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A269990
Primes fractility of n.
1
1, 2, 2, 3, 4, 4, 4, 5, 6, 6, 6, 7, 8, 8, 9, 9, 6, 10, 10, 10, 12, 12, 11, 12, 15, 7, 14, 15, 12, 17, 16, 13, 17, 15, 13, 18, 18, 16, 18, 23, 16, 20, 21, 14, 22, 23, 19, 23, 22, 20, 27, 26, 16, 24, 26, 21, 28, 27, 20, 29, 32, 18, 30, 33, 27, 35, 33, 27, 29
OFFSET
2,2
COMMENTS
In order to define (primes) 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 = (1/2, 1/3, 1/5, 1/7, 1/11, ... ) gives primes fractility.
binary fractility: A269570
factorial fractility: A269982
harmonic fractility: A270000
odds fractility: A269989
EXAMPLE
NI(1/11) = (5,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,...)
NI(2/11) = (3,1,1,2,1,1,3,1,3,1,1,1,2,2,1,1,1,2,1,1,1,4,4,1,2,10,1,1,1,1,1,1,1,1,1,2,1,1,8,1,1,1,1,1,2,1,2,1,1,1,1,1,2,1,4,1,1,3,1,8,1,1,1,1,1,1,...)
NI(3/11) = (2,1,2,1,1,1,1,1,4,1,1,1,1,1,3,2,9,1,1,1,2,1,2,2,2,2,1,1,1,4,1,1,1,1,1,1,1,1,1,1,11,1,2,4,1,4,3,1,1,1,1,1,1,1,1,1,1,3,1,1,1,1,1,1,1,2,...)
NI(4/11) = (1,8,2,4,1,2,1,1,1,1,1,2,1,3,2,1,5,1,1,8,1,4,1,1,1,1,1,1,1,2,3,3,1,3,1,1,1,1,1,5,2,3,2,4,2,1,8,2,1,1,2,2,106,2,3,1,1,1,1,1,1,2,2,6,1,,...)
NI(5/11) = (1,3,1,1,2,1,1,3,1,3,1,1,1,2,2,1,1,1,2,1,1,1,4,4,1,2,10,1,1,1,1,1,1,1,1,1,2,1,1,8,1,1,1,1,1,2,1,2,1,1,1,1,1,2,1,4,1,1,3,1,8,1,1,1,1,1,1,...)
NI(6/11) = (1,2,1,1,1,1,1,4,1,1,1,1,1,3,2,9,1,1,1,2,1,2,2,2,2,1,1,1,4,1,1,1,1,1,1,1,1,1,1,11,1,2,4,1,4,3,1,1,1,1,1,1,1,1,1,1,3,1,1,1,1,1,1,1,2,1,1,1,...);
NI(7/11) = (1,1,3,1,1,2,1,1,3,1,3,1,1,1,2,2,1,1,1,2,1,1,1,4,4,1,2,10,1,1,1,1,1,1,1,1,1,2,1,1,8,1,1,1,1,1,2,1,2,1,1,1,1,1,2,1,4,1,1,3,1,8,1,1,1,1,1,1,...);
NI(8/11) = (1,1,1,5,3,1,1,1,2,1,3,1,1,1,1,1,2,1,11,1,1,1,1,1,1,1,1,2,1,1,1,2,1,8,1,1,2,3,1,1,1,6,1,2,1,4,1,1,1,1,1,1,34,1,8,1,3,1,1,5,1,1,1,1,1,4,1,...);
NI(9/11) = (1,1,1,1,5,3,1,1,1,2,1,3,1,1,1,1,1,2,1,11,1,1,1,1,1,1,1,1,2,1,1,1,2,1,8,1,1,2,3,1,1,1,6,1,2,1,4,1,1,1,1,1,1,34,1,8,1,3,1,1,5,1,1,1,1,1,4,...);
NI(10/11) = (1,1,1,1,1,2,1,1,1,1,4,3,1,1,1,1,1,1,2,1,1,1,1,6,1,1,1,1,3,2,1,1,1,1,5,7,1,3,2,1,3,1,1,1,1,1,1,1,1,3,1,1,2,2,4,2,1,1,1,1,1,1,1,1,1,1,1,6,...); there are 6 equivalence classes: {1/11}, {2/11,5/11,7/11},{3,11,6/11},{4/11},{8/11,9/11},{10/11}, so that a(11) = 6.
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
STATUS
approved