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.
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
Offset: 1
Examples
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. (The array formed by these rows is A278492.) 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.
References
- V. Brun, Un procédé qui ressemble au crible d'Ératosthène, Analele Stiintifice Univ. "Al. I. Cuza", Iasi, Romania, Sect. Ia Matematica, 1965, vol. 11B, pp. 47-53.
- Problems 107, 116, Nord. Mat. Tidskr. 5 (1957), 114-116.
- 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), 301-307.
- L. Carlitz and Chr. U. Jansen, Solution to Problem 115 and 116, Nord. Mat. Tidskr. 5 (1957), 159-161.
- V. Gardiner, R.Lazarus, N. Metropolis and S. Ulam, On certain sequences of integers defined by sieves, Math. Mag., 29 (1955), 117-119.
- H. Killingbergtro, Solution to Problem 107, Nord. Mat. Tidskr. 5 (1957), 203-205.
- D. Wilson et al., Interesting sequence, SeqFan list, Nov. 2016
- Index entries for sequences related to the Josephus Problem
- Index entries for sequences generated by sieves
Crossrefs
Cf. A000012, A002491, A000959, A003309, A099259, A112557, A112558, A113742, A113743, A113744, A113745, A113746, A113747, A113748; A113749, A278484.
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. A278169 (characteristic function).
Programs
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Haskell
a000960 n = a000960_list !! (n-1) 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 -
Maple
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, 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 *)
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PARI
a(n)=local(A=n,D);for(i=1,n-1,D=n-i;A=D*ceil(A/D+1));return(A) \\ Paul D. Hanna, Oct 10 2005
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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] j+=1 return L flavius(100) # Robert FERREOL, Nov 08 2015
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 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 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 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, 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
a(n) = 2*A073359(n-1) + 1, cf. link to posts on the SeqFan list. - M. F. Hasler, Nov 23 2016
a(n) = 1 + A278484(n-1). - Antti Karttunen, Nov 23 2016, after David W. Wilson's posting on SeqFan list Nov 22 2016
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
Comments