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Let f(2n+1) = A000265(3n+2) be defined as in A159885. Then a(n) is the least number k of iterations such that either f^k(2n+1) < 2n+1 or A000120(f^k(2n+1)) < A000120(2n+1).

Using induction, one can prove that the Collatz (3x+1)-conjecture follows from the finiteness of a(n) for every n. - Vladimir Shevelev, May 05 2009

Let f be the operation defined in A159885, namely f(2n+1) = A075677(n+1), and f^k its k-fold iteration.

Then a(n) is the smallest k such that either f^k(2n+1)< 2n+1 or A006694((f^k(2n+1)-1)/2) < A006694(n).

Note that the Collatz conjecture is assumed. Otherwise there may exist (very large) odd numbers for which no finite dropping time exists

This sequence is obtained from the residue classes modulo 256 of the entries a(1) to a(19): 27, 31, 47, 63, 71, 91, 103, 111, 127, 155, 159, 167, 191, 207, 223, 231, 239, 251, 255. See the Klee-Wagon reference where the function C is C(2*n+1) = A075677(n+1) = A000265(3*n+2), for n >= 0, and dropping time is called there stopping time (on p. 225 Exercise 4 should be Exercise 5 given on p. 229, with hints on p. 309; also in the first line after (1) 'less than 5' should be replaced by 'not exceeding 5').

The values of (a(j)-1)/2, for j = 1..19, are 13, 15, 23, 31, 35, 45, 51, 55, 63, 77, 79, 83, 95, 103, 111, 115, 119, 125, 127.

The dropping times are 5 for the seven residue classes modulo 256 of 39, 79, 95, 123, 175, 199, and 219. See Klee-Wagon, Exercise 5 (a), p. 229.

The dropping times for a(n) are given in A324249(n).

The odd numbers with dropping time >= 6 under reduced Collatz iteration are given in A324248. Note that the dropping times do not follow the modulo 256 pattern of A324248.

Note that the Collatz conjecture is assumed. Otherwise there may exist (very large) odd numbers for which no finite dropping time exists.

Conjecture. For every n>=1, there exists a finite value of a(n). It is easy to see that this conjecture is equivalent to the well known Collatz 3n+1 conjecture.

Let f be defined as in A159885. Then a(n) is the least k such that either f^k(2n+1))<2n+1 or A000120(f^k(2n+1)) < A000120(2n+1) or A006694((f^k(2n+1)-1)/2) < A006694(n).

In connection with A160198, A160267, A160322 we pose a new (3x+1)-problem: does there exist a finite number of sequences A_i(n), i=1,...,T, such that: 1) A_i(0)=0 and A_i(n)>0 for n>=1; 2) if B_i(n) denotes the least k for which A_i(n)>A_i((f^k(2n+1)-1)/2), then B(n)=min_{i=1,...,T}B_i(n)=1 for every n>=1? Note that this problem is weaker than (3x+1)-Collatz problem. Indeed, if the Collatz conjecture is true, then there exist nonnegative sequences A(n) for which A(0)=0 and A(n)>A((f(2n+1)-1)/2) for every n>=1 (see A160348). - Vladimir Shevelev, May 15 2009

If the (3x+1)-Collatz conjecture is true, then this sequence is a permutation of the nonnegative integers.

A unimodal Collatz glide sequence is successive rises x -> (3x+1)/2 followed by successive falls x -> x/2 until dropping below its starting x.

After n increases, there are ceiling(n*log(3)/log(2) - n) decreases to drop below the initial value.

The trajectory starts with a minimum for odd n and with a maximum (see A351974) for even n (>=2). Since the trajectory always stops at 1 (a minimum) assuming the Collatz conjecture holds, a(n) is odd if n is odd and vice versa.