2

I heard here that Collatz conjecture was checked at least for every first $5 \cdot 10^{18}$ natural numbers, but I cannot find any source or actual information about this. Can anyone help to find out up to which number (or factor of 10) we can now say that Collatz Conjecture is definitely true?

Kusavil
  • 566
  • I t would be better to include the term "consecutive numbers" in the title; I think this is more explanative because it is simple to refer to classes of infinitely many numbers for which the Collatz conjecture is true: if you have an odd number $n_0$ for which the conjecture is true, then iteratively for all $n_{k+1}=4n_k+1$ the conjecture is also true. Written as bitstring we can begin with $n_0=1$, then $n_1=101,n_2=10101,n_3=1010101,...$ are also numbers for which the conjecture is true -but the set is sparse. Thus the benefit of Oliveira's tables is, that they cover contiguous intervals. – Gottfried Helms May 28 '14 at 05:19
  • The same question in 2019: https://math.stackexchange.com/questions/3314430/how-far-has-collatz-conjecture-been-computationally-verified – DaBler Aug 07 '19 at 14:41

3 Answers3

3

The following published paper gives $20\times 2^{58}\approx 5.7646\times 10^{18}$:

Tomás Oliveira e Silva, "Empirical Verification of the 3x+1 and Related Conjectures." In "The Ultimate Challenge: The 3x+1 Problem," (edited by Jeffrey C. Lagarias), pp. 189-207, American Mathematical Society, 2010.

The author's website suggests ongoing work, but the sublink is down.

vadim123
  • 82,796
2

I think that is a few orders of magnitude too low. According to this discussion board, the number is at least $2 \cdot 10^{21}$.

Fengyang Wang
  • 1,784
2

The official number Wolfram Mathworld gives is $5.48×10^{18}$.

ThreeFx
  • 280