The snake and the ladder 2

using generating functions it is a simple one-liner in Mathematica

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Coefficient[
 Expand[Sum[(x + x^2 + x^3 + x^4 + x^5 + x^6)^t, {t, 1, 35}]], x^35]

To break it down, after t throws of the die, the number of ways of arriving at step n is given by the expansion of (x+x^2+x^3+x^4+x^5+x^6)^t

After 35 moves he is going to be on or past step step 35, then we need to sum the above for t=1 to 35. After that we simply find the coefficient of x^35.

let $f(m)$ be the number of ways for the snake to reach $m$ steps.

We can see that : $f(m+6) = f(m+5) + f(m+4) + f(m+3) + f(m+2) + f(m+1) + f(m)$

Here is a solution in python 3.4:

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def solve(n):
    " The snake and the ladder 2 "
    fm = [1]
    for i in range(n):
        fm.append(sum(fm[-6:]))
    return fm[n]

print(solve(35))