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Forum: And now for something completely different 2021-06-24, 06:49
Replies: 47
Views: 20,888
Posted By sweety439
A new interest for "Divides Phi"

3*2^n+1 divides Phi(3*2^(n-1),2) for n = 5, 6, 8, 12, and the cofactor of them are all primes (this is the cofactor for n=12 (http://factordb.com/index.php?id=1100000000017315254)), are there any...
Forum: And now for something completely different 2020-06-03, 16:45
Replies: 47
Views: 20,888
Posted By sweety439
You can search 6*p^n+1 for all primes p, if p !=...

You can search 6*p^n+1 for all primes p, if p != 1 mod 7 and p != 34 mod 35, then there should be infinitely many primes of the form 6*p^n+1, but there is not always an easy prime, e.g. for p = 409,...
Forum: And now for something completely different 2020-05-30, 14:02
Replies: 47
Views: 20,888
Posted By sweety439
Are there any factorization status for Phi(n,2)...

Are there any factorization status for Phi(n,2) for n<=2^16

e.g. for n=3005, the factorization status for Phi(n,2) is


385549595471 (12 digits) * 13122095873179566186720736445243053751 (38...
Forum: And now for something completely different 2020-05-30, 13:58
Replies: 47
Views: 20,888
Posted By sweety439
The thread seems to be "factorization of Phi(n,2)...

The thread seems to be "factorization of Phi(n,2) for very large n", e.g. 2 ยท 10859^87905 + 1 is a factor of Phi(n,2) for n=10859^87905

Also, Phi(n,2) is completely factored for all n<=1206
Forum: And now for something completely different 2020-05-30, 13:57
Replies: 47
Views: 20,888
Posted By sweety439
All odd prime p divides Phi(n,2) for an integer...

All odd prime p divides Phi(n,2) for an integer n, in fact, n = ord_p(2)


p, n
3, 2
5, 4
7, 3
11, 10
13, 12
17, 8
Forum: And now for something completely different 2019-03-07, 18:57
Replies: 47
Views: 20,888
Posted By sweety439
2*3^152529+1 is 6*3^152528+1 (with k=6, b=3),...

2*3^152529+1 is 6*3^152528+1 (with k=6, b=3), 2*3^6225+1 is 18*3^6223+1 (with k=18, b=3), thus they also divide Phi(b^n,2).
Forum: And now for something completely different 2019-03-06, 16:04
Replies: 47
Views: 20,888
Posted By sweety439
This is not "Divides Phi(695^94625,2)", i.e....

This is not "Divides Phi(695^94625,2)", i.e. k*b^n+1 does not divide Phi(b^n,2), and not belong to this category.
Forum: And now for something completely different 2019-03-05, 20:48
Replies: 47
Views: 20,888
Posted By sweety439
Um... Currently I know what you means, not only...

Um... Currently I know what you means, not only 2*b^n+1 with b ends with E in duodecimal, but also some numbers k*b^n+1 with k>2 also divides Phi(b^n,2) ... but these b's seem to be all primes .........
Forum: And now for something completely different 2019-03-05, 20:46
Replies: 47
Views: 20,888
Posted By sweety439
Well, I am already searched 2*n^k+1 and 2*n^k-1...

Well, I am already searched 2*n^k+1 and 2*n^k-1 for all bases n up to duodecimal 1000 (decimal 1728), and the exponent k are also searched to k=duodecimal 1000 (decimal 1728), there are only few...
Forum: And now for something completely different 2019-03-05, 14:48
Replies: 47
Views: 20,888
Posted By sweety439
all n's and k's are written in duodecimal, only...

all n's and k's are written in duodecimal, only consider n ends with E (i.e. n = 11 mod 12), searched to duodecimal 1000 (decimal 1728).
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