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%I #21 Apr 11 2022 12:55:08
%S 1,0,1,1,0,1,1,0,0,1,2,0,1,0,1,3,1,2,0,0,1,5,3,0,1,1,0,1,1,1,3,3,0,0,
%T 0,1,5,0,3,3,2,2,1,0,1,3,2,6,5,3,1,1,0,0,1,4,1,7,5,1,3,0,0,1,0,1,0,9,
%U 8,4,4,3,3,3,2,0,0,1,5,1,4,6,1,3,5,3,2,1,1,0,1
%N Array read by ascending antidiagonals. A(n, k) = F(k, n) mod n for n >= 1 and k >= 0, where F(n, k) = A352744(n, k) are the Fibonacci numbers, A(0, k) = 1 for k >= 0.
%C This array aims the study of the divisibility properties of the Fibonacci numbers A352744. The identity F(n, k) = (-1)^k*F(1 - n, -k) from A352744 shows that negative indices do not add to the divisibility properties of F(n, k).
%C All rows A(n, .) are pure periodic sequences. The length of the periods is given by (1, A270313). For n > 0 the length of the period of row A(n, .) is <= n.
%C The period length is 1 for n in (1, A023172) and n for n in (1, A074215), as observed by _Robert Israel_ in A270313. In particular, if n is a power of 2 or a prime (A174090), then the period length is n.
%C The indices of the zero-free rows are in A353280. A zero-free row A(n, .) means that n will not divide F(k, n) whatever value k takes. For that it is sufficient to check that period(A(n, .)) is zero-free.
%C If period(A(n, .)) = [k | 0 <= k < n] we call n a 'Fibonacci friend'. In other words, in this case F(k, n) mod n = k for 0 <= k < n. A Fibonacci friend does not have to be prime (since 1 is a Fibonacci friend), but if it is prime then it is congruent to {1, 4} mod 5 (A045468), and all such primes are Fibonacci friends.
%C To say that n is a Fibonacci friend is equivalent to saying that A(n, n) = 0 and that n divides F(n, n). Fibonacci friends are the indices of the zeros in A002752.
%C Integers n > 0 that divide Sum{k=0..n-1} (F(k, n) mod n) are congruent to {0, 1, 3, 5} mod 6 (A301729).
%F A(n, 0) = A(n, n) = A002752(n).
%F Clearly 0 <= A(n, k) < n for all k and n > 0.
%e Array starts (periods are indicated with () ):
%e [n\k] 0 1 2 3 4 5 6 7 8 9 10 11 12
%e ----------------------------------------------------------
%e [ 0] (1), 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, ...
%e [ 1] (0), 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, ...
%e [ 2] (1, 0), 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, ...
%e [ 3] (1, 0, 2), 1, 0, 2, 1, 0, 2, 1, 0, 2, 1, ...
%e [ 4] (2, 1, 0, 3), 2, 1, 0, 3, 2, 1, 0, 3, 2, ...
%e [ 5] (3), 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, ...
%e [ 6] (5, 1, 3), 5, 1, 3, 5, 1, 3, 5, 1, 3, 5, ...
%e [ 7] (1, 0, 6, 5, 4, 3, 2), 1, 0, 6, 5, 4, 3, ...
%e [ 8] (5, 2, 7, 4, 1, 6, 3, 0), 5, 2, 7, 4, 1, ...
%e [ 9] (3, 1, 8, 6, 4, 2, 0, 7, 5), 3, 1, 8, 6, ...
%e [10] (4, 9), 4, 9, 4, 9, 4, 9, 4, 9, 4, 9, 4, ...
%e [11] (0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10), 0, 1, ...
%e [12] (5), 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, ...
%p f := n -> combinat:-fibonacci(n + 1):
%p F := proc(n, k) option remember; (n-1)*f(k-1) + f(k) end:
%p A := (n, k) -> ifelse(n = 0, 1, modp(F(k, n), n)):
%p for n from 0 to 12 do seq(A(n, k), k = 0..10) od;
%t F[n_, k_] := (n - 1)*Fibonacci[k] + Fibonacci[k + 1];
%t A[n_, k_] := If[n == 0, 1, Mod[F[k, n], n]];
%t Table[A[n, k], {n, 0, 12}, {k, 0, 10}] // TableForm
%o (SageMath)
%o def F(n, k): return (n - 1)*fibonacci(k) + fibonacci(k + 1)
%o def A(n,k): return mod(F(k, n), n)
%o for n in range(13): print([A(n,k) for k in range(13)])
%Y Cf. A352744, A002752, A045468, A074215, A088209, A174090, A212804, A270313, A301729, A353280.
%K nonn,tabl,easy
%O 0,11
%A _Peter Luschny_, Apr 08 2022
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