Detailed mathematical approach

Problem Summary

Let \(E_n=(\mathbb{Z}_{>0})^n\). For two positive-integer vectors \(u=(u_1,\dots,u_n)\) and \(v=(v_1,\dots,v_n)\), define

$\langle u,v\rangle=\sum_{i=1}^{n}u_iv_i,\qquad u\star v=\prod_{i=1}^{n}(u_i+v_i).$

The quantity \(R_n(M)\) is the sum of \(u\star v\) over all ordered pairs \((u,v)\in E_n\times E_n\) satisfying \(\langle u,v\rangle=M\). The target is

$R_6(10000!) \pmod{10^9+7}.$

A direct enumeration of six-dimensional vector pairs is hopeless, so the solution converts the problem into coefficient extraction from a generating function and then reduces that generating function with quasimodular-form identities.

Mathematical Approach

The computation rests on two ideas: first, each coordinate contributes independently through a one-variable divisor sum; second, the sixth power of the resulting generating series can be rewritten in a weight-12 quasimodular basis whose coefficients are easy to evaluate at \(10000!\).

Step 1: Collapse One Coordinate to a Divisor Sum

For \(n=1\), the condition \(\langle u,v\rangle=M\) is simply \(uv=M\). Every divisor \(d\mid M\) gives the ordered pair \((d,M/d)\), whose contribution to \(u\star v\) is

$d+\frac{M}{d}.$

Therefore

$R_1(M)=\sum_{d\mid M}\left(d+\frac{M}{d}\right)=2\sum_{d\mid M}d=2\sigma_1(M),$

where \(\sigma_1(M)\) is the usual sum-of-divisors function.

Step 2: Turn Six Coordinates into an Additive Convolution

For a general pair of vectors, set

$m_i=u_iv_i \qquad (1\le i\le n).$

Then

$m_1+\cdots+m_n=M,$

and for a fixed decomposition \(M=m_1+\cdots+m_n\), the contributions from the coordinates multiply. Hence

$R_n(M)=\sum_{\substack{m_1+\cdots+m_n=M\\m_i\ge 1}}R_1(m_1)\cdots R_1(m_n).$

If we define the generating series

$G(q)=\sum_{m\ge 1}R_1(m)q^m,$

then this is exactly the \(n\)-fold additive convolution identity

$R_n(M)=[q^M]\,G(q)^n.$

Step 3: Replace the Generating Series by Eisenstein Series

The normalized quasimodular Eisenstein series \(E_2\) satisfies

$E_2(q)=1-24\sum_{m\ge 1}\sigma_1(m)q^m.$

Since \(R_1(m)=2\sigma_1(m)\), we obtain

$G(q)=2\sum_{m\ge 1}\sigma_1(m)q^m=\frac{1-E_2(q)}{12}.$

For the required dimension \(n=6\), this gives

$R_6(M)=[q^M]\left(\frac{1-E_2(q)}{12}\right)^6.$

Because \(M=10000!>0\), the constant term contributes nothing, so the task becomes the extraction of the positive-\(q\) coefficients of powers of \(E_2\).

Step 4: Reduce Powers of \(E_2\) in Weight 12

Write

$D=q\frac{d}{dq}.$

For any series \(f(q)=\sum_{m\ge 0}a_mq^m\), we have

$[q^M]D^r f=M^r a_M.$

The implementation uses the standard quasimodular reductions

$E_2^2=E_4+12DE_2,$

$E_2^3=E_6+9DE_4+72D^2E_2,$

$E_2^4=E_8+8DE_6+\frac{216}{5}D^2E_4+288D^3E_2,$

$E_2^5=E_{10}+\frac{15}{2}DE_8+\frac{240}{7}D^2E_6+144D^3E_4+864D^4E_2,$

$E_2^6=E_{12}-\frac{4608}{24185}\Delta+\frac{36}{5}DE_{10}+30D^2E_8+\frac{720}{7}D^3E_6+\frac{2592}{7}D^4E_4+\frac{10368}{5}D^5E_2.$

Here \(\Delta(q)=\sum_{m\ge 1}\tau(m)q^m\) is the discriminant cusp form. The positive-\(q\) coefficients of the Eisenstein series are

$[q^M]E_2=-24\sigma_1(M),\qquad [q^M]E_4=240\sigma_3(M),\qquad [q^M]E_6=-504\sigma_5(M),$

$[q^M]E_8=480\sigma_7(M),\qquad [q^M]E_{10}=-264\sigma_9(M),\qquad [q^M]E_{12}=\frac{65520}{691}\sigma_{11}(M),$

and

$[q^M]\Delta=\tau(M).$

So every coefficient needed for \((1-E_2)^6\) becomes a linear combination of \(M^r\sigma_{2k-1}(M)\) and \(\tau(M)\).

Step 5: Evaluate the Arithmetic Data at \(M=10000!\)

Let \(N=10000\). For every prime \(p\le N\), Legendre's formula gives the exponent of \(p\) in \(N!\):

$v_p(N!)=\sum_{t\ge 1}\left\lfloor\frac{N}{p^t}\right\rfloor.$

Thus

$10000!=\prod_{p\le 10000}p^{v_p(10000!)}.$

For every odd \(r\in\{1,3,5,7,9,11\}\), multiplicativity gives

$\sigma_r(10000!)=\prod_{p\le 10000}\left(1+p^r+\cdots+p^{r\,v_p(10000!)}\right)=\prod_{p\le 10000}\frac{p^{r(v_p(10000!)+1)}-1}{p^r-1}.$

The Ramanujan tau function is multiplicative as well, so

$\tau(10000!)=\prod_{p^e\parallel 10000!}\tau(p^e),$

with the prime-power recurrence

$\tau(p^e)=\tau(p)\tau(p^{e-1})-p^{11}\tau(p^{e-2}).$

To obtain the prime values \(\tau(p)\), the implementation first builds the sequence \(\tau(n)\) up to \(10000\) from

$\tau(1)=1,\qquad (n-1)\tau(n)=-24\sum_{k=1}^{n-1}\sigma_1(k)\tau(n-k)\qquad (n\ge 2).$

Worked Example: \(R_1(10)\) and \(R_2(3)\)

For \(M=10\), the divisor pairs are \((1,10)\), \((2,5)\), \((5,2)\), and \((10,1)\). Their contributions are \(11\), \(7\), \(7\), and \(11\), so

$R_1(10)=11+7+7+11=36=2(1+2+5+10).$

Now use convolution for \(n=2\) and \(M=3\). The only decompositions into positive parts are \(3=1+2\) and \(3=2+1\). Since \(R_1(1)=2\) and \(R_1(2)=2(1+2)=6\), we get

$R_2(3)=R_1(1)R_1(2)+R_1(2)R_1(1)=2\cdot 6+6\cdot 2=24.$

This small case shows exactly why generating functions turn the problem into a power of one basic series.

How the Code Works

The C++, Python, and Java implementations all follow the same pipeline. First they sieve the primes up to \(10000\) and compute every \(v_p(10000!)\). From those exponents they evaluate \(\sigma_1,\sigma_3,\sigma_5,\sigma_7,\sigma_9,\sigma_{11}\) multiplicatively modulo \(10^9+7\), and they also compute \(10000! \bmod (10^9+7)\) so that the factors \(M,M^2,\dots,M^5\) needed after differentiation are available.

Next the implementation builds \(\sigma_1(n)\) for all \(1\le n\le 10000\), uses the recurrence above to obtain \(\tau(n)\) up to \(10000\), and then extends from prime values to prime powers through the second-order recurrence for \(\tau(p^e)\). A \(2\times 2\) matrix power is used to evaluate that recurrence efficiently for each prime power.

Finally the program substitutes all divisor sums, the \(\tau\) term, and the powers of \(10000!\) into the weight-12 reduction of \((1-E_2)^6\). The last step is division by \(12^6=2985984\), implemented modulo \(10^9+7\) by multiplying with the modular inverse.

Complexity Analysis

Let \(N=10000\). The prime sieve and the factorial valuations cost \(O(N\log\log N)\) time. Building the table of \(\sigma_1(n)\) by divisor accumulation costs \(O(N\log N)\). The recurrence for \(\tau(n)\) is quadratic, \(O(N^2)\), and this is the dominant step in the actual implementations. The remaining prime-product evaluations and prime-power recurrences are lower-order, roughly \(O(\pi(N)\log N)\). Memory usage is \(O(N)\).

Footnotes and References

1. Problem page: https://projecteuler.net/problem=851
2. Divisor function: Wikipedia - Divisor function
3. Eisenstein series: Wikipedia - Eisenstein series
4. Quasimodular form: Wikipedia - Quasimodular form
5. Ramanujan tau function: Wikipedia - Ramanujan tau function
6. Legendre's formula: Wikipedia - Legendre's formula