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A136119 Limiting sequence when we start with the positive integers (A000027) and delete in step n >= 1 the term at position n + a(n).

%I A136119 #61 Jul 04 2022 12:21:45
%S A136119 1,3,4,5,7,8,10,11,13,14,15,17,18,20,21,22,24,25,27,28,29,31,32,34,35,
%T A136119 37,38,39,41,42,44,45,46,48,49,51,52,54,55,56,58,59,61,62,63,65,66,68,
%U A136119 69,71,72,73,75,76,78,79,80,82,83,85,86,87,89,90,92,93,95,96,97,99,100
%N A136119 Limiting sequence when we start with the positive integers (A000027) and delete in step n >= 1 the term at position n + a(n).
%C A136119 Apparently a(n) = A001953(n-1)+1 = floor((n-1/2)*sqrt(2))+1 (confirmed for n < 20000) and a(n+1) - a(n) = A001030(n). From the definitions these conjectures are by no means obvious. Can they be proved? - _Klaus Brockhaus_, Apr 15 2008 [For an affirmative answer, see the Cloitre link.]
%C A136119 This is the s(n)-Wythoff sequence for s(n)=2n-1; see A184117 for the definition. Complement of A184119. - _Clark Kimberling_, Jan 09 2011
%D A136119 B. Cloitre, The golden sieve, preprint 2008
%H A136119 Alois P. Heinz, <a href="/A136119/b136119.txt">Table of n, a(n) for n = 1..1000</a>
%H A136119 D. X. Charles, <a href="http://pages.cs.wisc.edu/~cdx/Sieve.pdf">Sieve Methods</a>, July 2000, University of Wisconsin.
%H A136119 Benoit Cloitre, <a href="/A136119/a136119.txt">On the proof of Klaus Brockhaus's conjectures</a>
%H A136119 R. Eismann, <a href="http://arxiv.org/abs/0711.0865">Decomposition of natural numbers into weight X level + jump and application to a new classification of prime numbers</a>, arXiv:0711.0865 [math.NT], 2007-2010.
%H A136119 M. C. Wunderlich, <a href="http://matwbn.icm.edu.pl/ksiazki/aa/aa16/aa1614.pdf">A general class of sieve generated sequences</a>, Acta Arithmetica XVI,1969, pp.41-56.
%H A136119 <a href="/index/Si#sieve">Index entries for sequences generated by sieves</a>
%F A136119 a(n) = ceiling((n-1/2)*sqrt(2)). This can be proved in the same way as the formula given for A099267. There are some generalizations. For instance, it is possible to consider "a(n)+K*n" instead of "a(n)+n" for deleting terms where K=0,1,2,... is fixed. The constant involved in the Beatty sequence for the sequence of deleted terms then depends on K and equals (K + 1 + sqrt((K+1)^2 + 4))/2. K=0 is related to A099267. 1+A001954 is the complement sequence of this sequence A136119. - _Benoit Cloitre_, Apr 18 2008
%F A136119 a(n) = floor(1 + 2*sqrt(T(n-1))), with triangular numbers T(). - _Ralf Steiner_, Oct 23 2019
%F A136119 Lim_{n->inf}(a(n)/(n - 1)) = sqrt(2), with {a(n)/(n - 1)} decreasing. - _Ralf Steiner_, Oct 24 2019
%e A136119 First few steps are:
%e A136119 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,...
%e A136119 n = 1; delete term at position 1+a(1) = 2: 2;
%e A136119 1,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,...
%e A136119 n = 2; delete term at position 2+a(2) = 5: 6;
%e A136119 1,3,4,5,7,8,9,10,11,12,13,14,15,16,17,18,19,20,...
%e A136119 n = 3; delete term at position 3+a(3) = 7: 9;
%e A136119 1,3,4,5,7,8,10,11,12,13,14,15,16,17,18,19,20,...
%e A136119 n = 4; delete term at position 4+a(4) = 9: 12;
%e A136119 1,3,4,5,7,8,10,11,13,14,15,16,17,18,19,20,...
%e A136119 n = 5; delete term at position 5+a(5) = 12: 16;
%e A136119 1,3,4,5,7,8,10,11,13,14,15,17,18,19,20,...
%e A136119 n = 6; delete term at position 6+a(6) = 14: 19;
%e A136119 1,3,4,5,7,8,10,11,13,14,15,17,18,20,...
%t A136119 f[0] = Range[100]; f[n_] := f[n] = Module[{pos = n + f[n-1][[n]]}, If[pos > Length[f[n-1]], f[n-1], Delete[f[n-1], pos]]]; f[1]; f[n = 2]; While[f[n] != f[n-1], n++]; f[n] (* _Jean-François Alcover_, May 08 2019 *)
%t A136119 T[n_] := n (n + 1)/2; Table[1 + 2 Sqrt[T[n-1]] , {n, 1, 71}] // Floor (* _Ralf Steiner_, Oct 23 2019 *)
%o A136119 (Haskell)
%o A136119 import Data.List (delete)
%o A136119 a136119 n = a136119_list !! (n-1)
%o A136119 a136119_list = f [1..] where
%o A136119    f zs@(y:xs) = y : f (delete (zs !! y) xs)
%o A136119 -- _Reinhard Zumkeller_, May 17 2014
%o A136119 (Magma) [Ceiling((n-1/2)*Sqrt(2)): n in [1..100]]; // _Vincenzo Librandi_, Jul 01 2019
%o A136119 (PARI) apply( {A136119(n)=sqrtint(n*(n-1)*2)+1}, [1..99]) \\ _M. F. Hasler_, Jul 04 2022
%Y A136119 Cf. A000027, A001953 (floor((n+1/2)*sqrt(2))), A001030 (fixed under 1 -> 21, 2 -> 211), A136110, A137292.
%Y A136119 Cf. A000959, A099267.
%Y A136119 Cf. A242535.
%Y A136119 Cf. A000217 (T).
%K A136119 easy,nonn
%O A136119 1,2
%A A136119 _Ctibor O. Zizka_, Mar 16 2008
%E A136119 Edited and extended by _Klaus Brockhaus_, Apr 15 2008
%E A136119 An incorrect g.f. removed by _Alois P. Heinz_, Dec 14 2012