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.

A224166 Number of halving and tripling steps to reach the last number of the cycle in the Collatz (3x+1) problem for the negative integers (the initial term is counted).

Original entry on oeis.org

2, 2, 5, 3, 5, 6, 5, 4, 12, 5, 7, 7, 10, 5, 10, 5, 18, 13, 8, 5, 24, 8, 19, 8, 18, 11, 8, 6, 11, 11, 22, 6, 29, 18, 14, 14, 18, 9, 14, 6, 18, 25, 9, 9, 23, 20, 17, 9, 27, 18, 12, 12, 17, 9, 18, 7, 30, 12, 12, 12, 18, 23, 20, 7, 38, 30, 21, 18, 20, 15, 20, 15
Offset: 1

Views

Author

Michel Lagneau, Apr 01 2013

Keywords

Examples

			a(10) = 5 because the trajectory of -10 is -10 -> -5 -> -14 -> -7 -> -20 -> -10 and -10 is the last term of the cycle, hence 5 iterations where the first term -10 is counted.

Crossrefs

Cf. A006577, A224183.

Programs

  • Mathematica
    Collatz[n_] := NestWhileList[If[EvenQ[#], #/2, 3 # + 1] &, n, UnsameQ, All]; Table[s = Collatz[-n]; len = Length[s] - 2; If[s[[-1]] == 2, len = len - 1]; len+1, {n, 1, 100}]

Formula

a(n) = A224183(n) + 1.

Extensions

a(1) changed to 2 by Pontus von Brömssen, Jan 24 2021

A224254 Full cycle lengths in the Collatz (3x+1) problem when the negative integers are used.

Original entry on oeis.org

2, 2, 2, 2, 5, 2, 5, 2, 5, 5, 2, 2, 5, 5, 2, 2, 18, 5, 5, 5, 18, 2, 18, 2, 18, 5, 5, 5, 2, 2, 18, 2, 18, 18, 5, 5, 18, 5, 2, 5, 18, 18, 2, 2, 18, 18, 5, 2, 18, 18, 5, 5, 2, 5, 18, 5, 2, 2, 2, 2, 18, 18, 5, 2, 2, 18, 18, 18, 2, 5, 2, 5, 18, 18, 5, 5, 2, 2, 2, 5, 5
Offset: 1

Views

Author

Michel Lagneau, Apr 02 2013

Keywords

Comments

There are other cycles of lengths 2, 5 and 18 if negative integers are used. In Z, it is conjectured that the five values of cycle are 1, 2, 3, 5 and 18 (see A121510).

Examples

			a(1) = 2 because the cycle -1 -> -2 -> -1... contains 2 distinct terms;
a(5) = 5 because the cycle -5 -> -14 -> -7->-20 -> -5 ... contains 5 distinct terms;
a(17) = 18 because the cycle -17 -> -50 -> -25->-74 -> -37 -> -110 -> -55->-164 -> -82 -> -41 -> -122->-61 -> -182 -> -91 -> -272->-136 -> -68 -> -34 -> -17... contains 18 distinct terms.

Links

Crossrefs

Cf. A121510, A224166, A224183.

Programs

  • Python
    import sympy
def A224254(n):
  return next(sympy.cycle_length(lambda x:3*x+1 if x%2 else x//2,-n))[0] # Pontus von Brömssen, Jan 24 2021

Extensions

Data corrected by Pontus von Brömssen, Jan 24 2021

A346369 a(n) is the number of steps the Collatz trajectory of -n takes to reach a cycle, or -1 if no cycle is ever reached.

Original entry on oeis.org

0, 0, 3, 1, 0, 4, 0, 2, 7, 0, 5, 5, 5, 0, 8, 3, 0, 8, 3, 0, 6, 6, 1, 6, 0, 6, 3, 1, 9, 9, 4, 4, 11, 0, 9, 9, 0, 4, 12, 1, 0, 7, 7, 7, 5, 2, 12, 7, 9, 0, 7, 7, 15, 4, 0, 2, 28, 10, 10, 10, 0, 5, 15, 5, 36, 12, 3, 0, 18, 10, 18, 10, 7, 0, 5, 5, 13, 13, 13, 2, 18
Offset: 1

Views

Author

Felix Fröhlich, Jul 14 2021

Keywords

Comments

The analog of A139399 for negative starting values.
Is a(n) > -1 for all n?

Examples

			For n = 5: The trajectory of -5 starts -5, -14, -7, -20, -10, -5, -14, -7, -20, -10, -5, ..., with -5 already being part of the cycle {-5, -14, -7, -20, -10}, so a(5) = 0.
For n = 6: The trajectory of -6 starts -6, -3, -8, -4, -2, -1, -2, -1, -2, -1, -2, ..., reaching a term of the cycle {-2, -1} after 4 steps, so a(6) = 4.

Crossrefs

Cf. A139399, A224166, A224183, A224254.

Programs

  • PARI
    a006370(n) = if(n%2==0, n/2, 3*n+1)
trajectory(n, terms) = my(v=[n]); while(#v < terms, v=concat(v, a006370(v[#v]))); v
a(n) = my(t, v=[]); for(k=1, oo, t=trajectory(-n, k); for(x=1, #t, if(x < #t && t[x]==t[#t], return(x-1))))

Formula

a(n) = A224166(n) - A224254(n) = 1 + A224183(n) - A224254(n). - Pontus von Brömssen, Jul 24 2021
Showing 1-3 of 3 results.

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

Content is available under The OEIS End-User License Agreement.