PRE-NUMBERED NEW SEQUENCE A139393 FROM M. F. Hasler

[Maximilian Hasler]

Sorry for flooding the list with ill-composed submissions. Obviously I
made some copy-paste errors.
Neil, in order to avoid that this pre-numbered new sequence interferes
with the existing one, please delete my preceding message (subject:
"Re: COMMENT ....").
Here is the correct version:

%I A139393
%S A139393 0, 1, 1, 2, 1, 11, 1, 3, 2, 11, 1, 12, 1, 11, 11, 4, 1, 12,
1, 12, 11, 11, 1, 13, 2, 11, 3, 12, 1, 111, 1, 5, 11, 11, 11, 22, 1,
11, 11, 13, 1, 111, 1, 12, 12, 11, 1, 14, 2, 12, 11, 12, 1, 13, 11,
13, 11, 11, 1, 112, 1, 11, 12, 6, 11, 111, 1, 12, 11, 111, 1, 23, 1,
11, 12, 12, 11, 111, 1, 14, 4, 11, 1, 112, 11, 11, 11, 13, 1, 112, 11,
12, 11, 11, 11, 15, 1, 12, 12
%N A139393 Sum( e[i] 10^(m-i), i=1..m ) where e[1]<=...<=e[m] are the
nonzero exponents in the prime factorization of n: a representation of
the prime signature of n.
%C A139393 The sorted sequence of (nonzero) exponents in the prime
factorization of a number is called its prime signature. Here this is
"approximated" by multiplying them by powers of 10. Up to 2^10 this
coincides with the concatenation of these exponents written in base 10
(but that sequence would be "base" specific). For n>=1024 one should
use a modified definition, replacing 10 by 10^floor(log2(n)/10), to
avoid ambiguity of the representation.
%H A139393 Wikipedia, <a
href="http://en.wikipedia.org/wiki/Prime_signature">Prime
signature</a>.
%H A139393 E.W.Weisstein, <a
href="http://mathworld.wolfram.com/PrimeSignature.html">Prime
signature</a> on mathworld.wolfram.com.
%o A139393 (PARI) A139393(n)=sum(i=1,#n=vecsort(factor(n)[,2]),10^(#n-i)*n[i])
%K A139393 nonn,easy
%Y A139393 Cf. A037916.
%O A139393 1,4
%A A139393 M. F. Hasler (www.univ-ag.fr/~mhasler), Apr 17 2008




cb> From seqfan-owner at ext.jussieu.fr  Wed Apr 16 03:07:57 2008
cb> Date: Tue, 15 Apr 2008 18:07:36 -0700
cb> From: "Christian G. Bower" <bowerc at usa.net>
cb> To: "seqfan" <seqfan at ext.jussieu.fr>
cb> Subject: Re: conjecture: if p is prime, the no. of rings of order p^2 is 11
cb> ...
cb> 
cb> John Conway did a proof of a(p^2)=11 in that discussion from 1998. I can post
cb> it if anyone is interested.
cb> 
cb> Christian

There is also the statement

"This completes the discussion and we see that there are in 
all 3+2+2+3+1=11 mutually nonisomorphic rings of order p^2"

on page 228 of 
R. Raghavendran, Finite associative rings,
www.ams.org/mathscinet-getitem?mr=246905
www.numdam.org/item?id=CM_1969__21_2_195_0
http://www.numdam.org/numdam-bin/browse?j=CM&sl=0

--
Richard