# Ponder This

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## February 2017 - Solution

February 2017 Challenge

Trying to run the program yields very large numbers, but note that since we are only interested in s modulo 7, we can make the computations modulo 7.
We are actually implementing a finite automaton. From the polynomial, we get that transition as

```Current s	0 1 2 3 4 5 6
New s if b=0	6 3 5 3 6 3 1
New s if b=1	4 0 4 0 4 4 4
```
The states actually represent:

0 last 3 bits are 001
1 last 3 bits are 100
2 initial state
3 last 2 bits are 00
4 last bit is 1
5 last bit is 0
6 last 2 bits are 10

And what this finite automaton is counting is the number of times the 1001 pattern appears, minus the number of times the pattern 0010 appears.
So we are looking for the number of binary sequences where these two patterns appear a different number of times.
This can be computed using dynamic programming, which produces the result 3271652365264 for 42 bits.
As for the bonus: we get 8296 for 14 bits, but when solving for the complementary problem (the number of sequences where these two patterns appear the same number of times), we get 8088, which is the record breaking number of patents IBM received in 2016.