COMMENT FROM M. F. Hasler RE A037916

[franktaw at netscape.net]

It's a subtle distinction, but I have generally seen prime signature 
used to mean
the multiset of exponents in the prime factorization, rather than the 
sequence of
same.  This agrees with the Wikipedia definition -- note the words 
"sorted by size"
in that definition.

Franklin T. Adams-Watters

-----Original Message-----
From: Maximilian Hasler <Maximilian.Hasler at martinique.univ-ag.fr>

%I A037916
%S A037916 
0,1,1,2,1,11,1,3,2,11,1,21,1,11,11,4,1,12,1,21,11,11,1,31,2,11,3,21,1
%C A037916 The sequence of (nonzero) exponents in the prime
factorization of a number is called its prime signature.
%H A037916 Wikipedia, <a 
href="http://en.wikipedia.org/wiki/Prime_signature">
   Prime signature</a>.
%o A037916 (PARI) A037916(n)=if( n>1, 
eval(concat(concat([""],factor(n)[,2]~))))
%o A037916 (PARI) primesig(n)=sum(i=1,#n=factor(n)[,2],10^(#n-i)*n[i])
\\ the same result up to 2^10
%K A037916 nonn,easy
%O A037916 1,4
%A A037916 M. F. Hasler (www.univ-ag.fr/~mhasler), Apr 17 2008





------ Original Message ------

> On 4/15/08, Christian G. Bower <bowerc at usa.net> wrote:
> >
> >
> > John Conway did a proof of a(p^2)=11 in that discussion from 1998. I can
post
> > it if anyone is interested.
> >
> 
> I'm interested :-)
> 
> Andrew
> 

Here it is

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

 From John Conway 6/23/1998

Part 1

we can call its elements  0,a,2a,..., and all we need to know is
which of these is  a^2 : call it  Ka.  Now if  k  is prime to  N,
we can replace  a  by the  b  for which  kb = a,  so we have
(kb)^2 = Kkb,  b^2 = (K/k)b.  This allows us to suppose that  K
divides  N.  The rings of this form are non-isomorphic, though,
being distinguished by the size  N/K of their sets of products.

d(p^2) = 3 rings whose additive structure is cyclic, defined by:




makes the ring a vector space over the field of order  p.  Now
Then every ring element has the form  ma + na^2,  and the structure
is determined by knowing what  a^3  is - I suppose that we have
a^3 = Pa^2 + Qa = 0.  

of the form    a = Ra^3 + Sa^2 = (Ra^2 + Sa)a,  which shows that
Ra^2 + Sa  is a unit,  which I'll call 1,  and we can rewrite 
the generic ring element as  ma + n, and need only know what
a^2 is.  There are three cases:








of the form  a^2 = ka.  I see at once


If the  k  for a given  a  is non-zero,  then  ka  is an
idempotent.  So in all remaining cases, we have some idempotents
and everything is either a multiple of one of these or a 
nilpotent.  Since the going seems to be getting a bit hard, I'll


Part 2

nilpotent  (a^2 = a)  or a multiple of an idempotent, and there
were some of the latter.

Then  every element has the form  ma + nb,  and in particular


Q = 0.  Postmultiplying by  b  shows that also  P = 0, and similarly
R  and  S  must be  0,  and we obtain the zero ring again.

which we'll call  a  if it exists, and will suppose  b  is an
idempotent.  Write  ab  and  ba  as above;  then we can still prove
that  Q  and  S  must be  0, so that


which we can multiply by  b  on the appropriate sides to get


gives us 4 cases for the  a,b  multiplication table:


here, but since I don't want to forget any case, I'll postpone their
discussion.  Any further ring can contain no nilpotent, so is
generated by idempotents  a  and  b.  Write  ab  and  ba  as above,
then premultiplying the left equation  by  a  gives  ab = Pa + Qab,
when  Q is  0.  Postmultiplying it by  b  gives  ab = Pab + Qb,
both 0, or one is 1 and the other 0, giving  



or


Part 3



first.  We know that  (a+b)^2  has to be a multiple of
a+b,  which shows that  ab + ba  can only  be  0  or  a+b,
reducing us to three cases


of which the last two are isomorphic.  What about things
like  2a+b,  whose square  4aa + 2ab + 2ba + bb  must be
a multiple of itself?  It becomes

in the two cases, and therefore the first can only work if p = 2,
which I'll temporarily suppose it aint.  The other case does
appear to work (though it must later be checked): 




in which the square of  0 + b + 0/a + 0/a  can only be  a+b,


or its opposite ring.

11) (and see why the opposite of  11)  isn't a new case).



of the leftmost one by what I'll call the height  h(Ra+Sb) = R+S  
of the rightmost.  This is easily checked to be distributive
and associative once one's realised that the height map is a 
homomorphism.  So 10) does define a ring.  However, the elements 
of height 0 are nilpotents, which contradicts one of the assumptions 
made in getting to  10).  



which gives a similar "height" rule   xy = h(x).y,  where now
h(Pa+Qb) = Q.  

those given by the two equations



where  h  is any non-trivial linear function from the space
to the field of order  p.



was only disproved for odd  p.  Let's see what happens to the
general product


This does define a ring, which is a sort of "internal direct
product" of the subrings generated by  a  and  b.  It might
help to see its multiplication table:


gone into a different case.  So indeed we don't have to consider
it for any  p.

11 rings.


Part 4








where  h  is any non-zero linear function from the additive
group to the groundfield.

form.  In the 6th case, the description

>     6)  It's a square - but then we can make it  x^2,  and what
>        we have is the field of order  p  extended by a quadratic
>        unipotent  a  with  a^2 = 1.


to know so much.









------ Original Message ------

> Is A027623(n) a function of the prime signature of n?
> 
> 

No

REFERENCES  R. Ballieu [ Math. Rev. 9, 267; see also Math. Rev. 51#5655 ]

Hence a(8)=52, a(27)=59, a(125)=65, ...

Christian