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 Showing results 1 to 10 of 10 Search took 0.01 seconds. Search: Posts Made By: sweety439
 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...
 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,...
 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...
 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
 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
 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).
 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.
 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 .........
 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...
 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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