Detailed mathematical approach

Problem Summary

For each odd prime \(p \lt 10^6\), Problem 826 leads to a single problem-specific quantity from the birds-on-a-wire setting. The C++, Python, and Java implementations do not simulate wire configurations directly. Instead, they rely on the already derived closed form

$B(p)=\frac{7p+15}{18(p+1)}.$

The final answer is the arithmetic mean of \(B(p)\) over all odd primes below \(10^6\). Once that formula is known, the job becomes purely computational: generate the relevant primes, evaluate the rational expression for each one, and average the results.

Mathematical Approach

Let

$\mathcal P=\{p: 3 \le p \lt 10^6,\ p \text{ is prime}\}.$

The required value is

$A=\frac{1}{|\mathcal P|}\sum_{p\in\mathcal P} B(p).$

The mathematics used by the implementations is the reduction of the original combinatorial quantity to this prime-indexed rational function.

Step 1: Start from the Closed Formula

For every admissible prime \(p\), the quantity attached to the problem is already known in exact form:

$B(p)=\frac{7p+15}{18(p+1)}.$

This is the decisive simplification. There is no remaining state space to enumerate, no random process to simulate, and no recursive structure to evaluate. The whole problem collapses to averaging one rational expression over a specified set of primes.

Step 2: Describe the Averaging Set Precisely

The implementations average only over odd primes strictly smaller than \(10^6\). That is why the set begins at \(3\) and why \(p=2\) is skipped.

So the answer is not a weighted sum, and it is not an average over all integers. It is the simple arithmetic mean

$A=\frac{B(p_1)+B(p_2)+\cdots+B(p_k)}{k},$

where \(p_1,p_2,\dots,p_k\) are exactly the primes in \(\mathcal P\).

Step 3: Separate the Main Term from the Correction Term

A useful algebraic rewrite is

$B(p)=\frac{7p+15}{18(p+1)}=\frac{7(p+1)+8}{18(p+1)}=\frac{7}{18}+\frac{4}{9(p+1)}.$

This makes the structure transparent. Most of the value comes from the constant term \(7/18\), while the dependence on \(p\) is entirely contained in the smaller correction term \(4/(9(p+1))\).

Because \(p\ge 3\), the correction is always positive and at most \(1/9\). Therefore

$\frac{7}{18} \lt B(p)\le \frac{1}{2}.$

The upper endpoint occurs at the smallest allowed prime \(p=3\), and larger primes push the value downward toward \(7/18\).

Step 4: Rewrite the Final Average

Substituting the decomposition into the mean gives

$A=\frac{1}{|\mathcal P|}\sum_{p\in\mathcal P}\left(\frac{7}{18}+\frac{4}{9(p+1)}\right)=\frac{7}{18}+\frac{4}{9|\mathcal P|}\sum_{p\in\mathcal P}\frac{1}{p+1}.$

This identity is mathematically simple but conceptually useful. It shows that the answer is a fixed baseline \(7/18\) plus the average of a decaying reciprocal correction. In particular, the result must lie only slightly above \(7/18\), because \(1/(p+1)\) becomes small very quickly.

Worked Example: Averaging over \(p=3,5,7\)

If we temporarily replace the full prime set by the smaller set \(\{3,5,7\}\), then

$B(3)=\frac{7\cdot 3+15}{18\cdot 4}=\frac{36}{72}=\frac{1}{2},$

$B(5)=\frac{7\cdot 5+15}{18\cdot 6}=\frac{50}{108}=\frac{25}{54},$

$B(7)=\frac{7\cdot 7+15}{18\cdot 8}=\frac{64}{144}=\frac{4}{9}.$

Their arithmetic mean is

$\frac{1}{3}\left(\frac{1}{2}+\frac{25}{54}+\frac{4}{9}\right)=\frac{1}{3}\cdot\frac{76}{54}=\frac{38}{81}\approx 0.4691358025.$

The same procedure is what the implementations apply to the full set of odd primes below \(10^6\); only the size of the prime set changes.

How the Code Works

The C++, Python, and Java implementations all follow the same numerical pipeline. First they build a sieve of Eratosthenes up to \(10^6\), which marks exactly which integers are prime.

They then iterate through the valid odd primes, evaluate

$\frac{7p+15}{18(p+1)}$

for each one, add that value to a running sum, and increment a counter. After the loop finishes, they divide the accumulated sum by the number of processed primes to obtain the mean.

One implementation also includes a small checkpoint at \(p=3\), verifying numerically that the formula gives \(1/2\). Finally, all three implementations print the result with ten digits after the decimal point.

Complexity Analysis

Let \(n=10^6\). Building the sieve costs \(O(n\log\log n)\) time and \(O(n)\) memory. The averaging pass is linear in the number of integers scanned, or equivalently \(O(\pi(n))\) arithmetic evaluations over the primes themselves. Since the closed formula is constant-time to evaluate, the sieve dominates the running time, and the overall memory usage stays \(O(n)\).

Footnotes and References

1. Problem page: https://projecteuler.net/problem=826
2. Sieve of Eratosthenes: Wikipedia — Sieve of Eratosthenes
3. Prime number: Wikipedia — Prime number
4. Arithmetic mean: Wikipedia — Arithmetic mean
5. Rational function: Wikipedia — Rational function