A100002 Start with a sequence of 1's, then replace every other 1 with a 2; then replace every third of the remaining 1's with a 3 and every third of the remaining 2's with a 3; then replace every fourth remaining 1, 2 or 3 with a 4; and so on. The limiting sequence is shown here.

1, 2, 1, 2, 3, 3, 1, 2, 4, 4, 3, 4, 1, 2, 5, 5, 3, 5, 1, 2, 4, 5, 3, 4, 6, 6, 1, 2, 6, 3, 7, 7, 6, 4, 7, 7, 5, 6, 1, 2, 5, 3, 8, 8, 7, 4, 8, 8, 1, 2, 6, 7, 3, 6, 5, 8, 4, 8, 5, 6, 9, 9, 1, 2, 9, 3, 10, 10, 9, 4, 10, 10, 7, 8, 9, 5, 7, 10, 1, 2, 9, 7, 3, 4, 9, 6, 11, 11, 10, 11

Offset: 1

Author

David A. Madore, Oct 25 2004

Keywords

• easy
• nice
• nonn

Comments

The position of the 1's is given by A000960. - T. D. Noe, Oct 26 2004

Examples

			Here are the first 6 stages in the construction:
  1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1...
  1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2...
  1 2 1 2 3 3 1 2 1 2 3 3 1 2 1 2 3 3 1 2 1 2 3 3 1 2 1 2 3 3...
  1 2 1 2 3 3 1 2 4 4 3 4 1 2 1 2 3 3 1 2 4 4 3 4 1 2 1 2 3 3...
  1 2 1 2 3 3 1 2 4 4 3 4 1 2 5 5 3 5 1 2 4 5 3 4 1 2 1 2 3 3...
  1 2 1 2 3 3 1 2 4 4 3 4 1 2 5 5 3 5 1 2 4 5 3 4 6 6 1 2 6 3...
  ...
		

Links

• T. D. Noe, Table of n, a(n) for n = 1..10000
• T. D. Noe, Plot of first 5000 terms
• A post on sci.math.research newsgroup.

Crossrefs

Cf. A100287 (first occurrence of n).

Programs

C

#define MAXVAL 2048 /* Large enough... */
unsigned int counts[MAXVAL][MAXVAL]; /* Initialized at all 0 */ unsigned int seq_value (void) /* Successive calls return values in the sequence, in order. */ { unsigned int value; unsigned int i; value = 1; for ( i=2; i= i ) { counts[i][value] = 0; value = i; } return value; }

Mathematica

nn=100; t=Table[1, {nn}]; done=False; k=1; While[ !done, k++; cnt=Table[0, {k-1}]; Do[If[t[[i]]T. D. Noe *)
a[n_] := Fold[Function[{b1, b2},Fold[Function[{a1, a2},ReplacePart[a1, Pick[Position[a1, a2], Take[Flatten[Array[{Array[0 &, b2 - 1], 1} &, Length[a1]]], Length[Position[a1, a2]]], 1] -> b2]], b1, Range[b2 - 1]]], Array[1 &, n], Range[2, 2 Sqrt[n/Pi] + 1]]; a[100] (* Birkas Gyorgy, Feb 06 2011 *)

Formula

a(1, j)=1 for all j>=1; a(n, j)=a(n-1, j) except when #{i<=j s.t. a(n-1, i)=a(n-1, j)} is multiple of n, in which case a(n, j)=n; a(j) is the limit of the (stationary) a(n, j) when n tends to infinity.

It appears that the maximal value among the first n terms grows like sqrt(4n/3).

Note that the first occurrence of n is bounded by A000960; that is, A100287(n) <= A000960(n), with equality only for n=1. - T. D. Noe, Nov 12 2004