# What are the rules here?

1, 5, 1, 3, 1, 1, 1,...

14, 2, 10, 2, 6, 2, 2, 2,...

2, 14, 2, 10, 2, 6, 2, 2, 2,...

9, 3, 3, 3, 3,...

26, 61, 26, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0, 0, 0,...

• What's the source of this puzzle? Commented Jun 12, 2018 at 18:28
• @NathanHinchey No source, just created it myself. Commented Jun 13, 2018 at 6:05

A minor refinement on athin's answer to account for the last sequence:

Denote the last two numbers in the sequence are $A$ and $B$ (initially there are only two numbers).

If $A < B$, append $A$ to the sequence;
else, append the greater of $A-2B$ and 0.

• But, for $1 1$, it should be $1$ not $0$ on the first sequence. Commented Jun 12, 2018 at 0:25
• Good point! My answer does not account for all of them! Commented Jun 12, 2018 at 18:27

It's possible that the rule is:

Denote the last two numbers in the sequence are $A$ and $B$ (initially there are only two numbers).

If $A < B$, append $A$ to the sequence;
else, append $|A-2B|$.

• In this case, the last sequence is wrong! Commented Jun 11, 2018 at 13:25
• you need a second "if": if $A>2B$, append $|A-2B|$, else append $A - 2$ (or $B - 1$?). Commented Jun 11, 2018 at 14:53
• Oh, and if $A=B$, append $A$ (or $B$). Commented Jun 11, 2018 at 14:54
• Whoops I just realized this.. and uh tbh it's rather hacky to have 3 ifs btw :( Commented Jun 12, 2018 at 0:22