gives a way to speed up simulation of the sequence. To jump ahead steps on each iteration (using the function from that section), break up the current number into two parts, (the least significant bits, interpreted as an integer), and (the rest of the bits as an integer). The result of jumping ahead is given by . The values of (or better ) and can be precalculated for all possible -bit numbers , where is the result of applying the function times to , and is the number of odd numbers encountered on the way. For example, if , one can jump ahead 5 steps on each iteration by separating out the 5 least significant bits of a number and using (0...31, 5) = { 0, 3, 2, 2, 2, 2, 2, 4, 1, 4, 1, 3, 2, 2, 3, 4, 1, 2, 3, 3, 1, 1, 3, 3, 2, 3, 2, 4, 3, 3, 4, 5 }, (0...31, 5) = { 0, 2, 1, 1, 2, 2, 2, 20, 1, 26, 1, 10, 4, 4, 13, 40, 2, 5, 17, 17, 2, 2, 20, 20, 8, 22, 8, 71, 26, 26, 80, 242 }. This requires precomputation and storage to speed up the resulting calculation by a factor of , a space–time tradeoff. Modular restrictions For the special purpose of searching for a counterexample to the Collatz conjecture, this precomputation leads to an even more important acceleration, used by Tomás Oliveira e Silva in his computational confirmations of the Collatz conjecture up to large values of . If, for some given and , the inequality holds for all , then the first counterexample, if it exists, cannot be modulo . For instance, the first counterexample must be odd because , smaller than ; and it must be 3 mod 4 because , smaller than . For each starting value which is not a counterexample to the Collatz conjecture, there is a for which such an inequality holds, so checking the Collatz conjecture for one starting value is as good as checking an entire congruence class. As increases, the search only needs to check those residues that are not eliminated by lower values of . Only an exponentially small fraction of the residues survive. For example, the only surviving residues mod 32 are 7, 15, 27, and 31. Syracuse function If is an odd integer, then is even, so with