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.

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

Offset: 1

Author

Amarnath Murthy, Apr 28 2004

Keywords

• nonn

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]

Examples

			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].
		

Crossrefs

Cf. A001511.

Programs

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

Extensions

Edited & corrected by M. F. Hasler, Apr 10 2009