A127521 a(1)=1. n-th run, after the initial 1, of a(n) integers is the (1st, 2nd, 3rd,.., a(n)th) integer from among those positive integers which don't occur earlier in the sequence (including earlier in the same run).

1, 2, 3, 5, 4, 7, 9, 6, 10, 12, 14, 16, 8, 13, 17, 19, 11, 18, 21, 23, 25, 27, 29, 15, 22, 26, 30, 32, 34, 36, 38, 40, 20, 28, 33, 37, 41, 43, 24, 35, 42, 45, 47, 49, 51, 53, 55, 57, 31, 44, 48, 52, 56, 59, 61, 63, 65, 67, 69, 71, 39, 50, 58, 62, 66, 70, 73, 75, 77, 79, 81, 83

Offset

1,2

Comments

This sequence is a permutation of the positive integers.

Links

John Tyler Rascoe, Table of n, a(n) for n = 1..10000

Example

The beginning of the sequence grouped by runs:
1,(2),(3,5),(4,7,9),(6,10,12,14,16),(8,13,17,19),(11,18,21,23,25,27,29),..
The n-th run, after the initial 1, has a(n) terms.
The 5th run has a(5)=4 terms. The positive integers which don't occur before this run are 8,11,13,15,17,18,19,20,... The first term of the 5th run is the first term of these integers, which is 8.
Since 8 has just occurred in the sequence, the 2nd term in the fifth run is not 11, but is instead 13. Now the positive integers that have yet to occur are 11,15,17,18,19,...For the 3rd term of the 5th run we want the 3rd of these, which is 17. And finally we want the 4th positive integer which has yet to occur in the sequence, which is 19.

Mathematica

f[l_List] := Block[{r = {}, s, c, k, m}, m = Flatten[l][[Length[l]]]; Do[ s = Flatten[Append[l, r]]; c = i; k = 0; While[c > 0, k++; While[MemberQ[s, k], k++ ]; c--; ]; AppendTo[r, k]; , {i, m}]; Append[l, r]]; Flatten@Nest[f, {{1}}, 11] (* Ray Chandler, Jan 22 2007 *)

Crossrefs

Cf. A127522.

Sequence in context: A326777 A193973 A245057 * A350583 A102399 A118318

Adjacent sequences: A127518 A127519 A127520 * A127522 A127523 A127524

Keyword

easy,nonn

Author

Leroy Quet, Jan 17 2007

Extensions

Extended by Ray Chandler, Jan 22 2007

Status

approved