Rotavirus

Treat the rotavirus as a labelled graph on $N+1$ vertices. By the matrix tree theorem, the number of spanning trees of the graph is given by

$\begin{vmatrix} 3 & -1 & 0 & 0 & \cdots & 0 & -1 \\ -1 & 3 & -1 & 0 & \cdots & 0 & 0 \\ 0 & -1 & 3 & -1 & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\ -1 & 0 & 0 & 0 & \cdots & -1 & 3 \\ \end{vmatrix}$

If we call the above determinant $A_N$ , then we get, after "some" algebraic manipulations, $A_N = L_{2N} - 2$ where $L_k$ is the $k$ th Lucas number. Thus

$A_{100} = \left ( \frac{1+\sqrt{5}}{2} \right )^{200} + \left ( \frac{1-\sqrt{5}}{2} \right )^{200} - 2$ which has $42$ digits.

Honestly, this is Discrete Mathematics.

My solution is much simpler with all respect to @typical beam and @K T . The answer of the universe is 42, so the answer of this problem is 42 as well. Brilliant!

After finding 1,5,16,45, I checked the OEIS A004146 . It said: 'This is the r=5 member in the r-family of sequences S_r(n) defined in A092184 Number of spanning trees of the wheel W_n on n+1 vertices. ....' And under A092184, under FORMULA: 'S_r type sequences are defined by a(0)=0, a(1)=1, a(2)=r and a(n-1)*a(n+1) = (a(n)-1)^2.'

This formula rewrites as $a_n= \frac{(a_{n-1}-1)^2}{a_{n-2}}$ and led me to coding it (see below) with output

S_5(0)=0

S_5(1)=1

S_5(2)=5

S_5(3)=16

...

S_5(100)=627376215338105766356982006981782561278125

The last entry has 42 digits

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def S_r_array(r, upto):
    s=[0,1,r]
    for n in range(3,upto):
        numer = (s[n-1]-1) * (s[n-1]-1)
        denom = s[n-2]
        if numer % denom ==0 :
            newterm=numer//denom
        else:
            newterm=numer/denom
        s.append(newterm)
    return s

s = S_r_array(5,101)
for n in range(101):
    print("S_{}({})={}".format(5,n,s[n]))