|
%I #7 Jan 30 2017 21:46:57
%S 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,
%T 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,
%U 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
%N Irregular triangle read by rows. Row n gives the orders of the primes of row n of the irregular triangle A279399 modulo A033949(n).
%C The length of row n is given by A279400(n).
%C See the A279399 comments.
%C The entries in row n are proper divisors of phi(A033949(n)), where phi(n) = A000010(n).
%C This is because no A033949 number has a primitive root.
%F T(n, k) = order(A279399(n, k)) (mod A033949(n)), n >= 1, k = 1..A279400(n).
%e The irregular triangle T(n, k) begins (here N = A033949(n)):
%e n, N \ k 1 2 3 4 5 6 7 8 9 10 ...
%e 1, 8: 2 2 2
%e 2, 12: 2 2 2
%e 3, 15: 4 4 2 4
%e 4, 16: 4 4 2 4 4
%e 5, 20: 4 4 2 4 4 2
%e 6, 21: 6 6 6 2 6 6
%e 7, 24: 2 2 2 2 2 2 2
%e 8, 28: 6 6 6 2 6 6 6
%e 9, 30: 4 2 4 4 2 4 2
%e 10, 32: 8 8 4 8 8 2 8 4 8 2
%e 11, 33: 10 10 10 10 10 10 2 10 5
%e 12, 35: 12 12 3 4 12 6 12 2 6
%e 13, 36: 6 6 6 3 2 2 6 6 6
%e 14, 39: 12 4 12 12 6 12 6 6 4 12
%e 15, 40: 4 4 2 4 4 2 4 2 2 4
%e ...
%e The sequence of phi(N) begins: 4, 4, 8, 8, 8, 12, 8, 12, 8, 16, 20, 24, 12, 24, 16, ...
%e 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)).
%Y Cf. A000010, A033949, A279399, A279400.
%K nonn,tabf
%O 1,1
%A _Wolfdieter Lang_, Jan 30 2017
|
