|
%I #42 Nov 22 2018 20:20:57
%S 0,0,1,0,0,1,1,1,2,0,0,1,0,0,1,1,1,2,1,1,2,1,1,3,2,2,3,0,0,1,0,0,1,1,
%T 1,2,0,0,1,0,0,1,1,1,2,1,1,2,1,1,3,2,2,3,1,1,2,1,1,3,2,2,3,1,1,2,1,1,
%U 4,3,3,5,2,2,4,2,2,5,3,3,4,0,0,1,0,0,1
%N a(n) is the number of arithmetic triples n<p<q (three numbers in arithmetic progression) such that p and q contain no 2's in their ternary representation.
%C This is a recursive sequence that gives the number of times n is rejected from A005836, if n is the smallest member of an arithmetic triple whose final two terms are contained in A005836.
%C This is similar to both A002487, which has a similar recurrence relation and counts hyperbinary representations of n, and A000119, which counts representations of n as a sum of distinct Fibonacci numbers.
%C For n<k (choose the smallest k), a(n)=0, a(k)=0, a(n)=A262097(k), a(n+1)=A262097(k-1), a(n+2)=A262097(k-2)... a(k)=A262097(n).
%C Indices of maxima between a(n) and a(k) appear to converge to (3/4)(k-n) and (11/12)(k-n).
%H Max Barrentine, <a href="/A273513/b273513.txt">Table of n, a(n) for n = 0..19683</a> (terms 1 through 10000 from Robert Israel)
%H Robert Israel, <a href="/A273513/a273513.png">Plot of first 10^5 terms</a>
%F a(0)=0, a(n)=a(3n)=a(3n+1);
%F if a(n+1)=0, a(3n+2)=1+a(n), otherwise a(3n+2)=a(n)+a(n+1).
%p f:= proc(n) option remember; local m;
%p m:= floor(n/3);
%p if n mod 3 <> 2 then procname(m)
%p elif procname(m+1)=0 then 1 + procname(m)
%p else procname(m) + procname(m+1)
%p fi
%p end proc:
%p f(0):= 0:
%p map(f, [$0..100]); # _Robert Israel_, Jun 16 2016
%Y Cf. A000119, A002487, A005836, A262097, A273514.
%K nonn,base,look
%O 0,9
%A _Max Barrentine_, May 23 2016
|
