A290601 - OEIS

%I #20 Sep 05 2017 08:05:14

%S 2,1,1,1,3,2,3,2,2,1,1,1,1,1,2,1,1,2,1,1,2,1,1,1,1,1,1,1,1,1,1,1,1,1,

%T 4,2,4,2,4,2,4,1,2,1,2,1,1,1,2,1,2,1,2,1,1,2,1,2,1,2,1,1,2,1,1,1,1,1,

%U 1,1,1,1,1,1,1,1,1,1,1,1,1,1,3,1,2,3,1,1,3,2,3,1,1,3,2,1,3,2,2,2,2

%N Irregular triangle read by rows: T(n, k) gives the least positive integer imin such that the sequence {A290600(n, k)^i}_{i >= imin} (mod A002808(n)) is periodic.

%C The length of row n is A290599(n).

%C The corresponding period lengths are given in A290602(n, k).

%C Conjecture:There exists always an imin >= 1 such that the power sequence A290600(n, k)^i (mod A002808(n)) is periodic for i >= imin. Otherwise T(n, k) is undefined, and one could use T(n, k) = -1. See below for a proof.

%C This entry resulted from finding the correct initial exponent i for the power sequence considered in triangle A057593 for composite n and gcd(n, k) not equal to 1.

%C It is clear that the sequence {A290600(n, k)^i}_{i >= 1} is never congruent to 1 (mod A002808(n)) for k = 1..A290599(n). Proof by contradiction. Therefore one can replace 'least positive integer' with 'least nonnegative integer' in the name of this sequence.

%C The values of these powers of A290600(n, k) modulo A290599(n) are 0 or any of the row entries of A290600(n, k).

%C To prove periodicity of {A290600(n, k)^i (mod A290599(n))} of the type (imin,P) (starting at i = imin with period length P) one has to solve, with given K = A290600(n, k) and N = A002808(n), the congruence  K^imin * (K^P - 1) == 0 (mod N).

%C The above conjecture is trivially true because {K^i}_{i >= 0} (mod N) becomes (or is) always periodic (K and N positive integer). This follows from the fact that at most N values are possible, namely {0, 1, ..., N-1} and for i >= N one of these values has to appear for the second time, leading to periodicity. Thus the above sequence where gcd(N, K) is not 1 is also defined. - _Wolfdieter Lang_, Sep 05 2017

%e The irregular triangle T(n, k) begins (N(n) = A002808(n)):

%e n   N(n) \ k  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 ...

%e 1   4         2

%e 2   6         1  1  1

%e 3   8         3  2  3

%e 4   9         2  2

%e 5   10        1  1  1  1  1

%e 6   12        2  1  1  2  1  1  2

%e 7   14        1  1  1  1  1  1  1

%e 8   15        1  1  1  1  1  1

%e 9   16        4  2  4  2  4  2  4

%e 10  18        1  2  1  2  1  1  1  2  1  2  1

%e 11  20        2  1  1  2  1  2  1  2  1  1  2

%e 12  21        1  1  1  1  1  1  1  1

%e 13  22        1  1  1  1  1  1  1  1  1  1  1

%e 14  24        3  1  2  3  1  1  3  2  3  1  1  3  2  1  3

%e 15  25        2  2  2  2

%e ...

%e T(4, 1) = 2 because A290600(4, 1) = 3, A002808(4) = 9 and {3^i}_{i>=2} (mod 9) = {repeat(0)}, but 3^0 == 1 (mod 9) and  3^1 == 3 (mod 9).

%e T(4, 2) = 2 because A290600(4, 2) = 6 and  {6^i}_{i>=2} (mod 9) == {repeat(0)}, and 6^0 (mod 9) = 1, 6^1 (mod 9) = 6.

%e Example for a proof of periodicity of type (1,6) for K = A290600(10, 1) = 2, N = A002808(10) = 18: 2^imin*(2^P - 1) == 0 (mod 18). gcd(2^imin, 18) = 2, and with imin = 1 one has  2^P == 1 (mod 9). Since gcd(2, 9) = 1, a solution exists (Euler): P = phi(9) = 6.

%Y Cf. A002808, A057593, A290599, A290600, A290602.

%K nonn,tabf

%O 1,1

%A _Wolfdieter Lang_, Aug 30 2017