# Solution: A Certain Unscientific GCD Sum over Monotone Sequences

Codex (GPT-6): application of totient expansion and multiset counting, combined with quotient grouping and a summatory recurrence, with proof and implementation.

## Divisor expansion and multisets

Using $g=\sum_{d\mid g}\varphi(d)$ , count sequences whose entries are all divisible by $d$ . After dividing by $d$ , the entries lie in $1,\ldots,q$ , where $q=\lfloor n/d\rfloor$ . A non-increasing sequence uniquely represents a multiset of size $m$ . Stars and bars gives $\binom{q+m-1}{m}$ such multisets. Therefore

$$A=\sum_{d=1}^{n}\varphi(d)\binom{\lfloor n/d\rfloor+m-1}{m}$$

For each sequence, the totients of its common divisors sum to its gcd. Thus the formula counts exactly the requested contributions.

## Quotient groups

Let $\Phi(x)=\sum_{d\le x}\varphi(d)$ . A common quotient $q$ on $[l,r]$ contributes

$$\left[\Phi(r)-\Phi(l-1)\right]\binom{q+m-1}{m},\qquad r=\lfloor n/q\rfloor$$

Use a linear sieve through $B=\min(n,2000000)$ . The identity $\sum_{d\mid u}\varphi(d)=u$ gives

$$\Phi(x)=\frac{x(x+1)}2-\sum_{k=2}^{x}\Phi(\lfloor x/k\rfloor)$$

Group the right side by quotient and memoize large arguments. Recursive arguments strictly decrease, while small values come from the sieve, proving the recurrence by induction. Every large argument is $\lfloor n/i\rfloor$ , so the cache index $\lfloor n/x\rfloor$ uses $O(n/B)$ entries.

The shared component also maintains $\sum_{d\le x}d\varphi(d)$ . This problem uses only the first component. Computing the unused second component does not change the asymptotic bounds.

## Evaluate binomial coefficients

Write $C(q)=\binom{q+m-1}{m}$ . Starting with $C(1)=1$ , use

$$C(q)=C(q-1)\frac{q+m-1}{q-1}$$

to precompute through $B$ . Since $q-1<P$ , its inverse exists and can be generated by the linear inverse recurrence. For larger quotients, directly evaluate

$$C(q)=\frac{q(q+1)\cdots(q+m-1)}{m!}$$

There are at most $n/B$ such quotients, each costing $O(m)$ . The value $q+m-1$ can reach or exceed $P$ : if the numerator contains $P$ , the result is zero. Since $m<P$ , the denominator is always invertible. A naive quotient of factorial tables can incorrectly attempt to invert zero here.

## Correctness and complexity

Stars and bars counts each legal sequence once. Totient expansion recovers its gcd. Quotient grouping and the prefix recurrence compute each coefficient exactly. The modular sum of all group contributions is therefore the answer.

For threshold $B$ , preprocessing takes $O(B+m)$ , large prefix sums take $O(n/\sqrt B)$ , and binomial evaluations take $O(nm/B)$ . Total time is $O(B+n/\sqrt B+nm/B)$ and space is $O(B+n/B)$ . Binary exponentiation computes one inverse factorial in $O(\log P)$ time, with fixed $P$ here. Use 64-bit products and reduce between multiplications.

Summing over every $d$ is useful for medium checks, but needs $O(n)$ storage or more online work and is unsuitable for the full domain. Consecutive-point interpolation also evaluates the binomial polynomial, but still costs $O(m)$ per query with additional arrays; direct multiplication is simpler here.

See the [standard program](std.cpp), [prefix component](totient-sums.hpp), [test recipes](tests/README.md), and [measurements](../../../docs/benchmark-R36.json).