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# Problem 277

A Modified Collatz sequence

A modified Collatz sequence of integers is obtained from a starting value a1 in the following way:

an+1 = an/3 if an is divisible by 3. We shall denote this as a large downward step, “D”.

an+1 = (4an + 2)/3 if an divided by 3 gives a remainder of 1. We shall denote this as an upward step, “U”.

an+1 = (2an - 1)/3 if an divided by 3 gives a remainder of 2. We shall denote this as a small downward step, “d”.

The sequence terminates when some an = 1.

Given any integer, we can list out the sequence of steps.
For instance if a1=231, then the sequence {an}={231,77,51,17,11,7,10,14,9,3,1} corresponds to the steps “DdDddUUdDD”.

Of course, there are other sequences that begin with that same sequence
“DdDddUUdDD….”.
For instance, if a1=1004064, then the sequence is
DdDddUUdDDDdUDUUUdDdUUDDDUdDD.
In fact, 1004064 is the smallest possible a1 > 106 that begins with the sequence
DdDddUUdDD.

What is the smallest a1 > 1015 that begins with the sequence
“UDDDUdddDDUDDddDdDddDDUDDdUUDd”?

an+1 = an/3，如果an能被3整除。这是一个较大的下降，我们记这一步操作为“D”。

an+1 = (4an + 2)/3，如果an除以3余1。这是一个较大的上升，我们记这一步操作为“U”。

an+1 = (2an - 1)/3，如果an除以3余2。这是一个较小的下降，我们记这一步操作为“d”。