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%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
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