Detailed mathematical approach

Problem Summary

The snowflake construction leads to two integer sequences \(A(n)\) and \(B(n)\). For each \(n\ge 3\) we define

$G(n)=\gcd(A(n),B(n)),$

and the goal is to evaluate

$\sum_{n=3}^{10^7} G(n).$

The closed forms are easy to write down, but the raw values grow exponentially, so directly building the integers and then taking gcds would be far too expensive. The key is to reduce every gcd to one involving only numbers of size about \(7n\).

Mathematical Approach

The implementation starts from explicit formulas for the two snowflake counts and then transforms the gcd into a much smaller modular problem.

Step 1: Closed Forms from the Snowflake Construction

The formulas used by the implementation are

$A(n)=3\cdot 4^{n-1}-2\cdot 3^{n-1},$

$B(n)=\frac{(9n-69)\cdot 4^{n-1}+2(4n+26)\cdot 3^{n-1}}{2}.$

These are exact integer expressions for the two quantities whose gcd is needed. The problem is not the formulas themselves, but the fact that \(4^{n-1}\) and \(3^{n-1}\) are enormous when \(n\) approaches \(10^7\).

Step 2: Factor Out the Common Multiple \(6\)

Write

$U=4^{n-2},\qquad V=3^{n-2}.$

Then both closed forms simplify cleanly:

$A(n)=6(2U-V),$

$B(n)=6\big((3n-23)U+(2n+13)V\big).$

If we define

$S=2U-V,\qquad T=(3n-23)U+(2n+13)V,$

then

$G(n)=\gcd(A(n),B(n))=6\gcd(S,T).$

This already removes one fixed factor from every term in the sum.

Step 3: Use a Determinant Identity to Force a Small Divisor

The pair \((S,T)\) is a linear transformation of \((U,V)\). Two useful combinations are

$ (2n+13)S+T=(7n+3)U, $

$ 2T-(3n-23)S=(7n+3)V. $

Therefore any common divisor \(d\) of \(S\) and \(T\) must divide both \((7n+3)U\) and \((7n+3)V\).

But \(U=4^{n-2}\) and \(V=3^{n-2}\) are powers of coprime bases, so

$\gcd(U,V)=1.$

Hence every common divisor of \(S\) and \(T\) must divide

$7n+3.$

Equivalently, the determinant of the coefficient matrix

$\begin{pmatrix}2 & -1 \\ 3n-23 & 2n+13\end{pmatrix}$

is exactly

$2(2n+13)-(-1)(3n-23)=7n+3,$

and that determinant controls the gcd.

Step 4: Prove the Reduction Is Exact

We now show that no information is lost when replacing \(T\) by \(7n+3\). Suppose \(d\) divides both \(S\) and \(7n+3\). Since \(S=2U-V\), we have

$V\equiv 2U \pmod d.$

Substituting this into \(T\) gives

$T\equiv (3n-23)U+(2n+13)(2U)=(7n+3)U\equiv 0 \pmod d.$

So every common divisor of \(S\) and \(7n+3\) is also a divisor of \(T\). Combining this with Step 3 yields

$\gcd(S,T)=\gcd(S,7n+3).$

Therefore the exact formula becomes

$\boxed{G(n)=6\gcd\left(7n+3,\ 2\cdot 4^{n-2}-3^{n-2}\right).}$

Step 5: Replace Huge Powers by Modular Powers

Let

$M=7n+3.$

The gcd only depends on the second argument modulo \(M\), so we only need the residue

$R\equiv 2\cdot 4^{n-2}-3^{n-2}\pmod M,\qquad 0\le R<M.$

Then

$G(n)=6\gcd(M,R).$

This is the crucial speedup: instead of manipulating exponentially large integers, we compute two modular powers and one gcd for each \(n\).

Step 6: Final Summation Formula

The entire problem is therefore reduced to

$\sum_{n=3}^{10^7} 6\gcd(M,R),\qquad M=7n+3,\qquad R\equiv 2\cdot 4^{n-2}-3^{n-2}\pmod M.$

That expression is exactly what the implementations evaluate.

Worked Example: \(n=11\)

For \(n=11\), the reduced modulus is

$M=7\cdot 11+3=80.$

Now compute the two powers only modulo \(80\):

$4^9\equiv 64 \pmod{80},\qquad 3^9\equiv 3 \pmod{80}.$

So

$R\equiv 2\cdot 64-3=125\equiv 45 \pmod{80}.$

Hence

$G(11)=6\gcd(80,45)=6\cdot 5=30.$

This agrees with the exact values

$A(11)=3027630,\qquad B(11)=19862070,$

whose gcd is indeed \(30\).

How the Code Works

The C++, Python, and Java implementations all follow the same loop from \(n=3\) to \(10^7\). For each \(n\), they first form the modulus \(7n+3\). They then use binary exponentiation to evaluate \(4^{n-2}\bmod (7n+3)\) and \(3^{n-2}\bmod (7n+3)\) efficiently, combine those residues into the reduced value \(R\), compute \(\gcd(7n+3,R)\), multiply by \(6\), and add the result to the running total.

The C++ implementation adds a practical optimization: it splits the interval into disjoint blocks and lets several threads process different ranges in parallel, then combines the partial sums at the end. The Python and Java implementations keep the same mathematics but use a direct single-loop version of the algorithm.

Complexity Analysis

For one value of \(n\), the dominant work is modular exponentiation with exponent \(n-2\), which costs \(O(\log n)\) time by repeated squaring. Summing over all \(n\le N\) gives total time

$O\!\left(\sum_{n=3}^{N}\log n\right)=O(N\log N).$

The memory usage is \(O(1)\) for the serial versions. The threaded C++ version uses \(O(T)\) extra memory for \(T\) partial sums and worker state, which is still constant with respect to \(N\).

Footnotes and References

1. Problem page: Project Euler 570
2. Greatest common divisor: Wikipedia - Greatest common divisor
3. Euclidean algorithm: Wikipedia - Euclidean algorithm
4. Modular exponentiation: Wikipedia - Modular exponentiation
5. Exponentiation by squaring: Wikipedia - Exponentiation by squaring