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

1,1

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.

Links

Table of n, a(n) for n=1..105.

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

Cf. A000010, A033949, A279399, A279400.

Sequence in context: A036263 A307378 A339049 * A168514 A326353 A060447

Adjacent sequences: A279398 A279399 A279400 * A279402 A279403 A279404

Keyword

nonn,tabf

Author

Wolfdieter Lang, Jan 30 2017

Status

approved