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A352747
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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.
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4
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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, 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, 8, 4, 4, 3, 3, 3, 2, 0, 0, 1, 5, 1, 4, 6, 1, 3, 5, 3, 2, 1, 1, 0, 1
(list;
table;
graph;
refs;
listen;
history;
text;
internal format)
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OFFSET
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0,11
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COMMENTS
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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).
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.
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.
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.
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.
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).
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LINKS
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FORMULA
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Clearly 0 <= A(n, k) < n for all k and n > 0.
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EXAMPLE
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Array starts (periods are indicated with () ):
[n\k] 0 1 2 3 4 5 6 7 8 9 10 11 12
----------------------------------------------------------
[ 0] (1), 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, ...
[ 1] (0), 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, ...
[ 2] (1, 0), 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, ...
[ 3] (1, 0, 2), 1, 0, 2, 1, 0, 2, 1, 0, 2, 1, ...
[ 4] (2, 1, 0, 3), 2, 1, 0, 3, 2, 1, 0, 3, 2, ...
[ 5] (3), 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, ...
[ 6] (5, 1, 3), 5, 1, 3, 5, 1, 3, 5, 1, 3, 5, ...
[ 7] (1, 0, 6, 5, 4, 3, 2), 1, 0, 6, 5, 4, 3, ...
[ 8] (5, 2, 7, 4, 1, 6, 3, 0), 5, 2, 7, 4, 1, ...
[ 9] (3, 1, 8, 6, 4, 2, 0, 7, 5), 3, 1, 8, 6, ...
[10] (4, 9), 4, 9, 4, 9, 4, 9, 4, 9, 4, 9, 4, ...
[11] (0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10), 0, 1, ...
[12] (5), 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, ...
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MAPLE
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f := n -> combinat:-fibonacci(n + 1):
F := proc(n, k) option remember; (n-1)*f(k-1) + f(k) end:
A := (n, k) -> ifelse(n = 0, 1, modp(F(k, n), n)):
for n from 0 to 12 do seq(A(n, k), k = 0..10) od;
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MATHEMATICA
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F[n_, k_] := (n - 1)*Fibonacci[k] + Fibonacci[k + 1];
A[n_, k_] := If[n == 0, 1, Mod[F[k, n], n]];
Table[A[n, k], {n, 0, 12}, {k, 0, 10}] // TableForm
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PROG
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(SageMath)
def F(n, k): return (n - 1)*fibonacci(k) + fibonacci(k + 1)
def A(n, k): return mod(F(k, n), n)
for n in range(13): print([A(n, k) for k in range(13)])
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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