|
%I #16 Jan 02 2023 12:30:51
%S 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 73,74,81,84,87,90,93,96,99,102,105,106,107,108,109,110,111,112,113,
%U 114,115,116,117,118,119,120,121,122,123,162,165,168,169,170,171,172
%N 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 Suggested by _Eric Angelini_, cf. link to SeqFan post.
%C This sequence has a nice self-similar graph.
%H M. F. Hasler, <a href="/A260139/b260139.txt">Table of n, a(n) for n = 0..999</a>
%H E. Angelini, <a href="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 a(n) <= 3n, with equality for indices of the form n = a(k) for some k.
%e The first term says that there are a(0) = 0 terms < 0.
%e Then it is not possible to go on with 1, since {0, 1} would be 2 terms < 3*1 = 3.
%e 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 Therefore, a(2) = 6 is the number of terms < 3*6 = 18, so there must be 3 more:
%e We have a(3) = 7 terms < 21, a(4) = 8 terms < 24, a(5) = 9 terms < 27.
%e 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 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 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 (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 Cf. A260107, A130011 and references therein; A037988, A094591 (analogs with 2n instead of 3n).
%K nonn,easy,look
%O 0,2
%A _M. F. Hasler_, Jul 16 2015
|
