

A308622


a(n) = nearest integer to f(n), where f(0) = f(1) = 1, f(n) = (nf(n1))/f(n2).


3



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
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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 xaxis 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[n1])/R[n2];
y:= (n  L[n1])/L[n2];
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



