

A000960


Flavius Josephus's sieve: Start with the natural numbers; at the kth sieving step, remove every (k+1)st term of the sequence remaining after the (k1)st sieving step; iterate.
(Formerly M2636 N1048)


58



1, 3, 7, 13, 19, 27, 39, 49, 63, 79, 91, 109, 133, 147, 181, 207, 223, 253, 289, 307, 349, 387, 399, 459, 481, 529, 567, 613, 649, 709, 763, 807, 843, 927, 949, 1009, 1093, 1111, 1189, 1261, 1321, 1359, 1471, 1483, 1579, 1693, 1719, 1807, 1899, 1933, 2023
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OFFSET

1,2


REFERENCES

V. Brun, Un procédé qui ressemble au crible d'Eratosthene, Analele Stiintifice Univ. "Al. I. Cuza", Iasi, Romania, Sect. Ia Matematica, 1965, vol. 11B, pp. 4753.
L. Carlitz, Solution to Problem 115, Nord. Mat. Tidskr. 5 (1957), 159160.
Problems 107, 116, Solutions, Nord. Mat. Tidskr. 5 (1957), 114116, 160161 and 203205.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).


LINKS

T. D. Noe, Table of n, a(n) for n=1..10000
M. E. Andersson, Das Flaviussche Sieb, Acta Arith., 85 (1998), 301307.
V. Gardiner, R.Lazarus, N. Metropolis and S. Ulam, On certain sequences of integers defined by sieves, Math. Mag., 29 (1955), 117119.
Index entries for sequences generated by sieves


FORMULA

Let F(n) = number of terms <= n. Andersson, improving results of Brun, shows that F(n) = 2 sqrt(n/Pi) + O(n^(1/6)). Hence a(n) grows like Pi n^2 / 4.
To get nth term, start with n and successively round up to next 2 multiples of n1, n2, ..., 1 (compare to Mancala sequence A002491). E.g.: to get 11th term: 11>30>45>56>63>72>80>84>87>90>91; i.e., start with 11, successively round up to next 2 multiples of 10, 9, .., 1.  Paul D. Hanna, Oct 10 2005
As in Paul D. Hanna's formula, start with n^2 and successively move down to the highest multiple of n1, n2, etc., smaller than your current number: 121 120 117 112 105 102 100 96 93 92 91, so a(11) = 91, from moving down to multiples of 10, 9, ..., 1.  Joshua Zucker, May 20 2006
Or, similarly for n = 5, begin with 25, down to a multiple of 4 = 24, down to a multiple of 3 = 21, then to a multiple of 2 = 20 and finally to a multiple of 1 = 19, so a(5) = 19.  Joshua Zucker, May 20 2006
This formula arises in A119446; the leading term of row k of that triangle = a(prime(k)/k) from this sequence.  Joshua Zucker, May 20 2006


EXAMPLE

Start with
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 ... (A000027) and delete every second term, giving
1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 ... (A005408) and delete every 3rd term, giving
1 3 7 9 13 15 19 21 25 27 ... (A056530) and delete every 4th term, giving
1 3 7 13 15 19 25 27 ... (A056531) and delete every 5th term, giving
.... Continue forever and what's left is the sequence.
For n = 5, 5^2 = 25, go down to a multiple of 4 giving 24, then to a multiple of 3 = 21, then to a multiple of 2 = 20, then to a multiple of 1 = 19, so a(5) = 19.


MAPLE

S[1]:={seq(i, i=1..2100)}: for n from 2 to 2100 do S[n]:=S[n1] minus {seq(S[n1][n*i], i=1..nops(S[n1])/n)} od: A:=S[2100]; # Emeric Deutsch, Nov 17 2004


MATHEMATICA

del[lst_, k_] := lst[[Select[Range[Length[lst]], Mod[ #, k] != 0 &]]]; For[k = 2; s = Range[2100], k <= Length[s], k++, s = del[s, k]]; s
f[n_] := Fold[ #2*Ceiling[ #1/#2 + 1] &, n, Reverse@Range[n  1]]; Array[f, 60] (* Robert G. Wilson v, Nov 05 2005 *)


PROG

(PARI) a(n)=local(A=n, D); for(i=1, n1, D=ni; A=D*ceil(A/D+1)); return(A) \\ Paul D. Hanna, Oct 10 2005
(Haskell)
a000960 n = a000960_list !! (n1)
a000960_list = sieve 1 [1..] where
sieve k (x:xs) = x : sieve (k+1) (flavius xs) where
flavius xs = us ++ flavius vs where (u:us, vs) = splitAt (k+1) xs
 Reinhard Zumkeller, Oct 31 2012
(Python)
def flavius(n):
....L=list(range(1, n+1)); j=2
....while j <= len(L):
........L=[L[i] for i in range(len(L)) if (i+1)%j!=0]
........j+=1
....return(L)
flavius(100)
# Robert FERREOL, Nov 08 2015


CROSSREFS

Cf. A056526, A056530, A056531, A100002.
Cf. A000012, A002491, A000960, A112557, A112558, A113742, A113743, A113744, A113745, A113746, A113747, A113748; A113749.
Cf. A119446 for triangle whose leading diagonal is A119447 and this sequence gives all possible values for A119447 (except A119447 cannot equal 1 because prime(n)/n is never 1).
Cf. A100617, A100618.
Sequence in context: A202117 A100458 A193765 * A147614 A171747 A031215
Adjacent sequences: A000957 A000958 A000959 * A000961 A000962 A000963


KEYWORD

nonn,easy,nice


AUTHOR

N. J. A. Sloane


EXTENSIONS

More terms and better description from Henry Bottomley, Jun 16 2000
Entry revised by N. J. A. Sloane, Nov 13 2004
More terms from Paul D. Hanna, Oct 10 2005


STATUS

approved



