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).

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

Author

Wolfdieter Lang, Jan 30 2017

Keywords

• nonn
• tabf

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.

Examples

			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

Cf. A000010, A033949, A279399, A279400.

Formula

T(n, k) = order(A279399(n, k)) (mod A033949(n)), n >= 1, k = 1..A279400(n).