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%I #12 Jan 25 2021 10:50:07
%S 1,2,1,1,3,1,2,1,1,1,2,1,1,1,2,1,1,4,1,2,1,1,3,1,2,1,1,1,2,1,1,1,2,1,
%T 1,1,2,1,1,3,1,2,1,1,1,2,1,1,1,2,1,1,1,2,1,1,3,1,2,1,1,1,2,1,1,1,2,1,
%U 1,1,2,1,1,3,1,2,1,1,1,2,1,1,1,2,1,1,5,1,2,1,1,3,1,2,1,1,1,2,1,1,1,2,1,1,4
%N At the n-th step, append the number n and n copies of the list of all preceding terms, starting with an empty list.
%C a(1)= 1, a(2) = 2. Let the index of first occurrence of n be k=A094294(n). Then from a(k+1) onwards the next n*(k-1) terms are the first (k-1) terms repeated n times, a(k+1) = a(1), a(k+2) = a(2) etc.
%C Let r be the index of the first occurrence of n-1 then the index of first occurrence of n is r+(n-1)*(r-1)+1 = (n+1)*r-n+2, cf. A094294. [Corrected by _M. F. Hasler_, Apr 09 2009]
%e a(5) = 3 and the first four terms are 1,2,1,1. hence the next 12 terms are 1,2,1,1,1,2,1,1,1,2,1,1 and a(18) = 4 (the first occurrence) and so on.
%e (Contribution by _M. F. Hasler_, start:) The sequence is created as follows:
%e First step: append 1 to the empty list: result = [1].
%e 2nd step: append 2 and two copies of the previous result, to get [1,2,1,1].
%e 3rd step: append 3 and three copies of [1,2,1,1], to get [1,2,1,1, 3, 1,2,1,1, 1,2,1,1, 1,2,1,1].
%o (PARI) A094293(n,a=[])={ for(k=1,1+n--, n<=(k+1)*#a & return(if(n>#a,a[1+(n-1)%#a],k)); a=concat(vector(k+2,j,if(j==2,[k],a))))} \\ _M. F. Hasler_, Apr 09 2009
%Y Cf. A001511.
%K nonn
%O 1,2
%A _Amarnath Murthy_, Apr 28 2004
%E Edited & corrected by _M. F. Hasler_, Apr 10 2009
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