A269990 Primes fractility of n.

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

Author

Clark Kimberling and Peter J. C. Moses, Mar 12 2016

Keywords

• nonn
• easy

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

Examples

			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

Cf. A000040, A269570, A269982, A269989, A270000.