Problem Summary

For each pair \((n,k)\) with \(1 \le k \le n\), the circle-of-coins problem has a counting function \(F(n,k)\) for the admissible binary coin configurations associated with that pair. The overall task is to evaluate

$$S(N)=\sum_{n=1}^{N}\sum_{k=1}^{n} F(n,k)\pmod{10^9+7}.$$

The key point in the implementations is that the combinatorics collapse to arithmetic data: only \(\gcd(n,k)\), the exponent of \(2\) dividing \(n\) and \(k\), and a totient-based regrouping are needed. That removes any need to enumerate coin states directly.

Mathematical Approach

The C++, Python, and Java implementations all use the same arithmetic decomposition. Once the fixed-pair formula is known, the remaining work is to reorganize the double sum so that equal contributions are collected together efficiently.

Step 1: Closed Form for a Fixed Pair

Let

$$g=\gcd(n,k),$$

and let \(v_2(x)\) denote the exponent of \(2\) in \(x\). The implementations use the closed form

$$F(n,k)=\begin{cases} 2^{n-g+1}, & v_2(k)\le v_2(n),\\ 2^{n-g}, & v_2(k)>v_2(n). \end{cases}$$

So the entire dependence on the original circle is compressed into two quantities: the common divisor \(g\), which controls the main exponent \(n-g\), and a single parity-sensitive comparison of 2-adic valuations, which decides whether there is one extra factor of \(2\).

Step 2: Separate the GCD from the Reduced Step

Write

$$n=t h,\qquad k=t r,\qquad t=\gcd(n,k),\qquad \gcd(r,h)=1.$$

Then \(t\) is exactly the gcd that appears in the fixed-pair formula, so

$$n-g=t h-t=t(h-1).$$

After removing the common factor \(t\), every pair \((n,k)\) is described by a reduced denominator \(h\) and a reduced residue \(r\) coprime to \(h\). The formula becomes

$$F(t h,t r)=\begin{cases} 2^{t(h-1)+1}, & v_2(r)\le v_2(h),\\ 2^{t(h-1)}, & v_2(r)>v_2(h). \end{cases}$$

This shows that, for fixed \(h\) and \(t\), all dependence on the reduced step is packed into how many coprime residues \(r\) satisfy the valuation test.

Step 3: Count the Reduced Residues for Fixed \(h\)

Now fix \(h \ge 2\) and sum over all \(r\) with \(1 \le r \le h\) and \(\gcd(r,h)=1\).

If \(h\) is even, every residue coprime to \(h\) must be odd. Hence \(v_2(r)=0\le v_2(h)\), so every reduced residue contributes the larger value \(2^{t(h-1)+1}\). Since there are \(\varphi(h)\) such residues, their total contribution is

$$\varphi(h)\cdot 2^{t(h-1)+1}=2\varphi(h)\,2^{t(h-1)}.$$

If \(h\) is odd, the reduced residues split evenly into odd and even values. Indeed, for every reduced residue \(r\), the paired residue \(h-r\) is also reduced and has opposite parity. Therefore exactly half of the \(\varphi(h)\) residues satisfy \(v_2(r)=0=v_2(h)\), while the other half fail the inequality.

Step 4: Derive the Coefficient \(c(h)\)

From the parity split above, the total contribution for one fixed \(h \ge 2\) and one fixed \(t\) is

$$c(h)\,2^{t(h-1)},$$

where

$$c(h)=\begin{cases} 2\varphi(h), & h \text{ even},\\ \dfrac{3\varphi(h)}{2}, & h \text{ odd}. \end{cases}$$

The odd case is just

$$\frac{\varphi(h)}{2}\cdot 2^{t(h-1)+1}+\frac{\varphi(h)}{2}\cdot 2^{t(h-1)} =\frac{3\varphi(h)}{2}\,2^{t(h-1)}.$$

This is the crucial compression step: once pairs are grouped by \(h\), all the messy dependence on \(k\) is replaced by a single totient-based coefficient.

Step 5: Rebuild the Whole Sum

The case \(h=1\) is special. Then \(k=n\), so \(g=n\) and the fixed-pair formula gives \(F(n,n)=2\). Summed over \(n=1,\dots,N\), this contributes

$$2N.$$

For every \(h \ge 2\), the remaining parameter is \(t\), and the condition \(n=t h \le N\) means

$$1\le t\le \left\lfloor\frac{N}{h}\right\rfloor.$$

Therefore the full sum becomes

$$\boxed{S(N)=2N+\sum_{h=2}^{N} c(h)\sum_{t=1}^{\lfloor N/h\rfloor} 2^{t(h-1)} \pmod{10^9+7}.}$$

Mathematically the inner sum is a finite geometric progression, but the implementations simply step through its exponents directly after precomputing powers of \(2\).

Worked Example: \(N=3\)

The special term is

$$2N=6.$$

For \(h=2\), we have \(\varphi(2)=1\) and \(c(2)=2\). Also \(\lfloor 3/2\rfloor=1\), so this block contributes

$$2\cdot 2^{1}=4.$$

For \(h=3\), we have \(\varphi(3)=2\) and \(c(3)=3\). Also \(\lfloor 3/3\rfloor=1\), so this block contributes

$$3\cdot 2^{2}=12.$$

Hence

$$S(3)=6+4+12=22,$$

which matches the checkpoint used by the implementation. A single-pair checkpoint is \(F(9,3)=2^{9-3+1}=128\), because \(\gcd(9,3)=3\) and \(v_2(3)\le v_2(9)\).

How the Code Works

The implementations first build Euler's totient values for all integers up to \(N\) with a linear sieve. They also precompute the table

$$2^0,2^1,\dots,2^N \pmod{10^9+7},$$

so every power lookup in the main sum is constant time.

After that, the algorithm starts with the special contribution \(2N\). It then iterates through \(h=2,3,\dots,N\), computes the coefficient \(c(h)\) from the parity of \(h\) and the totient value, and adds

$$c(h)\,2^{t(h-1)}$$

for each \(t=1,\dots,\lfloor N/h\rfloor\). All arithmetic is reduced modulo \(10^9+7\) after each addition. The C++, Python, and Java implementations follow exactly the same mathematical plan; the C++ version also includes small checkpoint assertions before producing the final large-input result.

Complexity Analysis

The totient sieve runs in \(O(N)\) time and uses \(O(N)\) memory. The double summation over blocks costs

$$\sum_{h=2}^{N}\left\lfloor\frac{N}{h}\right\rfloor=O(N\log N).$$

Therefore the total running time is \(O(N\log N)\), while the memory usage remains \(O(N)\).

Footnotes and References

1. Problem page: Project Euler 728 — Circle of Coins
2. Greatest common divisor: Wikipedia — Greatest common divisor
3. Euler's totient function: Wikipedia — Euler's totient function
4. \(p\)-adic valuation, including the special case \(v_2\): Wikipedia — \(p\)-adic valuation
5. Geometric progression: Wikipedia — Geometric progression