%I #16 Dec 26 2022 15:48:47
%S 0,1,0,1,4,0,1,0,3,4,0,1,2,0,4,0,2,2,6,4,0,2,0,0,12,0,3,2,0,8,5,0,4,2,
%T 0,4,0,12,0,3,2,3,8,0,6,7,0,2,6,3,4,5,0,7,8,0,0,6,3,8,0,6,7,8,0,0,4,3,
%U 4,10,0,7,8,9,0,2,4,0,8,5,0,14,0,9,10,0
%N Variant of the inventory sequence: Record the number of terms whose value occurs once thus far in the sequence, then the number of terms whose value occurs twice thus far, and so on; a row ends when a 0 that would repeat infinitely is reached.
%C Note that we are counting terms with repetition. For example, to find a(5)=4, we are looking for the number of terms that appear twice. 0 and 1 each occur twice, which is 2+2=4 (not 1+1=2). This means that each column contains only multiples of the number of occurences it is counting.
%e First few rows of irregular triangle:
%e 0;
%e 1, 0;
%e 1, 4, 0;
%e 1, 0, 3, 4, 0;
%e 1, 2, 0, 4, 0;
%e 2, 2, 6, 4, 0;
%e 2, 0, 0, 12, 0;
%e 3, 2, 0, 8, 5, 0;
%e 4, 2, 0, 4, 0, 12, 0;
%e 3, 2, 3, 8, 0, 6, 7, 0;
%e 2, 6, 3, 4, 5, 0, 7, 8, 0;
%Y Cf. A342585, A350768.
%K nonn,tabf
%O 1,5
%A _Neal Gersh Tolunsky_, Dec 11 2022