20200211, 03:41  #1 
Nov 2016
2,267 Posts 
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? 
20200211, 16:31  #2 
6809 > 6502
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Aug 2003
101×103 Posts
2^{6}×3^{3}×5 Posts 
Is this a puzzle or just an attempt to get someone else to do work for you?

20200211, 19:58  #3 
"Curtis"
Feb 2005
Riverside, CA
1000011100101_{2} Posts 
If it takes more than a CPUday, Sweety calls it a conjecture and hopes someone else does the work.

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