a(n) = sigma_2(a(n-1)): 10|a(n>3) ?

[Richard Guy]

The numbers of DISTINCT prime divisors are  1,1,2,3,6,8,11,20,...,
I believe --- I think you are quoting Omega instead of omega.  The
next term is perhaps 10751241448828145947377163154256928532205365443200
=2^7*3^4*5^2*7^3*13^3*19*31*37^3*41*43*53*61*127*137*157*307*601*1877*2521*135721
[E&OE]       R.

On Wed, 14 May 2008, franktaw at netscape.net wrote:

> Probably.  Very likely all are divisible by 2; only if n is of the form k^2 
> or
> 2 k^2 is sigma_2(n) odd.
>
> Divisibility by 5 is another matter.  It is pretty likely that a randomly 
> chosen
> (relatively) highly divisible number n will have 5 as a factor of sigma_2(n).
> For this not to be the case, 2 and 3 must be each be a factor an even
> number of times -- but this is only a constant probability.  It is not 
> unlikely
> that other primes p will have sigma_2(p) divisible by 5 -- this will be true 
> for
> any p = 3 or 7 (mod 10), which is half of all primes.  Certainly a(n) is on 
> the
> roughly on the order of 2^2^n (for any m,
> m^2 <= sigma_2(m) <= pi^2/6 m^2), so even if the number of distinct
> prime divisors of a(n) is only O(log n), the probability a term is not 
> divisible
> by 5 is O(1/2^n), and the number of such terms is probably finite (hence
> probably all remaining terms are thus divisible).
>
> For the first 7 terms (terms 6 and 7 are 1318281510000 and
> 2755017380686402917800000), the number of distinct prime divisors of
> a(n) is:
>
> 1,1,2,4,7,14,23.
>
> It thus appears that this grows at least linearly, perhaps even 
> exponentially.
> So very likely all remaining terms are divisible by 5.
>
> However, this is only a probabilistic argument; I don't see any way to 
> actually
> prove it.
>
> Note that by the same kind of argument, every prime probably divides all but
> finitely many members of this sequence.  In general, primes = 1 (mod 4) will
> start dividing a(n) sooner that those = 3 (mod 4).
>
> Franklin T. Adams-Watters
>
> -----Original Message-----
> From: zak seidov <zakseidov at yahoo.com>
>
> Rule:
> a(n>1)=sigma_2(a(n-1)) with a(1)=2.
> SEQ starts:
> 2,5,26,850,943950.
> My Q:
> Are all next terms divisible by 10?
> thanks, zak
>
>
>