cp's OEIS Frontend

A269989 Odds fractility of n.

%I A269989 #12 Feb 09 2018 11:43:28
%S A269989 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,
%T A269989 13,11,14,15,10,16,15,11,15,18,14,18,18,10,23,17,14,18,15,16,25,20,10,
%U A269989 20,24,15,25,23,16,27,27,14,23,24,21,26,27,25,26,29
%N A269989 Odds fractility of n.
%C A269989 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.
%C A269989 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.
%C A269989 binary fractility:  A269570
%C A269989 factorial fractility:  A269982
%C A269989 harmonic fractility:  A270000
%C A269989 primes fractility:  A269990
%e A269989 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,...)
%e A269989 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,...)
%e A269989 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,...)
%e A269989 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,...)
%e A269989 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,...)
%e A269989 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,...);
%e A269989 there are 4 equivalence classes:  {1/7,3/7},{2/7},{4,5},{6},so that a(7) = 4.
%Y A269989 Cf. A005408, A269570, A269982, A270000, A269990.
%K A269989 nonn,easy
%O A269989 2,2
%A A269989 _Clark Kimberling_ and _Peter J. C. Moses_, Mar 12 2016