%N A1 Related to sum of two squares: Puzzle(?)

[Joerg Arndt]

such that zeta(1/2+eps+I*C) == 0 where I==sqrt(-1) and -1/2<eps<+1/2.
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To: zakseidov at yahoo.com, seqfan at ext.jussieu.fr
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Subject: Re: a(n) = sigma_2(a(n-1)): 10|a(n>3) ?
Date: Wed, 14 May 2008 14:12:42 -0400
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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