A056533 -id:A056533 - cp's OEIS frontend

A119503 Increasing gaps in the even sieve (A056533) by lower term.

1, 5, 11, 27, 59, 85, 131, 163, 501, 603, 1145, 1381, 1563, 1937, 2763, 3953, 6731, 7665, 10741, 13787, 22001, 30555, 47953, 52745, 60201, 111541, 137913, 162499, 186163, 282593, 343483, 351163, 386805, 476051, 522129, 540497, 660409, 714643

Offset: 1

Robert G. Wilson v, May 26 2006

nonn

			The even sieve begins: 1, 3, 5, 9, 11, 17, 19, 25, 27, 35, 37 and between 1&3 there is a gap of 2;
the difference between 3&5 is also 2, but between 5&9 the difference or gap is 4; etc.

Cf. A056533.

lst = Range[10^6]; i = 2; While[i <= (len = Length[lst]), lst = Drop[lst, {i, len, i}]; i+=2]; l = {}; d = 0; Do[a = lst[[n + 1]] - lst[[n]]; If[a > d, d = a; AppendTo[l, lst[[n]]]], {n, Length@lst - 1}]; l

A247105 Variation of Flavius Josephus's sieve: Start with the natural numbers; at the k-th sieving step, make k passes removing every k-th term of the sequence remaining after the previous sieving step; iterate.

Original entry on oeis.org

1, 5, 25, 109, 385, 1373, 4645, 16009, 48817, 159757, 488377, 1571425, 4560901, 14482393, 43408013, 130394125, 380755429, 1118740741, 3326930413, 9931863461, 28466058257, 84243573797, 240453967777, 706827067045, 2009065808473, 5913933615149, 16711898903281

Offset: 1

Sergio Pimentel, Nov 18 2014

nonn

Starting with the natural numbers, make 2 passes removing every 2nd number, 3 passes removing every 3rd number, etc.
Is the limiting value of a(n+1)/a(n)=3?
Since 1/(1-1/n)^n converges to e (as n -> inf), a(n+1)/a(n) converges to e. - Hiroaki Yamanouchi, Nov 27 2014

			The 1st pass removes 2, 4, 6, 8, 10, etc. The 2nd pass (also with 2) removes 3, 7, 11, 15, 19, etc. Then there are 3 passes removing every 3rd number, of which the 1st pass removes 9, 21, 33, 45, ..., the 2nd removes 13, 29, 49, ..., and the 3rd removes 17, 41, 73, ...; then there are 4 passes with 4; 5 passes with 5; etc.

Hiroaki Yamanouchi, Table of n, a(n) for n = 1..1000
Index entries for sequences related to the Josephus Problem

Cf. A000960, A056533, A099204.

A247105 = Reap[Quiet @ For[n=1, n<28, n++, m = n; For[i=n, i >= 1, i--, For[j=1, j <= i, j++, t = Floor[(m*i)/(i-1)]; While[t - Floor[t/i] >= m, t -= 1];  om = m; m = t+1]]; Sow[om]]][[2, 1]] (* Jean-François Alcover, Nov 28 2014, translated and adapted from Hiroaki Yamanouchi's Python script *)

copydropmult(v,m)=vector(#v-#v\m,i,v[(i-1)*m\(m-1)+1])
alim(n)=my(r=vector(n,i,i),j=2,k=1);while(j<#r,r=copydropmult(r,j);if(k++>j,j++;k=1));r

for n in range(1, 101):
  m = n
  for i in range(n, 1, -1):
    for j in range(i):
      t = m * i // (i - 1)
      while t - t // i >= m:
        t -= 1
      m = t + 1
  print(f"{n} {m}") # Hiroaki Yamanouchi, Nov 28 2014

More values from Franklin T. Adams-Watters, Nov 21 2014
a(12)-a(20) from Alois P. Heinz, Nov 26 2014
a(21)-a(27) from Hiroaki Yamanouchi, Nov 27 2014

A361423 Start with natural numbers, for all positive integer periods p sieve out every p-th number p-1 times over.

Original entry on oeis.org

1, 3, 9, 27, 75, 225, 651, 1947, 5661, 15753, 44497, 128325, 357339, 1025029, 2881677, 8152327, 22251081, 62981541, 175699737, 491888331, 1353494089, 3827528649, 10655040429, 29413393659, 80737582089, 226955441541, 626061311481, 1745916338341, 4826531920159, 13166998285539

Offset: 1

Rok Cestnik, Jul 17 2023

nonn

Appears to grow as: a(n) ~ c n^n/(n-1)! where c is approximately 0.56...

The terms remaining after the p-th sieve-batch grow on average with slope p^(p-1)/(p-1)!.

			Start with naturals: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, ...
Sieve out every 1st number 0 times (do nothing)
Sieve out every 2nd number 1 times: 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, ...
Sieve out every 3rd number 2 times:
   first time: 1, 3, 7, 9, 13, 15, 19, 21, 25, 27, 31, 33, 37, 39, 43, ...
   second time: 1, 3, 9, 13, 19, 21, 27, 31, 37, 39, 45, 49, 55, 57, 63, ...
Sieve out every 4th number 3 times:
   first time: 1, 3, 9, 19, 21, 27, 37, 39, 45, 55, 57, 63, 73, 75, 81, ...
   second time: 1, 3, 9, 21, 27, 37, 45, 55, 57, 73, 75, 81, 93, 99, ...
   third time: 1, 3, 9, 27, 37, 45, 57, 73, 75, 93, 99, 109, 127, 129, ...
Sieve out every 5th number 4 times:
   first time: 1, 3, 9, 27, 45, 57, 73, 75, 99, 109, 127, 129, 153, 165, ...
   second time: 1, 3, 9, 27, 57, 73, 75, 99, 127, 129, 153, 165, 189, ...
   third time: 1, 3, 9, 27, 73, 75, 99, 127, 153, 165, 189, 201, 225, ...
   fourth time: 1, 3, 9, 27, 75, 99, 127, 153, 189, 201, 225, 261, 289, ...
Sieve out every 6th number 5 times:
   ...

Bert Dobbelaere, Table of n, a(n) for n = 1..600
Rok Cestnik, Sieve visualization
Index entries for sequences generated by sieves.

Cf. A000960 (sieve once each).

Cf. A000959, A111039, A007950, A056533, A099267, A003309.

def A361423(n):
    for p in range(n,1,-1):
        for k in range(p-1):
            n += (n-1)//(p-1)
    return n
# Bert Dobbelaere, Jul 21 2023

More terms from Bert Dobbelaere, Jul 21 2023

cp's OEIS Frontend

A119503 Increasing gaps in the even sieve (A056533) by lower term.

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A247105 Variation of Flavius Josephus's sieve: Start with the natural numbers; at the k-th sieving step, make k passes removing every k-th term of the sequence remaining after the previous sieving step; iterate.

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A361423 Start with natural numbers, for all positive integer periods p sieve out every p-th number p-1 times over.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Python

Extensions