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A269989
Odds fractility of n.
1
1, 2, 2, 3, 3, 4, 3, 3, 6, 6, 5, 6, 7, 6, 8, 7, 5, 8, 8, 8, 10, 10, 10, 10, 11, 5, 13, 11, 9, 15, 13, 11, 14, 15, 10, 16, 15, 11, 15, 18, 14, 18, 18, 10, 23, 17, 14, 18, 15, 16, 25, 20, 10, 20, 24, 15, 25, 23, 16, 27, 27, 14, 23, 24, 21, 26, 27, 25, 26, 29
OFFSET
2,2
COMMENTS
In order to define (odds) fractility of an integer n > 1, we first define nested interval sequences. Suppose that r = (r(n)) is a sequence satisfying (i) 1 = r(1) > r(2) > r(3) > ... and (ii) r(n) -> 0. For x in (0,1], let n(1) be the index n such that r(n+1) , x <= r(n), and let L(1) = r(n(1))-r(n(1)+1). Let n(2) be the index n such that r(n(1)+1) < x <= r(n(1)+1) + L(1)r(n), and let L(2) = (r(n(2))-r(r(n)+1))L(1). Continue inductively to obtain the sequence (n(1), n(2), n(3), ... ), the r-nested interval sequence of x.
For fixed r, call x and y equivalent if NI(x) and NI(y) are eventually identical. For n > 1, the r-fractility of n is the number of equivalence classes of sequences NI(m/n) for 0 < m < n. Taking r = (1/1, 1/3, 1/5, 1/7, 1/9, ... ) gives odds fractilily.
binary fractility: A269570
factorial fractility: A269982
harmonic fractility: A270000
primes fractility: A269990
EXAMPLE
NI(1/7) = (4,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,...)
NI(2/7) = (2,1,1,3,1,1,1,1,1,1,2,1,1,2,2,1,2,1,1,1,1,1,1,2,1,1,1,1,1,1,1,1,2,1,2,1,1,1,19,1,30,1,2,2,1,10,1,1,3,1,...)
NI(3/7) = (1,4,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,...)
NI(4/7) = (1,1,14,1,2,3,1,2,1,1,1,1,1,2,1,3,2,1,6,1,1,11,1,1,1,1,1,1,1,12,2,1,1,1,2,3,1,1,1,1,1,6,1,1,1,1,2,3,1,7,...)
NI(5/7) = (1,1,1,14,1,2,3,1,2,1,1,1,1,1,2,1,3,2,1,6,1,1,11,1,1,1,1,1,1,1,12,2,1,1,1,2,3,1,1,1,1,1,6,1,1,1,1,2,3,1,...)
NI(6/7) = (1,1,1,1,2,1,1,11,1,2,1,1,1,1,1,1,1,6,1,7,1,1,1,1,1,1,1,2,1,1,6,1,1,1,194,1,2,7,6,2,1,1,1,1,1,1,3,1,2,1,...);
there are 4 equivalence classes: {1/7,3/7},{2/7},{4,5},{6},so that a(7) = 4.
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
STATUS
approved