A051707 (prime signature)

[Christian G.Bower]

Yasutoshi:

> I rewrite the condition as follows. 
    
> Pair (i,j) must satisfy 0<i, 0<j, and if i=1 then j=1.
> The second condition is equivalence as that if (i,j) is not
> (1,1) then i is not 1. 
 
Implying that (1,1) is allowed as a factor.

That cannot be the case, otherwise there would be infinitely many
factorizations:

(2,2)
(2,2)*(1,1)
(2,2)*(1,1)*(1,1)
...

So a better restriction is that: 1<i, 0<j

Using that I get for the first several terms:

1 1 1 3 1 5 1 8 3 5 1 23 1 5 5 23 1 23 1 23 5 5 1 91 3 5 8 23 ...

This matches most of A051707
ID Number: A051707
URL:       http://www.research.att.com/projects/OEIS?Anum=A051707
Sequence:  1,1,1,3,1,5,1,8,3,5,1,21,1,5,5,23,1,21,1,21
Name:      Number of factorizations of (n,n) into pairs (k,l).

but is different for 12, 18 and 20

The 23 factorizations of 12 are as follows:

(12,12)
6 of the form (6,x)*(2,y)
6 of the form (4,x)*(3,y)
10 of the form (3,x)*(2,y)*(2,z)

I can send an update if Yasutoshi can confirm that those values are
errors and not the result of yet another rule.

> I want to know the formula.
> I wish some one tell me it.

I can help if you're not too picky about what a formula is.

What I did was apply the setprod transform I recently described to an
array of numbers a(m,n) such that a(1,n) = 0 and a(m,n) = 1 for all
m>=2, n>=1

The output array gives the number of factorizations of (m,n):

1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 
2 2 2 3 2 3 2 3 3 3 2 4 2 3 3 4 2 4 2 4 3 3 2 5 
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 
2 3 3 4 3 5 3 5 4 5 3 7 3 5 5 6 3 7 3 7 5 5 3 9 
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 
3 4 4 6 4 7 4 8 6 7 4 11 4 7 7 10 4 11 4 11 7 7 4 15 
2 2 2 3 2 3 2 3 3 3 2 4 2 3 3 4 2 4 2 4 3 3 2 5 
2 3 3 4 3 5 3 5 4 5 3 7 3 5 5 6 3 7 3 7 5 5 3 9 
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 
4 7 7 11 7 14 7 15 11 14 7 23 7 14 14 20 7 23 7 23 14 14 7 33 
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 
...

Then I take the diagonal b(n,n) to produce the sequence.

The mathematics of this is to treat the input array as a
2 dimensional Dirichlet generating function:

A(s,t) = SUM {m,n=1 to infinity} a(m,n)/(m^s*n^t)

apply setprod as follows

B(s,t) = exp(A(s,t)/1 + A(2*s,2*t)/2 + A(3*s,3*t)/3 + ...)

       = PROD {m,n=1 to infinity} (1/(1-a(m,n)/(m^s*n^t)))

If you want to know the algorithm I actually use or the code I
use to execute it, you should probably contact me off line unless
there are several on the list that are interested.

Christian