A350369 a(n) is the length of the longest sequence of consecutive tripling steps in the Collatz (3x+1) sequence beginning at n.

0, 0, 2, 0, 1, 2, 3, 0, 3, 1, 2, 2, 1, 3, 4, 0, 1, 3, 2, 1, 1, 2, 3, 2, 2, 1, 6, 3, 2, 4, 6, 0, 2, 1, 2, 3, 3, 2, 3, 1, 6, 1, 3, 2, 1, 3, 6, 2, 3, 2, 2, 1, 1, 6, 6, 3, 3, 2, 2, 4, 3, 6, 6, 0, 3, 2, 2, 1, 1, 2, 6, 3, 6, 3, 2, 2, 2, 3, 4, 1, 3, 6, 6, 1, 1, 3, 3

Offset: 1

Author

Kevin P. Thompson, Dec 27 2021

Keywords

• nonn
• easy

Comments

"Consecutive tripling steps" are repeated (3x+1)/2 operations that are not interrupted by a second division by 2.

This sequence attempts to measure the largest upward thrust in each Collatz sequence and so is correlated to some degree with the maximum value (A025586) and length (A006577) of Collatz sequences.

If n = 2^x * (2^y*z - 1), then a(n) >= y. - Charles R Greathouse IV, Oct 25 2022

Examples

			The Collatz sequence for n=7 has a streak of 3 consecutive tripling steps (at 7, 11, and 17), so a(7) = 3.
7, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1
^      ^       ^
		

Links

• Kevin P. Thompson, Table of n, a(n) for n = 1..10000
• Index entries for sequences related to 3x+1 (or Collatz) problem

Crossrefs

Cf. A006370, A006577, A006667, A025586, A213215, A221469, A347270, A347409.

Programs

PARI

a(n)=my(c,r); n>>=valuation(n,2); while(n>1, n+=(n+1)/2; if(n%2, c++, r=max(r,c+1); n>>=valuation(n,2); c=0)); max(r,c) \\ Charles R Greathouse IV, Oct 25 2022