How good is your arithmetic modulo 2017?

One solution is to find a general form for $B_n$ :

We have the characteristic polynomial for $B_n$ as $x^3-3x^2+4=0$ .

We may try a few integers as roots and find that $x = -1$ is a root. We reduce the polynomial to a quadratic via long division:

$x^3-3x^2+4 = (x+1)\frac{x^3-3x^2+4}{x+1} = (x+1)\frac{x^3+x^2-4x^2 - 4x + 4x + 4}{x+1}$

$= (x+1)\Bigg(\frac{x^3+x^2}{x+1} - \frac{4x^2+4x}{x+1} + \frac{4x+4}{x+1}\Bigg) = (x+1)(x^2-4x+4) = (x+1)(x-2)^2$

Due to the nature of the characteristic polynomial, this implies that

$B_n = a2^n+bn2^n+c(-1)^n$ for some constants $a, b, c$ . We are given initial values for $B_n$ , so we substitute:

$a2^0+b(0)2^0+c(-1)^0 = a + c = 6$

$a2^1+b(1)2^1+c(-1)^1 = 2a+2b-c = -1$

$a2^2+b(2)2^2+c(-1)^2 = 4a+8b + c = 17$

Solving yields $a = b = 1$ and $c = 5$ . So, the general form of $B_n$ is given by

$B_n = (n+1)2^n+5(-1)^n$

Now, we consider $\sum_{n = 1}^{2018} B_n$ :

$\sum_{n = 1}^{2018} = \sum_{n = 1}^{2018}2^n+n2^n+5(-1)^n = \sum_{n = 1}^{2018}2^n+\sum_{n = 1}^{2018}n2^n+\sum_{n = 1}^{2018}5(-1)^n$

Note that because 2018 is even, the third term is zero, leaving

$\sum_{n = 1}^{2018}B_n = \sum_{n = 1}^{2018}2^n+\sum_{n = 1}^{2018}n2^n$

Recall that, if $x\neq 1$ , then

$\sum_{n = 1}^{q}2^n = 2(2^q-1)$

and (as it can be shown with some simple calculus):

$\sum_{n = 1}^{q}n2^n = (q-1)2^{q+1} + 2$

So,

$\sum_{n = 1}^{2018}B_n = \sum_{n = 1}^{2018}2^n+\sum_{n = 1}^{2018}n2^n$

$=2\frac{2^{2018}-1} + (2017)2^{2019} + 2 = 2^{2019} - 2 + (2017)2^{2019}+2 = (2018)2^{2019}$

We now have that

$\sum_{n = 1}^{2018}B_n\equiv (2018)2^{2019} \equiv 2^{2019} \mod 2017$

Since 2017 is prime, we can invoke Fermat's little theorem:

$2^{2019} \equiv 2^{2017-1}2^3 \equiv 1\cdot 2^3 \equiv 8 \mod 2017$

Therefore, if $B_n$ is defined as above, then

$\sum_{n = 1}^{2018}B_n \equiv 8 \mod 2017$

Fun problem. I simply ran the following C# code:

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            int bn, bnm1, bnm2;
            int bnp1 = 0;
            bnm2 = 6;
            bnm1 = -1;
            bn = 17;
            int count = 3;
            int acc = bnm1+bn;
            while (count-1 != 2018)
            { 
                bnp1 = (3 * (bn) - 4 * (bnm2)) % 2017;
                if (bnp1 < 0) { bnp1 += 2017; }
                bnm2 = bnm1;
                bnm1 = bn;
                bn = bnp1;
                acc += bnp1;
                acc %= 2017;
                count++;
            }
            Console.WriteLine(acc);

The output of this code was 8 .