%I A000960 M2636 N1048
%S A000960 1,3,7,13,19,27,39,49,63,79,91,109,133,147,181,207,223,253,289,307,349,
%T A000960 387,399,459,481,529,567,613,649,709,763,807,843,927,949,1009,1093,
%U A000960 1111,1189,1261,1321,1359,1471,1483,1579,1693,1719,1807,1899,1933,2023
%N A000960 Flavius Josephus's sieve: Start with the natural numbers; at the k-th 
               sieving step, remove every (k+1)-st term of the sequence remaining 
               after the (k-1)-st sieving step; iterate.
%D A000960 N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, 
               Academic Press, 1995 (includes this sequence).
%D A000960 N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 
               (includes this sequence).
%D A000960 M. E. Andersson, Das Flaviussche Sieb, Acta Arith., 85 (1998), 301-307.
%D A000960 V. Brun, Un proc\'{e}d\'{e} qui ressemble au crible d'Eratostene, Analele 
               Stiintifice Univ. "Al. I. Cuza", Iasi, Romania, Sect. Ia Matematica, 
               1965, vol. 11B, pp. 47-53.
%D A000960 L. Carlitz, Solution to Problem 115, Nord. Mat. Tidskr. 5 (1957), 159-160.
%D A000960 V. Gardiner, R. Lazarus, N. Metropolis and S. Ulam, On certain sequences 
               of integers defined by sieves, Math. Mag., 29 (1955), 117-119.
%D A000960 Solutions to Problems 107, 116, Nord. Mat. Tidskr. 5 (1957), 114-116, 
               160-161 and 203-205.
%H A000960 T. D. Noe, <a href="b000960.txt">Table of n, a(n) for n=1..10000</a>
%H A000960 <a href="Sindx_Si.html#sieve">Index entries for sequences generated by 
               sieves</a>
%F A000960 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.
%F A000960 To get n-th term, start with n and successively round up to next 2 multiples 
               of n-1, n-2, ..., 1 (compare to Mancala sequence A002491). E.g.: 
               to get 11-th 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 (pauldhanna(AT)juno.com), Oct 10 2005
%F A000960 As in Paul D. Hanna's formula, start with n^2 and successively move down 
               to the highest multiple of n-1, n-2, 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 (joshua.zucker(AT)stanfordalumni.org), 
               May 20 2006
%F A000960 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 (joshua.zucker(AT)stanfordalumni.org), 
               May 20 2006
%F A000960 This formula arises in A119446; the leading term of row k of that triangle 
               = a(prime(k)/k) from this sequence. - Joshua Zucker (joshua.zucker(AT)stanfordalumni.org), 
               May 20 2006
%e A000960 Start with
%e A000960 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 ... (A000027) and delete every 
               second term, giving
%e A000960 1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 ... (A005408) and delete every 
               3rd term, giving
%e A000960 1 3 7 9 13 15 19 21 25 27 ... (A056530) and delete every 4th term, giving
%e A000960 1 3 7 13 15 19 25 27 ... (A056531) and delete every 5th term, giving
%e A000960 .... Continue for ever and what's left is the sequence.
%e A000960 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.
%p A000960 S[1]:={seq(i,i=1..2100)}: for n from 2 to 2100 do S[n]:=S[n-1] minus 
               {seq(S[n-1][n*i],i=1..nops(S[n-1])/n)} od: A:=S[2100]; (Emeric Deutsch 
               (deutsch(AT)duke.poly.edu), Nov 17 2004)
%t A000960 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
%t A000960 f[n_] := Fold[ #2*Ceiling[ #1/#2 + 1] &, n, Reverse@Range[n - 1]]; Array[f, 
               60] (from Robert G. Wilson v (rgwv(at)rgwv.com), Nov 05 2005)
%o A000960 (PARI) a(n)=local(A=n,D);for(i=1,n-1,D=n-i;A=D*ceil(A/D+1));return(A) 
               (Hanna)
%Y A000960 Cf. A056526, A056530, A056531, A100002.
%Y A000960 Cf. A000012, A002491, A000960, A112557, A112558, A113742, A113743, A113744, 
               A113745, A113746, A113747, A113748; A113749.
%Y A000960 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).
%Y A000960 Sequence in context: A102828 A117679 A100458 this_sequence A147614 A031215 
               A099957
%Y A000960 Adjacent sequences: A000957 A000958 A000959 this_sequence A000961 A000962 
               A000963
%K A000960 nonn,easy,nice
%O A000960 1,2
%A A000960 N. J. A. Sloane (njas(AT)research.att.com).
%E A000960 More terms and better description from Henry Bottomley (se16(AT)btinternet.com), 
               Jun 16 2000
%E A000960 Entry revised by N. J. A. Sloane (njas(AT)research.att.com), Nov 13 2004.
%E A000960 More terms from Paul D. Hanna (pauldhanna(AT)juno.com), Oct 10 2005