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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.

%I A260139 #17 Dec 23 2024 14:53:44
%S A260139 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,
%T A260139 73,74,81,84,87,90,93,96,99,102,105,106,107,108,109,110,111,112,113,
%U A260139 114,115,116,117,118,119,120,121,122,123,162,165,168,169,170,171,172
%N 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.
%C A260139 Suggested by _Eric Angelini_, cf. link to SeqFan post.
%C A260139 This sequence has a nice self-similar graph.
%H A260139 M. F. Hasler, <a href="/A260139/b260139.txt">Table of n, a(n) for n = 0..999</a>
%H A260139 E. Angelini, <a href="https://web.archive.org/web/*/http://list.seqfan.eu/oldermail/seqfan/2015-July/015057.html">Re: A130011 and the definition of "slowest increasing".</a>, SeqFan list, July 13, 2015
%F A260139 a(n) <= 3n, with equality for indices of the form n = a(k) for some k.
%e A260139 The first term says that there are a(0) = 0 terms < 0.
%e A260139 Then it is not possible to go on with 1, since {0, 1} would be 2 terms < 3*1 = 3.
%e A260139 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.
%e A260139 Therefore, a(2) = 6 is the number of terms < 3*6 = 18, so there must be 3 more:
%e A260139 We have a(3) = 7 terms < 21, a(4) = 8 terms < 24, a(5) = 9 terms < 27.
%e A260139 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:
%e A260139 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:
%e A260139 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).
%o A260139 (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))
%Y A260139 Cf. A260107, A130011 and references therein; A037988, A094591 (analogs with 2n instead of 3n).
%K A260139 nonn,easy,look
%O A260139 0,2
%A A260139 _M. F. Hasler_, Jul 16 2015