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a(n) = (# 1's)-(# 2's) in first n terms of A030341.
1

%I #15 Jun 23 2024 14:31:05

%S 0,1,0,0,1,2,3,2,3,3,2,3,2,1,0,0,0,1,2,2,3,2,2,3,3,4,5,6,7,8,7,8,9,9,

%T 8,9,10,9,10,9,8,9,9,9,8,9,9,8,7,7,6,6,7,6,7,8,7,6,7,6,6,5,4,5,4,3,2,

%U 1,0,0,0,0,1,2,2,2,3,2,2,2,3,3,4,4,5,6,7,7,8,7

%N a(n) = (# 1's)-(# 2's) in first n terms of A030341.

%C For k >= 1, a(3^k * (k - 1/2) + 1/2) = 1 and a(3^k * (k - 1/2) + 3/2) = a(3^k * (k - 1/2) + 5/2) = 0. - _Robert Israel_, Jun 23 2024

%H Robert Israel, <a href="/A030350/b030350.txt">Table of n, a(n) for n = 1..10000</a>

%p R:= 0: t:= 0:

%p for i from 1 to 100 do

%p L:= convert(i,base,3);

%p for j from 1 to nops(L) do

%p if L[j] = 1 then t:= t+1 elif L[j] = 2 then t:= t-1 fi;

%p R:= R,t;

%p od

%p od:

%p R; # _Robert Israel_, Jun 23 2024

%Y Cf. A030341.

%K nonn,look

%O 1,6

%A _Clark Kimberling_

%E a(1) = 0 inserted by _Robert Israel_, Jun 23 2024