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 A094293 At the n-th step, append the number n and n copies of the list of all preceding terms, starting with an empty list. 2
 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, 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, 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 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,2 COMMENTS 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. 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] LINKS EXAMPLE 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. (Contribution by M. F. Hasler, start:) The sequence is created as follows: First step: append 1 to the empty list: result = [1]. 2nd step: append 2 and two copies of the previous result, to get [1,2,1,1]. 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]. PROG (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] CROSSREFS Cf. A001511. Sequence in context: A194086 A164659 A057898 * A036036 A228531 A244316 Adjacent sequences:  A094290 A094291 A094292 * A094294 A094295 A094296 KEYWORD nonn AUTHOR Amarnath Murthy, Apr 28 2004 EXTENSIONS Edited & corrected by M. F. Hasler, Apr 10 2009 STATUS approved

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Last modified July 19 12:35 EDT 2019. Contains 325159 sequences. (Running on oeis4.)