A308622 a(n) = nearest integer to f(n), where f(0) = f(1) = 1, f(n) = (n-f(n-1))/f(n-2).

1, 1, 1, 2, 2, 2, 2, 3, 2, 2, 4, 3, 2, 3, 5, 3, 3, 4, 5, 3, 3, 5, 5, 3, 4, 6, 5, 4, 5, 7, 5, 4, 6, 7, 4, 5, 7, 7, 4, 5, 8, 6, 5, 6, 8, 6, 5, 7, 9, 6, 5, 8, 9, 6, 6, 9, 8, 5, 6, 10, 8, 6, 7, 10, 8, 6, 8, 10, 7, 6, 9, 10, 7, 6, 10, 10, 7, 7, 11, 10, 7, 7, 11

Offset

0,4

Comments

Comments from Jon Maiga, Aug 25 2019: (Start)

The function seems to follow the square root of n.

The first differences center around the x-axis with a similar look.

Changing the initial values to for example f(0)=1 and f(1)=3 creates a very different graph.

See the image for fractional versions with f(0)=f(1)=1 and f(0)=1, f(1)=3.

(End)

Links

Robert Israel, Table of n, a(n) for n = 0..10000

J. Maiga, A few good looking recursive functions, 2019.

Jon Maiga, Plots of f(n) for n up to 1000

Maple

L[0]:= 1: R[0]:= 1: A[0]:= 1:
L[1]:= 1: R[1]:= 1: A[1]:= 1:
for n from 2 to 100 do
  x:= (n - R[n-1])/R[n-2];
  y:= (n - L[n-1])/L[n-2];
  L[n]:= 10^(-100)*floor(x*10^100);
  R[n]:= 10^(-100)*ceil(y*10^100);
  if round(L[n]) <> round(R[n]) then printf("Oops: %d \n", n); break fi;
  A[n]:= round(L[n])
od:
seq(A[n], n=0..100); # Robert Israel, Aug 24 2019

Mathematica

f[0] = f[1] = 1;
f[n_] := f[n] = N[(n - f[n - 1])/f[n - 2]]
Round[Array[f, 83, 0]]

Crossrefs

Sequence in context: A237259 A235614 A127417 * A198897 A201375 A309865

Adjacent sequences: A308619 A308620 A308621 * A308623 A308624 A308625

Keyword

nonn,look

Author

Jon Maiga, Jun 11 2019

Status

approved