Project Euler Lab - Problem 624

#624 - Two Heads Are Better Than One

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An unbiased coin is tossed repeatedly until two consecutive heads are obtained. Suppose these occur on the \((M-1)\)th and \(M\)th toss.
Let \(P(n)\) be the probability that \(M\) is divisible by \(n\). For example, the outcomes HH, HTHH, and THTTHH all count towards \(P(2)\), but THH and HTTHH do not.

You are given that \(P(2) =\frac 3 5\) and \(P(3)=\frac 9 {31}\). Indeed, it can be shown that \(P(n)\) is always a rational number.

For a prime \(p\) and a fully reduced fraction \(\frac a b\), define \(Q(\frac a b,p)\) to be the smallest positive \(q\) for which \(a \equiv b q \pmod{p}\).
For example \(Q(P(2), 109) = Q(\frac 3 5, 109) = 66\), because \(5 \cdot 66 = 330 \equiv 3 \pmod{109}\) and \(66\) is the smallest positive such number.
Similarly \(Q(P(3),109) = 46\).

Find \(Q(P(10^{18}),1\,000\,000\,009)\).

This problem is taken from Project Euler, Problem 624.
Problem text © Project Euler, licensed under CC BY-NC-SA 4.0. Original: projecteuler.net/problem=624. Published Saturday, 7th April 2018, 01:00 pm. Solved by 772 members at time of mirroring.

Why this is useful

Probability Statistics. Expectation and state-based probability reasoning underpin pricing, risk, and statistical inference (Phases 7, 11, 13).

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