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.

User: David Seelmann

David Seelmann's wiki page.

David Seelmann has authored 3 sequences.

A268696 a(1)=1; thereafter a(n+1) = floor(m/2), where m = number of occurrences of a(n) in [a(1),...,a(n)].

Original entry on oeis.org

1, 0, 0, 1, 1, 1, 2, 0, 1, 2, 1, 3, 0, 2, 1, 3, 1, 4, 0, 2, 2, 2, 3, 1, 4, 1, 5, 0, 3, 2, 3, 2, 4, 1, 5, 1, 6, 0, 3, 3, 3, 4, 2, 4, 2, 5, 1, 6, 1, 7, 0, 4, 3, 4, 3, 5, 2, 5, 2, 6, 1, 7, 1, 8, 0, 4, 4, 4, 5, 3, 5, 3, 6, 2, 6, 2, 7, 1, 8, 1, 9, 0, 5, 4, 5, 4, 6, 3, 6, 3, 7, 2, 7, 2, 8, 1, 9, 1, 10, 0
Offset: 1

Views

Author

David Seelmann, Feb 11 2016

Keywords

Comments

It appears that a(n) <= ceiling(sqrt(n)).
Comments from N. J. A. Sloane, Feb 12 2016: (Start)
In fact it appears that a(n) <= floor(sqrt(n)) except when n belongs to the sequence S := [99, 120, 142, 167, 193, 222, 252, 285, 319, ...], which has second differences 1,3,1,3,1,3,... and is the sequence {99; A035608(k)+21*k+120, k>=0}. For these values of n it appears that a(n) = ceiling(sqrt(n)). The first example is a(99) = 10 = ceiling(sqrt(99)).
The zeros occur at positions [2, 3, 8, 13, 19, 28, 38, 51, 65, 82, 100, 121, 143, 168, 194, 223, 253, 286, 320, ...], which apart from the initial terms appears to be S+1.
Without the division by 2 in the definition (that is, if a(n+1)=m), we get A158416. (End)

Examples

			a(2) is equal to the number of times a(1) = 1 appears in the sequence before, divided by two, rounding down. Since 1 appears once before, a(2) = floor(1/2) = 0.
a(3) is  equal to the number of times 0 appears in the sequence before, which is again once, divided by two, rounding down. So a(3) = floor(1/2) = 0.
a(4) is the number of times 0 appears before (twice) divided by two, which gives us 1.

Links

Crossrefs

Cf. A035608, A158416.

Programs

  • Mathematica
    a = {1}; Do[AppendTo[a, Floor[Count[a, n_ /; n == a[[k - 1]]]/2]], {k, 2, 120}]; a (* Michael De Vlieger, Feb 11 2016 *)

Extensions

More terms from Michael De Vlieger, Feb 11 2016

A268690 A268642 with a(n+1) = 5*n if ceiling(a(n)/2) is already in the sequence.

Original entry on oeis.org

1, 5, 3, 2, 10, 50, 25, 13, 7, 4, 20, 100, 500, 250, 125, 63, 32, 16, 8, 40, 200, 1000, 5000, 2500, 1250, 625, 313, 157, 79, 395, 198, 99, 495, 248, 124, 62, 31, 155, 78, 39, 195, 98, 49, 245, 123, 615, 308, 154, 77
Offset: 1

Views

Author

David Seelmann, Feb 11 2016

Keywords

Comments

It is conjectured that this is a permutation of the integers, along with any Seelmann sequence in which a(n+1) = M*n if the divide by 2 rule is not permitted, for any integer M>1 and not of the form M = 2^N.

Crossrefs

Cf. A268642.

A268642 Seelmann's sequence: a(1) = 1; thereafter a(n + 1) = ceiling(a(n)/2) unless this is already in the sequence, in which case a(n + 1) = 3*a(n).

Original entry on oeis.org

1, 3, 2, 6, 18, 9, 5, 15, 8, 4, 12, 36, 108, 54, 27, 14, 7, 21, 11, 33, 17, 51, 26, 13, 39, 20, 10, 30, 90, 45, 23, 69, 35, 105, 53, 159, 80, 40, 120, 60, 180, 540, 270, 135, 68, 34, 102, 306, 153, 77, 231, 116, 58, 29, 87, 44, 22, 66, 198, 99, 50, 25, 75, 38
Offset: 1

Views

Author

Peter Kagey, Feb 09 2016, based on a posting by David Seelmann to the Reddit web site

Keywords

Comments

It is conjectured that this is a permutation of the positive integers, along with any Seelmann sequence in which a(n+1) = M*a(n) if the divide by 2 rule cannot be applied, for any integer M>1 and not of the form M = 2^N. [Corrected by Charlie Neder, Feb 06 2019]
Reminiscent of the 3x+1 or Collatz problem, cf. A006577. - N. J. A. Sloane, Feb 09 2016
The Reddit link contains what is claimed to be a proof that this sequence is a permutation. I don't know if it has been checked. - N. J. A. Sloane, Feb 11 2016

Links

Crossrefs

Cf. A006577, A050000 (with floor instead of ceiling).
For records see A268529, A268530. For inverse see A268531.

Programs

  • Mathematica
    a[1]=1; a[n_] := a[n] = Module[{an1, an}, an1 = a[n-1]; an = If[EvenQ[an1], an1/2, (an1+1)/2]; If[FreeQ[Array[a, n-1], an], an, 3*a[n-1]]]; Array[a, 100] (* Jean-François Alcover, Feb 27 2016 *)
Fold[Append[#1, If[FreeQ[#1, #3], #3, 3 #1[[-1]]]] & @@ {#1, #2, Ceiling[#1[[-1]]/2]} &, {1}, Range@ 63] (* Michael De Vlieger, Jan 13 2018 *)

Extensions

Title corrected by Charlie Neder, Feb 06 2019

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