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 1-5 of 5 results.

A073524 Number of steps to reach an integer starting with (n+1)/n and using the map x -> x*ceiling(x); or -1 if no integer is ever reached.

Original entry on oeis.org

0, 1, 2, 3, 18, 2, 3, 4, 6, 7, 26, 4, 9, 3, 4, 8, 6, 4, 56, 11, 3, 4, 42, 4, 33, 7, 5, 4, 38, 5, 79, 6, 4, 15, 14, 8, 200, 29, 13, 5, 36, 3, 4, 5, 7, 10, 11, 8, 6, 20, 47, 27, 43, 9, 41, 9, 10, 23, 37, 17, 18, 6, 7, 6, 32, 15, 225, 7, 73, 11, 20, 12, 182, 9, 16, 7, 10, 15, 196, 8
Offset: 1

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Author

N. J. A. Sloane, Aug 29 2002

Keywords

Comments

Computed by doing all computations over the integers (multiply by n) and by truncating modulo n^250. This avoids the explosion of the integers (of order 2^(2^k) after k iterations) and gives the correct answer if the final index i(n) is < 250 (or perhaps 249 or 248). If the algorithm does not stop before 245 one should increase precision (work with n^500 or even higher). - Roland Bacher
Always reaches an integer for n <= 100. - Roland Bacher, Aug 30 2002
Always reaches an integer for n <= 200. - N. J. A. Sloane, Sep 04 2002
Always reaches an integer for n <= 500 by comparing results with index 1000 and index 2500. - Robert G. Wilson v, Sep 11 2002
Always reaches an integer for n <= 3000. The Mathematica program automatically adjusts the modulus m required to find the first integral iterate. - T. D. Noe, Apr 10 2006
Always reaches an integer for n <= 5000. - Ben Branman, Feb 12 2011

Examples

			a(7) = 3 since 8/7 -> 16/7 -> 48/7 -> 48.

Links

Crossrefs

Cf. A073528, A073529, A068119, A001511, A072340, A075102, A073341.

Programs

  • Mathematica
    Table[{n, First[Flatten[Position[Map[Denominator, NestList[ # Ceiling[ # ] &, (n + 1)/n, 20]], 1]]]}, {n, 1, 20}]
f[n_] := Block[{k = (n + 1)/n, c = 0}, While[ !IntegerQ[k], c++; k = Mod[k*Ceiling[k], n^250]]; c]; Table[ f[n], {n, 1, 100}]
Table[lim=50; While[k=0; x=1+1/n; m=n^lim; While[kT. D. Noe, Apr 10 2006 *)

Extensions

a(5)-a(10), a(12)-a(18), a(20) = 11 from Ed Pegg Jr, Aug 29 2002
T. D. Noe also found a(5) and remarks that the final integer is 9.5329600...*10^57734. - Aug 29 2002
a(11) from T. D. Noe, who remarks that the final integer is 5.131986636061311...*10^13941166 - Aug 29 2002
a(19) and a(21) onwards from Roland Bacher, Aug 30 2002

A085785 Start at (2n+1)/4 and iterate the map x -> x*ceiling(x); sequence gives number of steps for denominator to drop to 1 or 2; or -1 if this never happens.

Original entry on oeis.org

1, 1, 3, 7, 1, 1, 2, 2, 1, 1, 2, 2, 1, 1, 6, 3, 1, 1, 4, 3, 1, 1, 2, 2, 1, 1, 2, 2, 1, 1, 3, 3, 1, 1, 5, 3, 1, 1, 2, 2, 1, 1, 2, 2, 1, 1, 3, 4, 1, 1, 3, 5, 1, 1, 2, 2, 1, 1, 2, 2, 1, 1, 6, 7, 1, 1, 3, 5, 1, 1, 2, 2, 1, 1, 2, 2, 1, 1, 5, 3, 1, 1, 5, 3, 1, 1, 2, 2, 1, 1, 2, 2, 1, 1, 3, 3, 1, 1, 8, 3, 1, 1, 2, 2, 1
Offset: 2

Views

Author

N. J. A. Sloane, Aug 16 2003

Keywords

Comments

We conjecture that the denominator always does drop.

Links

  • J. C. Lagarias and N. J. A. Sloane, Approximate squaring (pdf, ps), Experimental Math., 13 (2004), 113-128.

Crossrefs

A073341 gives number of steps until reach an integer. Cf. A085817, A085833.

A085817 Start at (2n+1)/4 and iterate the map x -> x*ceiling(x); sequence gives values of n for which the denominators in the orbit drop directly from 4 to 1, bypassing 2.

Original entry on oeis.org

4, 6, 7, 9, 12, 13, 14, 15, 20, 21, 22, 23, 24, 30, 31, 36, 38, 39, 41, 44, 45, 46, 47, 52, 53, 54, 55, 56, 62, 63, 69, 70, 71, 73, 76, 77, 78, 79, 81, 86, 87, 88, 94, 95, 96, 97, 101, 102, 103, 105, 108, 109, 110, 111, 112, 118, 119, 120, 126, 127, 129, 132, 133, 134, 135, 137
Offset: 1

Views

Author

N. J. A. Sloane, Aug 16 2003

Keywords

Comments

No formula is known for these numbers.

Links

  • J. C. Lagarias and N. J. A. Sloane, Approximate squaring (pdf, ps), Experimental Math., 13 (2004), 113-128.

Crossrefs

Complement of A085833.
A073341 gives number of steps until reach an integer. Cf. A085785.

A085833 Start at (2n+1)/4 and iterate the map x -> x*ceiling(x); sequence gives values of n for which the denominators in the orbit go from 4 to 2, instead of dropping directly to 1.

Original entry on oeis.org

2, 3, 5, 8, 10, 11, 16, 17, 18, 19, 25, 26, 27, 28, 29, 32, 33, 34, 35, 37, 40, 42, 43, 48, 49, 50, 51, 57, 58, 59, 60, 61, 64, 65, 66, 67, 68, 72, 74, 75, 80, 82, 83, 84, 85, 89, 90, 91, 92, 93, 98, 99, 100, 104, 106, 107, 113, 114, 115, 116, 117, 121, 122, 123, 124, 125, 128, 130
Offset: 1

Views

Author

N. J. A. Sloane, Aug 16 2003

Keywords

Comments

No formula is known for these numbers.

Links

  • J. C. Lagarias and N. J. A. Sloane, Approximate squaring (pdf, ps), Experimental Math., 13 (2004), 113-128.

Crossrefs

Complement of A085817.
A073341 gives number of steps until reach an integer. Cf. A085785.

A074735 Number of steps to reach an integer starting with (n+3)/4 and iterating the map x -> x*ceiling(x).

Original entry on oeis.org

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

Views

Author

Benoit Cloitre, Sep 05 2002

Keywords

Comments

Let S(n) = Sum_{k=1..n} a(k) then it seems that S(n) is asymptotic to 2n. S(n)=2n for many values of n, namely n=10,128,198,199,237,238,241,242,246,247,249,267,329... More generally, starting with (n+2^m-1)/2^m and iterating the same map seems to produce the same kind of behavior for a(n) (i.e., Sum_{k=1..n} a(k) is asymptotic to c(m)*n where c(m) depends on m and c(m) is a power of 2).

Crossrefs

Cf. A073524, A073341, A068119.

Programs

  • Mathematica
    Table[Length[NestWhileList[# Ceiling[#]&,(n+3)/4,!IntegerQ[#]&]]-1,{n,110}] (* Harvey P. Dale, Apr 11 2020 *)
  • PARI
    a(n)=if(n<0,0,s=(n+3)/4; c=0; while(frac(s)>0,s=s*ceil(s); c++); c)

Formula

Special cases: for k>= 0 a(4k+1) = 0, a(16k+10) = a(16k+11) = a(16k+12) = 1.

Extensions

Offset corrected by Sean A. Irvine, Jan 25 2025
Showing 1-5 of 5 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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