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

1,2

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

Table of n, a(n) for n=1..105.

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: A342723 A164659 A057898 * A338156 A335122 A334442

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