1,2

M. E. Andersson, Das Flaviussche Sieb, Acta Arith., 85 (1998), 301-307.

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.

L. Carlitz, Solution to Problem 115, Nord. Mat. Tidskr. 5 (1957), 159-160.

V. Gardiner, R. Lazarus, N. Metropolis and S. Ulam, On certain sequences of integers defined by sieves, Math. Mag., 29 (1955), 117-119.

Solutions to Problems 107, 116, Nord. Mat. Tidskr. 5 (1957), 114-116, 160-161 and 203-205.

T. D. Noe, Table of n, a(n) for n=1..10000

Index entries for sequences generated by sieves

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 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

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

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

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

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 for ever, 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.

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)

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] (from Robert G. Wilson v (rgwv(at)rgwv.com), Nov 05 2005)

(PARI) a(n)=local(A=n, D); for(i=1, n-1, D=n-i; A=D*ceil(A/D+1)); return(A) (Hanna)

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).

Adjacent sequences: A000957 A000958 A000959 this_sequence A000961 A000962 A000963

Sequence in context: A102828 A117679 A100458 this_sequence A031215 A099957 A086148

nonn,easy,nice

njas

More terms and better description from Henry Bottomley (se16(AT)btinternet.com), Jun 16 2000

Entry revised by njas, Nov 13 2004.

More terms from Paul D. Hanna (pauldhanna(AT)juno.com), Oct 10 2005