A136119 Limiting sequence when we start with the positive integers (A000027) and delete in step n >= 1 the term at position n + a(n).
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, 37, 38, 39, 41, 42, 44, 45, 46, 48, 49, 51, 52, 54, 55, 56, 58, 59, 61, 62, 63, 65, 66, 68, 69, 71, 72, 73, 75, 76, 78, 79, 80, 82, 83, 85, 86, 87, 89, 90, 92, 93, 95, 96, 97, 99, 100
Offset: 1
Examples
First few steps are: 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,... n = 1; delete term at position 1+a(1) = 2: 2; 1,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,... n = 2; delete term at position 2+a(2) = 5: 6; 1,3,4,5,7,8,9,10,11,12,13,14,15,16,17,18,19,20,... n = 3; delete term at position 3+a(3) = 7: 9; 1,3,4,5,7,8,10,11,12,13,14,15,16,17,18,19,20,... n = 4; delete term at position 4+a(4) = 9: 12; 1,3,4,5,7,8,10,11,13,14,15,16,17,18,19,20,... n = 5; delete term at position 5+a(5) = 12: 16; 1,3,4,5,7,8,10,11,13,14,15,17,18,19,20,... n = 6; delete term at position 6+a(6) = 14: 19; 1,3,4,5,7,8,10,11,13,14,15,17,18,20,...
References
- B. Cloitre, The golden sieve, preprint 2008
Links
- Alois P. Heinz, Table of n, a(n) for n = 1..1000
- D. X. Charles, Sieve Methods, July 2000, University of Wisconsin.
- Benoit Cloitre, On the proof of Klaus Brockhaus's conjectures
- R. Eismann, Decomposition of natural numbers into weight X level + jump and application to a new classification of prime numbers, arXiv:0711.0865 [math.NT], 2007-2010.
- M. C. Wunderlich, A general class of sieve generated sequences, Acta Arithmetica XVI,1969, pp.41-56.
- Index entries for sequences generated by sieves
Crossrefs
Programs
-
Haskell
import Data.List (delete) a136119 n = a136119_list !! (n-1) a136119_list = f [1..] where f zs@(y:xs) = y : f (delete (zs !! y) xs) -- Reinhard Zumkeller, May 17 2014
-
Magma
[Ceiling((n-1/2)*Sqrt(2)): n in [1..100]]; // Vincenzo Librandi, Jul 01 2019
-
Mathematica
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[n_] := n (n + 1)/2; Table[1 + 2 Sqrt[T[n-1]] , {n, 1, 71}] // Floor (* Ralf Steiner, Oct 23 2019 *) -
PARI
apply( {A136119(n)=sqrtint(n*(n-1)*2)+1}, [1..99]) \\ M. F. Hasler, Jul 04 2022
Formula
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
a(n) = floor(1 + 2*sqrt(T(n-1))), with triangular numbers T(). - Ralf Steiner, Oct 23 2019
Lim_{n->inf}(a(n)/(n - 1)) = sqrt(2), with {a(n)/(n - 1)} decreasing. - Ralf Steiner, Oct 24 2019
Extensions
Edited and extended by Klaus Brockhaus, Apr 15 2008
An incorrect g.f. removed by Alois P. Heinz, Dec 14 2012
Comments