• ## RE: [PrimeNumbers] AP with 385256-digit tau

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• ... Big enough tau no doubt. But a big enough k? Jon Perry perry@globalnet.co.uk http://www.users.globalnet.co.uk/~perry
Message 1 of 3 , Jan 29, 2002
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>I reckon this is big enough, Jon?

Big enough tau no doubt. But a big enough k?

Jon Perry
perry@...
http://www.users.globalnet.co.uk/~perry
http://www.users.globalnet.co.uk/~perry/maths
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-----Original Message-----
Sent: 29 January 2002 10:44
Subject: [PrimeNumbers] AP with 385256-digit tau

Primitive arithmetic progression with 385256-digit tau

Let tau(n) be the number of divisors of n.
Let p[k] be the k'th prime that is 1 mod 4.
Let a[k] be the k'th member of the sequence with
a[k+1] = a[k]*(a[k]+3) and a[1] = 1
which grows like
a[k] = O(c^(2^k)) with
c = 1.52652645702134522042544787...

Let
m = prod(k=1,20,p[k]^a[21-k])
q = 1516882248248112456504781063232686847
r = 3014843725938648456075753401380244321
n = m*q^2
d = m*(r^2-q^2)/2

Then we have the primitive equal-tau progression

tau(n) = tau(n+d) = tau(n+2*d) = a[21] - a[20]

with tau ~ 8.22215486298630539147326425 * 10^385255
ln(ln(ln(ln(ln(n))))) ~ 0.9420151824085290419150311

Proof:

? {p=[5,13,17,29,37,41,53,61,73,89,
97,101,109,113,137,149,157,173,181,193];
q=1516882248248112456504781063232686847;
r=3014843725938648456075753401380244321;
if(2*prod(k=1,20,p[k])^2==q^2+r^2,print(ok))}
ok

Primality testing 1516882248248112456504781063232686847
[N-1/N+1, Brillhart-Lehmer-Selfridge]
Calling N+1 BLS with factored part 100.00%
and helper 100.00% (400.00% proof)
1516882248248112456504781063232686847 is prime!

Primality testing 3014843725938648456075753401380244321
[N-1/N+1, Brillhart-Lehmer-Selfridge]
Calling N+1 BLS with factored part 100.00%
and helper 100.00% (400.00% proof)
3014843725938648456075753401380244321 is prime!

I reckon this is big enough, Jon?

David

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