%I
%S 1,1,1,1,1,1,1,1,1,1,1,1,1,0,1,1,1,1,1,1,1,1,2,2,1,1,1,1,1,1,
%T 1,1,1,1,1,1,2,2,2,1,3,3,3,3,1,4,4,1,5,5,5,5,5,5,5,5,5,4,4,4,4,4,
%U 4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,2,2,2,2,2,2
%N a(n) is such that A100827(n) = A082917(n  a(n))  1, or 1 if there is no corresponding term.
%C Most of the known terms of A100827 are 1 less than a term in A082917, and conversely. This sequences looks at the location in the sequence of the corresponding terms. Negative terms do not occur among the known terms of this sequence. When a(n+1) is different from a(n) (and both are nonnegative), there are a(n+1)a(n) terms in one of the sequences that aren't in the other. With some irregularities, this sequence generally gradually increases at first, reaching a(49)=5. Then there are 9 a(n)=5, followed by 20 a(n)=4, followed by 30 a(n)=3, and then a(n)=2 for n=108 to 229. What is the behavior of the rest of the sequence? Does it stay at a(n)=2?
%H Jud McCranie, <a href="/A287056/b287056.txt">Table of n, a(n) for n = 1..229</a>
%e Examples: A100827(6)=47, A082917(5)=47+1, so a(6) = 65 = 1. A100827(23)=779, A082917(21)=779+1, so a(23) = 2321 = 2.
%Y Cf. A082917, A100827.
%K sign
%O 1,23
%A _Jud McCranie_, May 18 2017
