various items

[N. J. A. Sloane]

1.  I will be traveling May 10 - 20, so no updates in that period.

2.  It turns out that divisor chains were probably invented by
Matthijs Coster, in the course of setting a problem for a Dutch competition 
- see
Matthijs Coster, Problem 2001/3-A of the Universitaire Wiskunde Competitie,
Nieuw Arch. Wisk. 5/3 (2002), 92-94.
The sequence giving the number of length n was already in the OEIS,
as A067957, so I have deleted A094096.

3.  There are now 2 recent puzzle sequences that remain unsolved:

%I A094177
%S A094177 4,3,4,9,21,51
%N A094177 Sequence from an aptitude test that I cannot work out!
%K A094177 nonn,unkn,new
%O A094177 1,1
%A A094177 Dominique Butterworth (dominique_butterworth(AT)hotmail.com), May 06 2004

%I A092827
%S A092827 3,10,16,17,64,14,121,11,25
%N A092827 From a quiz. The choices given for the next two terms are: i) 12,32; ii) 22,18; iii) 17,64; iv) 5,21; v) 13,19.
%K A092827 nonn,unkn,more
%O A092827 1,1
%A A092827 Madhav Narayanan (madhav.narayanan(AT)iiitb.ac.in), Apr 15 2004


4.  For the past year I have been having trouble with my cable modem at home,
which repeatedly interefered with maintaining the OEIS.
It turns out that squirrels had chewed through the cable, which was full of
water.


5.  I use a plain text-based email program on a Unix machine.
So please try to avoid html format, spread-sheets, doc files, etc.
when writing to me.  Plain text is the best!


6.  An old sequence which I cannot understand.  Could someone please
provide a clear definition?  

%I A084924
%S A084924 3,7,31,127,1279,3583,5119,6143,8191
%N A084924 Riesel primes that have an increasing order n for which one can find the next in the sequence easily with (2^n*3^m) for m=0, m<n.
%F A084924 R(p)= R(p-1)+(2^n*3^m) for m=0,m<n
%e A084924 If p=5 then Rp=1279, R(p-1)=R4=127 so 127+(2^7*3^2)=1279 1279+(2^8*3^2)=3583 3583+(2^9*3^1)=5119 and so on.
%Y A084924 Sequence in context: A016047 A003260 A051281 this_sequence A001348 A006515 A081093
%Y A084924 Adjacent sequences: A084921 A084922 A084923 this_sequence A084925 A084926 A084927
%K A084924 nonn,uned
%O A084924 2,1
%A A084924 Shane Findley (TTcreation(AT)aol.com), Jul 15 2003

6 (cont.)  An unknown correspondent wrote to me the other day, saying:
From: "divineprime1 at netzero.com" <divineprime1 at netzero.com>
Date: Mon, 3 May 2004 18:21:17 GMT
Subject: A084924

I have a correction to make for this sequence.

The algorithm:
(k*2^n-1) + (m*2^n)  = prime

For instance,
127=(1*2^7-1)
(1*2^7-1) + (1*2^7) = 255 composite
(1*2^7-1) + (3*2^7) = 511 composite
(1*2^7-1) + (5*2^7) =767 composite
(1*2^7-1) + (7*2^7) =1023 composite
(1*2^7-1) + (9*2^7)=1279 is prime
next
1279= (5*2^8-1)
(5*2^8-1) + (1*2^8) = 1535 composite
(5*2^8-1) + (3*2^8) = 2047 composite 
(5*2^8-1) + (5*2^8) = 2559 composite
(5*2^8-1) + (7*2^8) = 3071 composite
(5*2^8-1) + (9*2^8) = 3583 is prime
next
3583 = (7*2^9-1)
repeat etc..
 
As soon as the first prime is found, 
the algorithm is reset  with the new exponent.  Technically n is how many times 2 
appears as a factor of k, for each prime found during the algorithm.  
Those factors of 2 are passed over to their proper, exponent index.  
For instance, if we take the expression of odd m above, and replace it with an even k.

k & n =  prime
4  0 = 3
2  2 = 7
4  3 =  31
4  5 = 127
10  7 = 1279
14  8 = 3583
10  9 = 5119
6  10 = 6143
4   11 = 8191   two is a factor of k, 13 times.
The exponents of two, from k on the left, are added to n on the right,  
p = 2+1+2+2+1+1+1+1+2 = 13 
This is true of any candidate's exponent, gM or Mn for that matter.

I have written a program to easily test for these special primes.
http://www.15k.org/rma/

Here are more known gM:

1 2  
1 3  
1 5  
1 7  
5 8  
7 9  
5 10  
3 11  
1 13  
5 14  
1 17  
1 19  
7 21  
[... 1200 lines deleted ...]

6 (cont.).  I wrote to this person asking for a clearer
definition of the original sequence.

Here is the reply I received:

Fist define Riesel primes as k*2^n-1, with k < 2^n.
Now list the Riesel primes in the linear order of their size in digits.

R=3  7  11 23 31 47 79 127 191 223 239 383 479 607 863 1087 1151 1279

n=2  3_ 2  3  5_  4  4  7_   6   5   4   7   5   5   5   6    7    8_ 

You'll notice that Mersenne primes will always appear with the largest n_, up to that size.  
By default they are the main sequence of the algorithm.
But there are other primes with the largest n, up to that size.

Algorithm to find Riesel primes with the larges n, 
is really just a simple {expression change}, and is triggered by the first prime found:

3 = 4 * 2^0 -1  {Two divides k, twice
3 = 1 * 2^2 -1  {so n+2 

Searching for the first even k, producing a prime...

7 = 2 * 2^2 -1 {Two divides k, once
7 = 1 * 2^3 -1 {so we only add n+1

Searching for the first even k, producing a prime...

k = 2 composite
31 = 4 * 2^3 -1 {Two divides k, twice
31 = 1 * 2^5 -1 {so n+2

Searching for the first even k, producing a prime...

k = 2 composite
127 = 4 * 2^5 -1 {Two divides k, twice
127 = 1 * 2^7 -1 {so n+2
 
and so on...

This means that a Mersenne prime exponent n, 
is the sum of all the preceeding times, two has been factor k during the entire algorithm.
Any general Mersenne prime found with the algorithm, also has this property.

Riesel primes, such as 11,23,47,79,191,223,239,383,... 
cannot nessesarily be used as starting points, to find the largest exponent n.  
For instance from 191, the next prime found using it's k, and exponent, is 383.  
Although, they are still in an increasing order, and eventually merge back into the general sequence, 
if you start from them.  The algorithm also works similarly with Proth primes of the form k*2^n+1. 

In short when a prime is found, the expression is changed so that k, is odd.
Search through a field of even k, that are larger than the current value k.
The size of the field sieve should be somewhat proportional to the value n.
Repeat...

It could also be expressed as: The first prime, 
produced by an odd m:  (k*2^n-1) + (m*2^n)  = prime

---------------------------------------------------------

I am still confused!

Would someone please help?

Thanks
Neil