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A366601
a(0) = 0. For a(n-1) a novel term a(n) = a(A025480(a(n-1))), and if a(n-1) has occurred k (>1) times, a(n) = k-1 (the number of repetitions of a(n-1) in all prior terms).
1
0, 0, 1, 0, 2, 0, 3, 0, 4, 1, 1, 2, 1, 3, 1, 4, 1, 5, 0, 5, 1, 6, 0, 6, 1, 7, 0, 7, 1, 8, 2, 2, 3, 2, 4, 2, 5, 2, 6, 2, 7, 2, 8, 1, 9, 1, 10, 0, 8, 2, 9, 1, 11, 0, 9, 2, 10, 1, 12, 3, 3, 4, 3, 5, 3, 6, 3, 7, 3, 8, 3, 9, 3, 10, 2, 11, 1, 13, 0, 10, 3, 11, 2, 12, 1, 14, 0
OFFSET
0,5
COMMENTS
An experimental sequence in which fractal sequence A025480 is used to select the response to a novel term, whereas a repeat term is followed by the number of repeats of that term. Every number appears infinitely many times, suggesting the possibility of there being multiple proper subsequences identical to the original, although no specific example (expressible by formula) has been found. The scatterplot shows fractal like features and the records subsequence is A001477. Shelly's formula (using A001511) for A025480 has been used in the computation of terms. a(n) <= n (equality when n = 0).
LINKS
Michael De Vlieger, Scatterplot of a(n), n = 1..2^16.
Michael De Vlieger, Log log scatterplot of a(n), n = 1..2^20.
EXAMPLE
a(0) = 0 is a novel term, therefore a(1) = a(A025480(0)) = a(0) = 0.
Since 0 has now been repeated once, a(2) = 1, another novel term, so a(3) = a(A025480(a(2))) = a(A025480(1)) = a(0) = 0.
Now 0 has been repeated 2 times, so a(4) = 2.
The data can be shown as an irregular table in which each row begins with a record term:
0,0;
1,0;
2,0;
3,0;
4,1,1,2,1,3,1,4,1;
5,0,5,1;
6,0,6,1;
7,0,7,1;
8,2,2,3,2,4,2,5,2,6,2,7,2,8,1;
MATHEMATICA
nn = 120; f[x_] := Floor[x/2^IntegerExponent[2 (x + 1), 2]]; a[0] = 0; c[_] := 0; Do[a[n] = If[c[#] == 0, c[#]++; a[f[#]], c[#]++; c[#] - 1] &[a[n - 1]], {n, nn}]; Array[a, nn, 0]
CROSSREFS
KEYWORD
nonn
AUTHOR
STATUS
approved