A139391 Next odd term in Collatz trajectory with starting value n.
1, 1, 5, 1, 1, 3, 11, 1, 7, 5, 17, 3, 5, 7, 23, 1, 13, 9, 29, 5, 1, 11, 35, 3, 19, 13, 41, 7, 11, 15, 47, 1, 25, 17, 53, 9, 7, 19, 59, 5, 31, 21, 65, 11, 17, 23, 71, 3, 37, 25, 77, 13, 5, 27, 83, 7, 43, 29, 89, 15, 23, 31, 95, 1, 49, 33, 101, 17, 13, 35, 107, 9, 55, 37, 113, 19, 29
Offset: 1
Links
- Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
- Friedrich L. Bauer, Der (ungerade) Collatz-Baum, Informatik Spektrum 31 (Springer, April 2008).
- Eric Weisstein's World of Mathematics, Collatz Problem.
- Wikipedia, Collatz conjecture.
- Index entries for sequences related to 3x+1 (or Collatz) problem.
Programs
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Mathematica
a[n_]:=Select[NestWhileList[If[EvenQ[#],#/2,3#+1] &,n,#>1 &],OddQ]; Prepend[Table[If[EvenQ[n],a[n][[1]],a[n][[2]]],{n,2,77}],1] (* Jayanta Basu, May 27 2013 *) -
PARI
a(n) = my(x = if(n%2, 3*n+1, n/2)); x/2^valuation(x, 2); \\ Michel Marcus, Feb 27 2022
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Python
# first formula def A006370(n): return 3*n+1 if n%2 else n//2 def a(n): return x if (x := A006370(n))%2 else a(x) print([a(n) for n in range(1, 78)]) # Michael S. Branicky, Dec 15 2021
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Python
# fourth formula, uses A006370 above def A000265(n): while n%2 == 0: n //= 2 return n def a(n): return A000265(A006370(n)) print([a(n) for n in range(1, 78)]) # Michael S. Branicky, Dec 15 2021
Formula
a(n) = A006370(n) iff n mod 4 = 2;
a(A160967(n)) = 1. - Reinhard Zumkeller, May 31 2009
For odd n, a(n) = a(2*A350091((n-1)/2)+1). - Ruud H.G. van Tol, Dec 17 2021
Sum_{k=1..n} a(k) ~ n^2 / 3. - Amiram Eldar, Aug 26 2024