Hailstone collatz max sequence length upper bound of $260.5+x^{.43}$? Let the Collatz function be defined as if $x$ even $c(x)=x/2$, if $x$ odd then $c(x)=3x+1$ over the naturals. Each operation is defined as a step. For example $3$ goes $(3,10,5,16,8,4,2,1)$ and takes 8 steps to reach one. **Can anyone find an example for me where the number of steps for starting natural number $x$ to reach one is greater then $260.5+x^.43$?** Here is a graph with the max number of collatz length and the fitted upper bound }$) **I conjecture that there is no such number.** Is there any other works of this type?

I've searched up to $10^8$ and the closest you get to your bound is at $230631$ which appears in your first graph. Nothing else is even above $240 + x^{.43}$ let alone your bound. $$\\{x \in \Bbb{Z}|\quad0 240 + x^{.43} \\} = \\{230631\\}$$

As was mentioned in the comments. There are no known results of the type you are asking. I did the search to answer your query for an in-depth search.


#include
#include
#include
#define u64 long long unsigned

u64 *c;
u64 N = 100000000LL;
u64 collatz( u64 a )
{
if( a < N && c[a] ) return c[a];
u64 t = 1 + collatz( 1&a ? 3*a + 1 : a >> 1 );
return ( a < N ? c[a] = t : t );
}

int main()
{
c = calloc( N, sizeof(u64) );
c[1] = 1;
for( u64 i = 2; i < N; ++i )
if( collatz( i ) >= (u64) ceil( 240. + pow( i, 0.43 ) ) )
printf( "%llu\
", i );
}