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2020-02-11, 03:41   #1
sweety439

"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36

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Smallest k>1 such that Phi_n(k) is prime

I found the smallest k>=2 such that Phi_n(k) is (probable) prime (where Phi is the cyclotomic polynomial) for all 1<=n<=2500, see the text file (file format: "n k"). The k has been searched for special value of n's, see these OEIS sequences.

A066180 (for prime n)
A103795 (for n=2*p with p odd prime)
A056993 (for n=2^k with k>=1)
A153438 (for n=3^k with k>=2)
A246120 (for n=2*3^k with k>=1)
A246119 (for n=3*2^k with k>=1)
A298206 (for n=9*2^k with k>=1)
A246121 (for n=6^k with k>=1)
A206418 (for n=5^k with k>=2)
A205506 (for n=6*2^i*3^j with i,j>=0)
A181980 (for n=10*2^i*5^j with i,j>=0)

Let a(n) be the smallest k>=2 such that Phi_n(k) is prime, I found a(n) for all 1<=n<=2500, and according to these sequences, a(2^n) is known for all 0<=n<=21, a(3^n) is known for all 0<=n<=11, a(2*3^n) is known for all 0<=n<=10, etc. and the k's for some large n are a(2^21)=919444, a(3^12)=94259, a(2*3^11)=9087, etc. However, it seems that there is no project for finding a(n) for general n. (this a(n) is the OEIS sequence A085398)

Conjecture, for all n>=1, there exists k>=2 such that Phi_n(k) is prime. (if this conjecture is true, then there are infinitely many such k for all n>=1, besides, this conjecture is true if Bunyakovsky conjecture is true)

Can someone find the smallest k>=2 such that Phi_n(k) is (probable) prime (where Phi is the cyclotomic polynomial) for 2501<=n<=10000? Or larger n?
Attached Files
 least k such that phi(n,k) is prime.txt (22.0 KB, 204 views)

 2020-02-11, 16:31 #2 Uncwilly 6809 > 6502     """"""""""""""""""" Aug 2003 101×103 Posts 23×1,223 Posts Is this a puzzle or just an attempt to get someone else to do work for you?
 2020-02-11, 19:58 #3 VBCurtis     "Curtis" Feb 2005 Riverside, CA 28×19 Posts If it takes more than a CPU-day, Sweety calls it a conjecture and hopes someone else does the work.

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