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A351889 Table T(n,k) read by downward antidiagonals: period of n-step Fibonacci numbers mod k, n >= 1, k >= 1.

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%I A351889 #42 Feb 16 2025 08:34:03
%S A351889 1,1,1,1,3,1,1,8,4,1,1,6,13,5,1,1,20,8,26,6,1,1,24,31,10,104,7,1,1,16,
%T A351889 52,312,12,728,8,1,1,12,48,130,781,14,364,9,1,1,24,16,342,312,208,16,
%U A351889 80,10,1,1,60,39,20,2801,728,9372,18,91,11,1,1,10,124,78,24,342,728,195312
%N A351889 Table T(n,k) read by downward antidiagonals: period of n-step Fibonacci numbers mod k, n >= 1, k >= 1.
%H A351889 Chai Wah Wu, <a href="/A351889/b351889.txt">Table of n, a(n) for n = 1..226</a>
%H A351889 M. E. Waddill, <a href="https://web.archive.org/web/2024*/https://www.fq.math.ca/Scanned/16-4/waddill.pdf">Some properties of a generalized Fibonacci sequence modulo m</a>, The Fibonacci Quarterly, vol. 16, no. 4, pp. 344-353 (1978).
%H A351889 Marcellus E. Waddill, <a href="https://web.archive.org/web/2024*/https://www.fq.math.ca/Scanned/30-3/waddill.pdf">Some Properties of the Tetranacci Sequence Modulo m</a>, The Fibonacci Quarterly, vol. 30, no. 3, 232-238 (1992).
%H A351889 D. D. Wall, <a href="https://doi.org/10.1080/00029890.1960.11989541">Fibonacci Series Modulo m</a>, The American Mathematical Monthly, vol. 67, no. 6, pp. 525-532, (1960).
%H A351889 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/Fibonaccin-StepNumber.html">Fibonacci n-Step Number</a>
%H A351889 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/PisanoPeriod.html">Pisano Period</a>
%F A351889 T(1,k) = T(n,1) = 1.
%F A351889 T(2,k) = A001175(k).
%F A351889 T(3,k) = A046738(k).
%F A351889 T(4,k) = A106295(k) for k not a multiple of 563.
%F A351889 T(5,k) = A106303(k).
%F A351889 T(n,2) = n + 1 for n > 1.
%F A351889 T(n,3) = A337212(n).
%F A351889 T(n,n) = A351657(n).
%F A351889 T(n,p_1^e_1*...*p_m^e_m) = lcm(T(n,p_1^e_1),...,T(n,p_m^e_m)) for p_i distinct primes.
%F A351889 Conjecture 1: T(n,2^m) = (n+1)*2^(m-1) for n > 1.
%F A351889 Conjecture 2: For p prime, if T(n,p) != T(n,p^2) then T(n,p^k) = p^(k-1)T(n,p).
%F A351889 Conjecture 2 is true for n = 2, n = 3 and n = 4 (see [Wall, 1960], [Waddill, 1978] and [Waddill, 1992] resp.). It is easy to show that T(n,4) != n+1 for all n, and thus Conjecture 2 implies Conjecture 1.
%F A351889 Conjecture 3: T(p^m,p^k) = (p^(pm)-1)*p^(k-1)/(p^m-1) for p prime and k, m > 0.
%e A351889 Table T(n,k) starts:
%e A351889   1   1     1   1       1      1          1   1      1        1
%e A351889   1   3     8   6      20     24         16  12     24       60
%e A351889   1   4    13   8      31     52         48  16     39      124
%e A351889   1   5    26  10     312    130        342  20     78     1560
%e A351889   1   6   104  12     781    312       2801  24    312     4686
%e A351889   1   7   728  14     208    728        342  28   2184     1456
%e A351889   1   8   364  16    9372    728     137257  32   1092    18744
%e A351889   1   9    80  18  195312    720      13680  36    240   585936
%e A351889   1  10    91  20  488281    910    5764800  40    273  4882810
%e A351889   1  11  8744  22   19344  96184      19152  44  26232   212784
%e A351889   1  12  3851  24  406224  46212  109531200  48  11553   406224
%o A351889 (Python)
%o A351889 from functools import lru_cache
%o A351889 from math import lcm
%o A351889 from itertools import count
%o A351889 from sympy import factorint
%o A351889 @lru_cache(maxsize=None)
%o A351889 def A351889_T(n,k): # computes the period of the n-step Fibonacci sequence mod k
%o A351889     if len(fs := factorint(k)) <= 1:
%o A351889         a = b = (0,)*(n-1)+(1 % k,)
%o A351889         s = 1 % k
%o A351889         for m in count(1):
%o A351889             b, s = b[1:] + (s,), (s + s - b[0]) % k
%o A351889             if a == b:
%o A351889                 return m
%o A351889     else:
%o A351889         return lcm(*(A351889_T(n,p**e) for p, e in fs.items()))
%Y A351889 Cf. A000045, A001175, A046738, A106295, A106303, A337212, A351657 (diagonal).
%K A351889 nonn,tabl
%O A351889 1,5
%A A351889 _Chai Wah Wu_, Feb 24 2022