Detailed mathematical approach

Problem Summary

The problem asks for the tricoloured coin-fountain count \(T(20000)\) modulo \(10^9\). Exhaustive generation is completely impractical at this size, so the solution works with truncated generating functions and formal power-series algebra instead of exploring fountain states directly.

Mathematical Approach

The computation is organized around one auxiliary series \(F_2(x)\). After that series is known to the required degree, a second rational transformation produces the generating function whose coefficients are the desired values \(T(n)\).

Step 1: Start from the continued fraction

The auxiliary series is

$F_2(x)=\frac{1}{1-\frac{x^2}{1-\frac{x^3}{1-\frac{x^4}{\ddots}}}}.$

This continued fraction is the compact object behind the counting problem. Expanding it directly to degree \(20000\) would be awkward, so the implementations switch to an equivalent quotient of two explicit \(q\)-series.

Step 2: Replace it by a quotient of two \(q\)-series

Define the finite \(q\)-Pochhammer product

$(x;x)_m=\prod_{k=1}^{m}(1-x^k).$

The key identity used by the implementations is

$F_2(x)=\frac{\sum_{m\ge 0}(-1)^m\dfrac{x^{m(m+2)}}{(x;x)_m}}{\sum_{m\ge 0}(-1)^m\dfrac{x^{m(m+1)}}{(x;x)_m}}.$

So the continued fraction is converted into a numerator series and a denominator series. For a truncation to degree \(N\), only finitely many indices \(m\) contribute: once both \(m(m+1)\) and \(m(m+2)\) exceed \(N\), that index can no longer affect any coefficient up to \(x^N\).

Step 3: Expand \(1/(x;x)_m\) incrementally

The reciprocal products are updated one step at a time through

$\frac{1}{(x;x)_{m+1}}=\frac{1}{(x;x)_m}\cdot\frac{1}{1-x^{m+1}}=\frac{1}{(x;x)_m}\left(1+x^{m+1}+x^{2(m+1)}+\cdots\right).$

This means that once the truncated coefficients of \(1/(x;x)_m\) are known, the next one is obtained by adding shifted copies spaced by \(m+1\). The code therefore carries a single evolving truncated series instead of rebuilding each reciprocal product from scratch.

Step 4: Recover coefficients by formal series division

Suppose

$A(x)=\frac{N(x)}{D(x)},\qquad D(x)=1+\sum_{i\ge 1} d_i x^i,\qquad N(x)=\sum_{n\ge 0} n_n x^n.$

If

$A(x)=\sum_{n\ge 0} a_n x^n,$

then comparing coefficients in \(A(x)D(x)=N(x)\) gives

$a_0=n_0,\qquad a_n=n_n-\sum_{i=1}^{n} d_i a_{n-i}\pmod{10^9}\quad(n\ge 1).$

Because the constant term of the denominator is \(1\), every new coefficient depends only on earlier ones. The implementations use this recurrence first to recover \(F_2(x)\), and later to recover the final target series.

Step 5: Transform \(F_2(x)\) into the target sequence

After \(F_2(x)\) has been computed to the required degree, the target series is

$t(x)=\frac{x(2F_2(x)-1)}{1-2x(2F_2(x)-1)}.$

The desired counting sequence is then extracted through

$T(n)=3\,[x^n]\,t(x),$

where \([x^n]\) denotes the coefficient of \(x^n\). So the whole computation is: build the two \(q\)-series for \(F_2\), divide them, form the numerator and denominator of \(t(x)\), divide again, read the coefficient of \(x^n\), and multiply by \(3\).

Worked Example: Recover \(T(4)\)

To degree \(4\), the continued fraction contributes only enough to give

$F_2(x)=1+x^2+x^4+O(x^5).$

Hence

$A(x)=2F_2(x)-1=1+2x^2+2x^4+O(x^5),$

$N(x)=xA(x)=x+2x^3+O(x^5),$

$D(x)=1-2xA(x)=1-2x-4x^3+O(x^5).$

Writing \(t(x)=\sum_{n\ge 0} t_n x^n\), the division recurrence yields

$t_0=0,\qquad t_1=1,\qquad t_2=2,\qquad t_3=6,\qquad t_4=16.$

Therefore

$T(4)=3\,t_4=48,$

which matches the low-degree checkpoint used by the implementations.

How the Code Works

The C++, Python, and Java implementations all follow the same mathematical pipeline. The truncated numerator and denominator of \(F_2(x)\) are accumulated term by term, with alternating signs, while a rolling truncated series stores the current reciprocal product \(1/(x;x)_m\). After the first formal division, the code builds the numerator and denominator of \(t(x)\) from the coefficients of \(F_2(x)\) and applies the same division recurrence again.

The final answer is the coefficient of \(x^{20000}\) in \(t(x)\), multiplied by \(3\) and reduced modulo \(10^9\). The implementations also perform small internal checks: they compare the \(q\)-series coefficients of \(F_2\) with a direct low-degree continued-fraction expansion, and they verify checkpoint values such as \(T(4)=48\) and \(T(10)=17760\). The Java implementation delegates the heavy computation to the compiled implementation of the same formulas.

Complexity Analysis

Let \(N\) be the truncation degree. Building the two \(q\)-series for \(F_2(x)\) takes \(O(N\sqrt{N})\) coefficient updates because only \(m\le O(\sqrt{N})\) contribute, and each contributing index touches \(O(N)\) coefficients. The dominant cost is the two formal series divisions, each of which is \(O(N^2)\). Therefore the overall running time is \(O(N^2)\), and the memory usage is \(O(N)\).

Footnotes and References

1. Problem page: https://projecteuler.net/problem=519
2. Formal power series: Wikipedia — Formal power series
3. \(q\)-Pochhammer symbol: Wikipedia — q-Pochhammer symbol
4. Continued fraction: Wikipedia — Continued fraction