cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-3 of 3 results.

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.

Original entry on oeis.org

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

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Author

Keywords

Comments

a(n) is never divisible by 2 or 5. - Thomas Anton, Nov 01 2018

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).

Crossrefs

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 (a left inverse), A100618.
Cf. A278169 (characteristic function).
Main diagonal of A278492, leftmost column of A278505, positions of zeros in A278528 & A278529.

Programs

  • 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 *)
  • 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
    
  • 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

A056530 Sequence remaining after third round of Flavius Josephus sieve; remove every fourth term of A047241.

Original entry on oeis.org

1, 3, 7, 13, 15, 19, 25, 27, 31, 37, 39, 43, 49, 51, 55, 61, 63, 67, 73, 75, 79, 85, 87, 91, 97, 99, 103, 109, 111, 115, 121, 123, 127, 133, 135, 139, 145, 147, 151, 157, 159, 163, 169, 171, 175, 181, 183, 187, 193, 195, 199, 205, 207, 211, 217, 219, 223, 229, 231
Offset: 1

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Author

Henry Bottomley, Jun 19 2000

Keywords

Comments

Numbers {1, 3, 7} mod 12: A017533, A017557, A017605 interleaved.

Crossrefs

We have A000027 after 0 rounds of sieving, A005408 after 1 round of sieving, A047241 after 2 rounds, A056530 after 3 rounds, A056531 after 4 rounds, A000960 after all rounds. After n rounds the remaining sequence comprises A002944(n) numbers mod A003418(n+1), i.e. 1/(n+1) of them.

Programs

  • Mathematica
    LinearRecurrence[{1,0,1,-1},{1,3,7,13},60] (* Harvey P. Dale, Oct 19 2022 *)

Formula

From Chai Wah Wu, Jul 24 2016: (Start)
a(n) = a(n-1) + a(n-3) - a(n-4) for n > 4.
G.f.: x*(5*x^3 + 4*x^2 + 2*x + 1)/(x^4 - x^3 - x + 1). (End)
a(n) = 4*n - (13 + 2*A131713(n))/3. - R. J. Mathar, Jun 22 2020

A278492 Square array where row n (n >= 0) gives the numbers remaining after the n-th round of the Flavius sieve, read by descending antidiagonals.

Original entry on oeis.org

1, 2, 1, 3, 3, 1, 4, 5, 3, 1, 5, 7, 7, 3, 1, 6, 9, 9, 7, 3, 1, 7, 11, 13, 13, 7, 3, 1, 8, 13, 15, 15, 13, 7, 3, 1, 9, 15, 19, 19, 19, 13, 7, 3, 1, 10, 17, 21, 25, 25, 19, 13, 7, 3, 1, 11, 19, 25, 27, 27, 27, 19, 13, 7, 3, 1, 12, 21, 27, 31, 31, 31, 27, 19, 13, 7, 3, 1, 13, 23, 31, 37, 39, 39, 39, 27, 19, 13, 7, 3, 1
Offset: 0

Views

Author

Antti Karttunen, Nov 23 2016, after David W. Wilson's posting on SeqFan-list Nov 22 2016

Keywords

Comments

The terms of square array A(row,col) are read by descending antidiagonals as A(0,0), A(0,1), A(1,0), A(0,2), A(1,1), A(2,0), ...

Examples

			The top left corner of the array:
1, 2, 3,  4,  5,  6,  7,  8,  9, 10 (row 0: start from A000027)
1, 3, 5,  7,  9, 11, 13, 15, 17, 19 (after the 1st round, A005408 remain)
1, 3, 7,  9, 13, 15, 19, 21, 25, 27 (after the 2nd, A047241)
1, 3, 7, 13, 15, 19, 25, 27, 31, 37
1, 3, 7, 13, 19, 25, 27, 31, 39, 43
1, 3, 7, 13, 19, 27, 31, 39, 43, 49
1, 3, 7, 13, 19, 27, 39, 43, 49, 61
1, 3, 7, 13, 19, 27, 39, 49, 61, 63
1, 3, 7, 13, 19, 27, 39, 49, 63, 67
1, 3, 7, 13, 19, 27, 39, 49, 63, 79
		

Crossrefs

One more than A278482.
Transpose: A278493.
Main diagonal: A000960.
Cf. A278507 (the numbers removed at each round).
Similarly constructed arrays for other sieves: A258207, A260717.

Programs

Formula

A(n,k) = 1 + A278482(n,k).
Showing 1-3 of 3 results.