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.

A006884 In the '3x+1' problem, these values for the starting value set new records for highest point of trajectory before reaching 1.

Original entry on oeis.org

1, 2, 3, 7, 15, 27, 255, 447, 639, 703, 1819, 4255, 4591, 9663, 20895, 26623, 31911, 60975, 77671, 113383, 138367, 159487, 270271, 665215, 704511, 1042431, 1212415, 1441407, 1875711, 1988859, 2643183, 2684647, 3041127, 3873535, 4637979, 5656191
Offset: 1

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Author

Keywords

Comments

Both the 3x+1 steps and the halving steps are counted.
Where records occur in A025586: A006885(n) = A025586(a(n)) and A025586(m) < A006885(n) for m < a(n). - Reinhard Zumkeller, May 11 2013

References

  • R. B. Banks, Slicing Pizzas, Racing Turtles and Further Adventures in Applied Mathematics, Princeton Univ. Press, 1999. See p. 96.
  • D. R. Hofstadter, Goedel, Escher, Bach: an Eternal Golden Braid, Random House, 1980, p. 400.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

A060409 gives associated "dropping times", A060410 the maximal values and A060411 the steps at which the maxima occur.

Programs

  • Haskell
    a006884 n = a006884_list !! (n-1)
    a006884_list = f 1 0 a025586_list where
       f i r (x:xs) = if x > r then i : f (i + 1) x xs else f (i + 1) r xs
    -- Reinhard Zumkeller, May 11 2013
    
  • Mathematica
    mcoll[n_]:=Max@@NestWhileList[If[EvenQ[#],#/2,3#+1]&,n,#>1&]; t={1,max=2}; Do[If[(y=mcoll[n])>max,max=y; AppendTo[t,n]],{n,3,705000,4}]; t (* Jayanta Basu, May 28 2013 *)
    DeleteDuplicates[Parallelize[Table[{n,Max[NestWhileList[If[EvenQ[#],#/2,3#+1]&,n,#>1&]]},{n,57*10^5}]],GreaterEqual[#1[[2]],#2[[2]]]&][[;;,1]] (* Harvey P. Dale, Apr 23 2023 *)
  • PARI
    A025586(n)=my(r=n); while(n>2, if(n%2, n=3*n+1; if(n>r, r=n)); n>>=1); r
    r=0; for(n=1,1e6, t=A025586(n); if(t>r, r=t; print1(n", "))) \\ Charles R Greathouse IV, May 25 2016

A060412 In the '3x+1' problem, these values for the starting value set new records for the "dropping time", number of steps to reach a lower value than the start.

Original entry on oeis.org

2, 3, 7, 27, 703, 10087, 35655, 270271, 362343, 381727, 626331, 1027431, 1126015, 8088063, 13421671, 20638335, 26716671, 56924955, 63728127, 217740015, 1200991791, 1827397567, 2788008987, 12235060455
Offset: 1

Views

Author

N. J. A. Sloane, Apr 06 2001; b-file added Nov 27 2007

Keywords

Comments

The (3x+1)/2 steps and the halving steps are counted. - Don Reble, May 13 2006
Where records occur in A102419 (could be prefixed by an initial 1). - N. J. A. Sloane, Oct 20 2012

Examples

			See A102419.
		

Crossrefs

A060413 gives associated "dropping times", A060414 the maximal values and A060415 the steps at which the maxima occur. See also A217934.

Programs

  • Mathematica
    dcoll[n_]:=Length[NestWhileList[If[EvenQ[#],#/2,3#+1]&,n,#>=n&]]; t={max=2}; Do[If[(y=dcoll[n])>max,max=y; AppendTo[t,n]],{n,3,1130000,4}]; t (* Jayanta Basu, May 28 2013 *)

A060445 "Dropping time" in 3x+1 problem starting at 2n+1 (number of steps to reach a lower number than starting value). Also called glide(2n+1).

Original entry on oeis.org

0, 6, 3, 11, 3, 8, 3, 11, 3, 6, 3, 8, 3, 96, 3, 91, 3, 6, 3, 13, 3, 8, 3, 88, 3, 6, 3, 8, 3, 11, 3, 88, 3, 6, 3, 83, 3, 8, 3, 13, 3, 6, 3, 8, 3, 73, 3, 13, 3, 6, 3, 68, 3, 8, 3, 50, 3, 6, 3, 8, 3, 13, 3, 24, 3, 6, 3, 11, 3, 8, 3, 11, 3, 6, 3, 8, 3, 65, 3, 34, 3, 6, 3, 47, 3, 8, 3, 13, 3, 6, 3, 8, 3
Offset: 0

Views

Author

N. J. A. Sloane, Apr 07 2001

Keywords

Comments

If the starting value is even then of course the next step in the trajectory is smaller (cf. A102419).
The dropping time can be made arbitrarily large: If the starting value is of form n(2^m)-1 and m > 1, the next value is 3n(2^m)-3+1. That divided by 2 is 3n(2^(m-1))-1. It is bigger than the starting value and of the same form - substitute 3n -> n and m-1 -> m, so recursively get an increasing subsequence of m odd values. The dropping time is obviously longer than that. This holds even if Collatz conjecture were refuted. For example, m=5, n=3 -> 95, 286, 143, 430, 215, 646, 323, 970, 485, 1456, 728, 364, 182, 91. So the subsequence in reduced Collatz variant is 95, 143, 215, 323, 485. - Juhani Heino, Jul 21 2017

Examples

			3 -> 10 -> 5 -> 16 -> 8 -> 4 -> 2, taking 6 steps, so a(1) = 6.
		

Crossrefs

A060565 gives the first lower number that is reached. Cf. A060412-A060415, A217934.
See A074473, A102419 for other versions of this sequence.
Cf. A122437 (allowable dropping times), A122442 (least k having dropping time A122437(n)).
Cf. A070165.

Programs

  • Haskell
    a060445 0 = 0
    a060445 n = length $ takeWhile (>= n') $ a070165_row n'
                where n' = 2 * n + 1
    -- Reinhard Zumkeller, Mar 11 2013
    
  • Mathematica
    nxt[n_]:=If[OddQ[n],3n+1,n/2]; Join[{0},Table[Length[NestWhileList[nxt, n,#>=n&]]-1, {n,3,191,2}]]  (* Harvey P. Dale, Apr 23 2011 *)
  • Python
    def a(n):
        if n<1: return 0
        n=2*n + 1
        N=n
        x=0
        while True:
            if n%2==0: n//=2
            else: n = 3*n + 1
            x+=1
            if nIndranil Ghosh, Apr 22 2017

Extensions

More terms from Jason Earls, Apr 08 2001 and from Michel ten Voorde Apr 09 2001
Still more terms from Larry Reeves (larryr(AT)acm.org), Apr 12 2001
Showing 1-3 of 3 results.