# [seqfan] Re: Q regarding comment in A000079 (Powers of 2)

Joerg Arndt arndt at jjj.de
Wed Jun 24 07:54:39 CEST 2009

* Hagen von EItzen <math at von-eitzen.de> [Jun 22. 2009 13:54]:
> [...]
>
> Indeed, the n=3 example seems to indicate that in general there at least
> the 2^n combinations of v and z; whatever the letter u may possibly add
> to this, the result will become >2^n as soon as u occurs.

I just submitted these changes in the comments:

Killed this one:
%C A000079 With a different offset, number of
n-permutations (n>=0) of 3 objects: u, v, z with repetition
allowed, containing exactly zero (0) or free u's. For example,
a(3)=8 because we have vvv, vvz, vzv, vzz, zvv, zvz, zzv, zzz. -
Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Jul 08 2008

%C A000079 Permutations of n+1 elements where no element is more
than one position left of its original place. For example, there
are 4 such permutations of three elements: 123, 132, 213, and
312. The 8 such permutations of four elements are 1234, 1243,
1324, 1423, 2134, 2143, 3124, and 4123. [From Joerg
Arndt (arndt(AT)jjj.de), June 24 2009]

>
> Several other comments are not top quality either (or are at least
> obfuscating):
>
>  > Let P(A) be the power set of an n-element set A. Then a(n) = the
> number of pairs of elements {x,y} of P(A)
>  > for which x = y.
>
> As x=y is required this just describes #P(A) (i.e. the first comment
> "Number of subsets of an n-set") in a less readable fashion ...
>
>  > 2^(n-1) is the largest number having n divisors; A005179
> <http://www.research.att.com/%7Enjas/sequences/A005179>(n) is the smallest.
>
> In general p^(n-1) has n divisors for prime p, hence there is no largest ...
>
>  > a(n) appears to match the number of divisors of the modified
> primorials (excluding 2,3and 5) Very limited range examined
>
> Trivially, any product of n distinct primes has exactly 2^n divisors.
>
> Hagen

I'd suggest to kill all of these, but leave it to someone else.

cheers,  jj