login
A279401
Irregular triangle read by rows. Row n gives the orders of the primes of row n of the irregular triangle A279399 modulo A033949(n).
1
2, 2, 2, 2, 2, 2, 4, 4, 2, 4, 4, 4, 2, 4, 4, 4, 4, 2, 4, 4, 2, 6, 6, 6, 2, 6, 6, 2, 2, 2, 2, 2, 2, 2, 6, 6, 6, 2, 6, 6, 6, 4, 2, 4, 4, 2, 4, 2, 8, 8, 4, 8, 8, 2, 8, 4, 8, 2, 10, 10, 10, 10, 10, 10, 2, 10, 5, 12, 12, 3, 4, 12, 6, 12, 2, 6, 6, 6, 6, 3, 2, 2, 6, 6, 6, 12, 4, 12, 12, 6, 12, 6, 6, 4, 12, 4, 4, 2, 4, 4, 2, 4, 2, 2, 4
OFFSET
1,1
COMMENTS
The length of row n is given by A279400(n).
See the A279399 comments.
The entries in row n are proper divisors of phi(A033949(n)), where phi(n) = A000010(n).
This is because no A033949 number has a primitive root.
FORMULA
T(n, k) = order(A279399(n, k)) (mod A033949(n)), n >= 1, k = 1..A279400(n).
EXAMPLE
The irregular triangle T(n, k) begins (here N = A033949(n)):
n, N \ k 1 2 3 4 5 6 7 8 9 10 ...
1, 8: 2 2 2
2, 12: 2 2 2
3, 15: 4 4 2 4
4, 16: 4 4 2 4 4
5, 20: 4 4 2 4 4 2
6, 21: 6 6 6 2 6 6
7, 24: 2 2 2 2 2 2 2
8, 28: 6 6 6 2 6 6 6
9, 30: 4 2 4 4 2 4 2
10, 32: 8 8 4 8 8 2 8 4 8 2
11, 33: 10 10 10 10 10 10 2 10 5
12, 35: 12 12 3 4 12 6 12 2 6
13, 36: 6 6 6 3 2 2 6 6 6
14, 39: 12 4 12 12 6 12 6 6 4 12
15, 40: 4 4 2 4 4 2 4 2 2 4
...
The sequence of phi(N) begins: 4, 4, 8, 8, 8, 12, 8, 12, 8, 16, 20, 24, 12, 24, 16, ...
n = 2, N = 12: 5^2 == 7^2 == 11^2 == 1 (mod 12), therefore 2 is the least positive power k for each of the three primes p of row 2 of A279399 which satisfies p^k == 1 (mod A033949(2)).
CROSSREFS
KEYWORD
nonn,tabf
AUTHOR
Wolfdieter Lang, Jan 30 2017
STATUS
approved