A352747 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.
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
Offset: 0
Examples
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, ...
Crossrefs
Programs
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Maple
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
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 -
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)])
Formula
A(n, 0) = A(n, n) = A002752(n).
Clearly 0 <= A(n, k) < n for all k and n > 0.
Comments