OFFSET
1,3
COMMENTS
T(n,k) is either 1 or a prime.
Since p is a prime factor of Phi_k(n) => either p == 1 (mod k) or p is the largest prime factor of k. As a result, T(n,k) = 1 if and only if all prime factors of Phi_k(n) are congruent to 1 modulo k.
LINKS
Jianing Song, Table of n, a(n) for n = 1..5050 (the first 100 antidiagonals)
Jianing Song, Proof for the first formula
FORMULA
For k > 1, let p be the largest prime factor of k, then T(n,k) = p if p does not divide n and k = p^e*ord(p,n) for some e > 0, where ord(p,n) is the multiplicative order of n modulo p. See my link above for the proof.
T(n,k) = T(n,k*p^a) for all a, where p is the largest prime factor of k.
T(n,k) = Phi_k(n)/A323748(n,k) for n >= 2, k != 2.
For prime p, T(n,p^e) = p if n == 1 (mod p), 1 otherwise.
For odd prime p, T(n,2*p^e) = p if n == -1 (mod p), 1 otherwise.
EXAMPLE
Table begins
n\k | 1 2 3 4 5 6 7 8 9 10 11 12
------------------------------------------
1 | 1 2 3 2 5 1 7 2 3 1 11 1
2 | 1 1 1 1 1 3 1 1 1 1 1 1
3 | 1 2 1 2 1 1 1 2 1 1 1 1
4 | 1 1 3 1 1 1 1 1 3 5 1 1
5 | 1 2 1 2 1 3 1 2 1 1 1 1
6 | 1 1 1 1 5 1 1 1 1 1 1 1
7 | 1 2 3 2 1 1 1 2 3 1 1 1
8 | 1 1 1 1 1 3 7 1 1 1 1 1
9 | 1 2 1 2 1 1 1 2 1 5 1 1
10 | 1 1 3 1 1 1 1 1 3 1 1 1
11 | 1 2 1 2 5 3 1 2 1 1 1 1
12 | 1 1 1 1 1 1 1 1 1 1 11 1
MATHEMATICA
A342255[n_, k_] := GCD[k, Cyclotomic[k, n]];
Table[A342255[n-k+1, k], {n, 15}, {k, n}] (* Paolo Xausa, Feb 09 2024 *)
PROG
(PARI) T(n, k) = gcd(k, polcyclo(k, n))
CROSSREFS
KEYWORD
AUTHOR
Jianing Song, Mar 07 2021
STATUS
approved