

A094293


At the nth 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
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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*(k1) terms are the first (k1) 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 n1 then the index of first occurrence of n is r+(n1)*(r1)+1 = (n+1)*rn+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+(n1)%#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



