-conjecture.
Let n be any number not divisible by 3. If n==1 (mod 3) and (n-1)/3 is not divisible by 3, then set n_1 = (n-1)/3. Otherwise set n_1 = 2*n.

Conjecture. There exists an iteration n_m = 1.
Example: 19->38->76->25->8->16->5->10->20->40->13->
4->1.

Then we have: (3*n+1)-conjecture =>
A261690 is a permutation of numbers not divisible by 3
=> (n-1)/3-conjecture.
Does (n-1)/3-conjecture imply (3*n+1)-conjecture? "

Best,
Vladimir

________________________________________
________________________________________
From: SeqFan [seqfan-bounces at list.seqfan.eu] on behalf of Max Alekseyev [maxale at gmail.com]
Sent: 29 August 2015 04:40
To: Sequence Fanatics Discussion list
Subject: [seqfan] Re: An old sequence from van der Poorten

Vladimir,
I do not see equivalence here. The 3x+1 conjecture indeed implies that
A261690 is a permutation, but in the opposite direction it is not obvious
at all. Can you prove it?
Max
On Aug 28, 2015 2:50 PM, "Vladimir Shevelev" <shevelev at bgu.ac.il> wrote:

> Dear Neil,
>
>  I submitted A261690 which is an
> analog of A109732 such that the
> statement ' the sequence is a
> permutation of the positive integers
> not divisible by 3' is equivalent to
> the (3*n+1)-conjecture on numbers
> not divisible by 3.
> So I think that Van der Poorten's
> question is in the same degree unprovable
> as the (3*n+1)-conjecture.
>
>
> Best regards,
> Vladimir
>
> ________________________________________
> From: SeqFan [seqfan-bounces at list.seqfan.eu] on behalf of Neil Sloane [
> njasloane at gmail.com]
> Sent: 27 August 2015 20:39
> To: Sequence Fanatics Discussion list
> Subject: [seqfan] An old sequence from van der Poorten
>
> Dear Seq Fans:
>
> An old sequence suggested by a posting by Alf van der Poorten is A109732:
> a(1) = 1; for n>1, a(n) is the smallest number not already present which is
> entailed by the rules (i) k present => 2k+1 present; (ii) 3k present => k
> present.
> The open question is whether every odd number appears.
>
> It seems that numbers of the form 2^k+1 take an exceptionally long time to
> appear - see A261414, which needs more terms. In particular, when does 1025
> appear in A109732?
>
>
> Best regards
> Neil
>
> Neil J. A. Sloane, President, OEIS Foundation.
> 11 South Adelaide Avenue, Highland Park, NJ 08904, USA.
> Also Visiting Scientist, Math. Dept., Rutgers University, Piscataway, NJ.
> Phone: 732 828 6098; home page: http://NeilSloane.com
> Email: njasloane at gmail.com
>
> _______________________________________________
>
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>
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> Seqfan Mailing list - http://list.seqfan.eu/
>

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