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## Consecutive numbers all factored

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• Just a quick thought I had -- I was wondering if anyone has heard of any results in this area... For a given n =2, what is the largest n-tuple of consecutive
Message 1 of 4 , Nov 21, 2002
Just a quick thought I had -- I was wondering if anyone has
heard of any results in this area...

For a given n>=2, what is the largest n-tuple of consecutive integers
for which the prime factorization of each member is known?

Is there any method for constructing arbitrarily large
n-tuples of this form? I'm guessing that there isn't, and
that all of the "large" such n-tuples known today depend on
very large primes.

For instance, for n==2, an obvious answer is:
(2^13466917-1,2^13466917).

For n==3, I can easily construct:

(33218925*2^169690-1,33218925*2^169690,33218925*2^169690+1)

For n==5, go find the largest known BiTwins and we get:

(928223645*2503#*2^2-2,
928223645*2503#*2^2-1,
928223645*2503#*2^2,
928223645*2503#*2^2+1,
928223645*2503#*2^2+2)

So I'm curious, what is the longest titanic factorized n-tuple?
The above-mentioned 5-tuple is titanic. Can someone give me
6 consecutive titanic numbers along with their prime factorizations?
• Hi Jack, ... The only way that occurs to me right now is about as hard as finding a CC5, for which the record is ~500 digits. This would be by finding a highly
Message 2 of 4 , Nov 22, 2002
Hi Jack,

> So I'm curious, what is the longest titanic factorized n-tuple?
> The above-mentioned 5-tuple is titanic. Can someone give me
> 6 consecutive titanic numbers along with their prime factorizations?

The only way that occurs to me right now is about as hard as finding a CC5,
for which the record is ~500 digits. This would be by finding a highly
composite N (eg k.2^n or p#) such that the following 5 numbers are all prime:

3N-1, 6N+1, 6N-1, 3N+1, 2N+1.

Then the factorisations of the 6 numbers

6N-2, 6N-1, 6N, 6N+1, 6N+2, 6N+3

could all be found.

Any technique that relies on just 4 primes fails because then there would have
to be a number that was divisible by 3 but for which there was no trivial
factorisation, I think.

Regards,

Paul.

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• ... And to prove the concept, I proved the following numbers prime: 98497038*300#-1 98497038*300#+1 98497038*300#/2-1 98497038*300#/2+1 98497038*300#/3+1
Message 3 of 4 , Nov 22, 2002
> 3N-1, 6N+1, 6N-1, 3N+1, 2N+1.
>
> Then the factorisations of the 6 numbers
>
> 6N-2, 6N-1, 6N, 6N+1, 6N+2, 6N+3
>
> could all be found.

And to prove the concept, I proved the following numbers prime:

98497038*300#-1
98497038*300#+1
98497038*300#/2-1
98497038*300#/2+1
98497038*300#/3+1

Giving complete factorisations for the 6 numbers

98497038*300#+k k=-2...+3

Regards,

Paul.

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• If we find such 33-digit n that (6*n)^6/2-1, (6*n)^6/3-1, (6*n)^6-5, (6*n)^6/6-1, (6*n)^6-7 are all prime, then 11 numbers of about 200 digits from (6*n)^6-9
Message 4 of 4 , Nov 22, 2002
If we find such 33-digit n that (6*n)^6/2-1, (6*n)^6/3-1, (6*n)^6-5, (6*n)^6/6-1, (6*n)^6-7 are all prime, then 11 numbers of about 200 digits from (6*n)^6-9 through (6*n)^6+1 might be easily factored with a good chance.

Best wishes,

Andrey

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