A145384 - OEIS

%I #4 Nov 11 2010 07:34:06

%S 0,1,2,3,2,3,2,2,6,6,0,3,1,3,2,3,2,4,4,3,0,3,5,0,2,2,3,2,1,2,3,1,2,3,

%T 5,1,1,4,2,0,1,3,1,3,3,2,2,2,4,2,1,2,4,2,0,1,2,3,1,1,1,3,0,3,1,0,3,1,

%U 1,4,2,2,1,3,3,1,2,0,3,2,5,1,1,3,6,2,4,1,0,5,2,2,2,2,3,2,3,3,0,1

%N The number of terms of A050791 bracketed by successive terms of A141326

%C A141326 is a simply generated subsequence of A050791 and by observation it forms a natural measure of the parent sequence. The first several hundred terms of the parent sequence not belonging to A141326 are bracketed into groups with a small integral number of terms ( including 0 ) by the successive terms of the subsequence, A141326.

%C a(107),a(108) are the first occurrence of 2 consecutive 0's and a(119),a(120),a(121) are the first occurrence of 3 consecutive 0's. This leads to the following conjecture:

%C <a(n)> -> 0 as n ->inf

%C where <a(n)> = ( sum m=1,n of a(m) )/n

%H Lewis Mammel, <a href="/A145384/b145384.txt">Table of n, a(n) for n = 1..122</a>

%F a(1) = A145383(1) - 1

%F a(n) = A145383(n) - A145383(n-1) - 1 ; n>1

%e 0 = number of terms of A050791 preceding the first term of A141326

%e 1 = number of terms of A050791 between the first and 2nd terms of A141326

%e 2 = number of terms of A050791 between the 2nd and 3rd terms of A141326

%Y Cf. A145383, A141326, A050791

%K nonn

%O 1,3

%A Lewis Mammel (l_mammel(AT)att.net), Oct 10 2008