A sequence with the recurrence f(n)=3*f(n-1)+f(n-2) starts with two 1-digit numbers. The sequence contains the 8-digit number ABCDAECD. A≠0, and A, B, C, D, and E are not necessarily distinct. Find all possible values of ABCDAECD.
I figured if there were an answer or two I could find them by a simple hand search and then share my result.
I was wrong. There are around 27 answers. I don't have time to share them at the moment but I'm shocked by how many there are.
The big question would be why?
Edit: I hope no one thinks the title is rude. The problem is very much reasonable. I was referencing the fact that it seems that the are more solutions than we would expect.
I'll also point out that the solution could occur in an term from 11 to 14 depending on the size of the starting values.
I just realized this accounts for a few of the numbers occurring twice: Sequences that begin for example with 2,2,8,... will be counted as starting with both 2,2 and 2,8.
Edited on January 17, 2017, 1:39 pm
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Posted by Jer
on 2017-01-17 10:48:55 |
Unreasonable
