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

[Hagen von EItzen]

Joerg Arndt schrieb:
> This comment needs corrections and clarifications:
>
> 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
>
> Firstly, "With a different offset" seems wrong,
> a(3)==8 in the example and indeed a(3)==8 in the seq.
>
> Also: "containing exactly zero (0) or free"
> free --> three?
>
> Examples for n==4 or n==5 may be instructive.
> >From what I see the comment seems to be just wrong.
>   

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.

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