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 A290601 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. 3
 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, 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, 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 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 COMMENTS The length of row n is A290599(n). The corresponding period lengths are given in A290602(n, k). 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. 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. 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. The values of these powers of A290600(n, k) modulo A290599(n) are 0 or any of the row entries of A290600(n, k). 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). 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 LINKS EXAMPLE The irregular triangle T(n, k) begins (N(n) = A002808(n)): n   N(n) \ k  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 ... 1   4         2 2   6         1  1  1 3   8         3  2  3 4   9         2  2 5   10        1  1  1  1  1 6   12        2  1  1  2  1  1  2 7   14        1  1  1  1  1  1  1 8   15        1  1  1  1  1  1 9   16        4  2  4  2  4  2  4 10  18        1  2  1  2  1  1  1  2  1  2  1 11  20        2  1  1  2  1  2  1  2  1  1  2 12  21        1  1  1  1  1  1  1  1 13  22        1  1  1  1  1  1  1  1  1  1  1 14  24        3  1  2  3  1  1  3  2  3  1  1  3  2  1  3 15  25        2  2  2  2 ... 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). 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. 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. CROSSREFS Cf. A002808, A057593, A290599, A290600, A290602. Sequence in context: A048821 A120221 A300655 * A344446 A094899 A057774 Adjacent sequences:  A290598 A290599 A290600 * A290602 A290603 A290604 KEYWORD nonn,tabf AUTHOR Wolfdieter Lang, Aug 30 2017 STATUS approved

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Last modified July 28 13:15 EDT 2021. Contains 346332 sequences. (Running on oeis4.)