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 A308622 a(n) = nearest integer to f(n), where f(0) = f(1) = 1, f(n) = (n-f(n-1))/f(n-2). 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 (list; graph; refs; listen; history; text; internal format)
 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:= 1: R:= 1: A:= 1: L:= 1: R:= 1: A:= 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 = f = 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

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Last modified August 4 10:11 EDT 2021. Contains 346447 sequences. (Running on oeis4.)