cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A130011 A self-describing sequence. Pick any integer n in the sequence; this n says: "There are n terms in the sequence that are <= 3n". This sequence is the slowest increasing one with this property.

Original entry on oeis.org

1, 4, 5, 12, 15, 16, 17, 18, 19, 20, 21, 36, 37, 38, 45, 48, 51, 54, 57, 60, 63, 64, 65, 66, 67, 68, 69
Offset: 1

Views

Author

Eric Angelini, Jun 15 2007

Keywords

Comments

See comments in A094591 and A037988.
It is not clear in what sense "slowest increasing" is meant in the description of this sequence. The definition requires that there be exactly a(k) terms <= 3 a(k), for any index k. Therefore, a(n+1) > 3n for all indices n of the form n = a(k). Thus, any such sequence has an infinite number of terms a(k) >= 3k-2. The lexicographically first variant A260107, which starts (1, 4, 5, 6, 13, 16, 19, 20, 21, 22, ...), also has all its terms a(k) <= 3k-2, so it cannot be said to increase faster. - M. F. Hasler, Jul 16 2015

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.

Original entry on oeis.org

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

Views

Author

M. F. Hasler, Jul 16 2015

Keywords

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).
		

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.
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