A260139 For any term a(k), there are exactly a(k) terms strictly smaller than 3*a(k); this is the lexicographically first increasing sequence of nonnegative integers with this property.

0, 2, 6, 7, 8, 9, 18, 21, 24, 27, 28, 29, 30, 31, 32, 33, 34, 35, 54, 55, 56, 63, 64, 65, 72, 73, 74, 81, 84, 87, 90, 93, 96, 99, 102, 105, 106, 107, 108, 109, 110, 111, 112, 113, 114, 115, 116, 117, 118, 119, 120, 121, 122, 123, 162, 165, 168, 169, 170, 171, 172

Offset: 0

Author

M. F. Hasler, Jul 16 2015

Keywords

• nonn
• easy
• look

Comments

Suggested by Eric Angelini, cf. link to SeqFan post.

This sequence has a nice self-similar graph.

Examples

			The first term says that there are a(0) = 0 terms < 0.
Then it is not possible to go on with 1, since {0, 1} would be 2 terms < 3*1 = 3.
Thus we must have a(1) = 2 terms < 3*2 = 6; and since we already have {0, 2}, the next must be at least 6.
Therefore, a(2) = 6 is the number of terms < 3*6 = 18, so there must be 3 more:
We have a(3) = 7 terms < 21, a(4) = 8 terms < 24, a(5) = 9 terms < 27.
Now, in view of a(2), the sequence goes on with a(6) = 18 terms < 3*18. This was the 7th term, in view of a(3) the next must be >= 21:
We have a(7) = 21 terms <= 3*21, a(8) = 24 terms <= 3*24, a(9) = 27 terms <= 3*27. Then we can increase by 1 up to index 18:
a(10) = 28 terms <= 3*28, ..., a(17) = 35 terms <= 3*35. This was the 18th term, in view of a(6) the following terms must be >= 3*18 = 54 =: a(18).
		

Links

• M. F. Hasler, Table of n, a(n) for n = 0..999
• E. Angelini, Re: A130011 and the definition of "slowest increasing"., SeqFan list, July 13, 2015

Crossrefs

Cf. A260107, A130011 and references therein; A037988, A094591 (analogs with 2n instead of 3n).

Programs

PARI

a=vector(100);a[i=2]=2;for(k=3,#a,a[k]=if(k>a[i],3*a[i++-1],a[k-1]+1))

Formula

a(n) <= 3n, with equality for indices of the form n = a(k) for some k.