Write \(P(n)\) for the least prime divisor of \(n>1\). The proposed formula has the simpler unified form \[ g(1)=0,\qquad g(n)=\pi(P(n))-1\quad(n\geq2). \] Indeed, an even \(n\) has \(P(n)=2\), so the second expression gives zero.

We prove the formula by strong induction. The heaps 1 and 2 have no legal split, since the sole positive split of 2 is \(1+1\), whose parts are equal. Hence both values are zero.

Suppose first that \(n>2\) is even. A legal split \(n=a+b\) has two odd summands, since two even summands would have a common factor 2. If one summand is 1, its value is zero and the other summand is an odd integer greater than 1 with positive value. If both exceed 1 and had equal Grundy values, the induction formula and the injectivity of the prime-counting index on primes would give \(P(a)=P(b)\). That prime would divide \(\gcd(a,b)\), contrary to legality. Thus \(g(a)\mathbin{\mathrm{xor}}g(b)\neq0\) for every option. Zero is absent from the option-value set, so its mex is zero.

Now let \(n>1\) be odd, put \(p=P(n)\), and set \(k=\pi(p)-1\). Every split has one odd summand \(m\) and one even summand \(n-m\). The induction result makes the even summand's value zero, and \[ \gcd(m,n-m)=\gcd(m,n). \] Consequently the option values are exactly the values \(g(m)\) for odd \(m\) with \(1\leq m\leq n-2\) and \(\gcd(m,n)=1\).

The option \(m=1\) realizes value 0. For each \(j\) with \(1\leq j<k\), let \(q\) be the \((j+1)\)-st prime. Then \(q<p\), so \(q\leq p-2\leq n-2\) and \(\gcd(q,n)=1\). The induction formula gives \(g(q)=\pi(q)-1=j\). Every value below \(k\) therefore occurs.

Value \(k\) cannot occur. If a legal odd summand \(m>1\) had \(g(m)=k\), induction would give \(\pi(P(m))=\pi(p)\), hence \(P(m)=p\). Then \(p\) would divide both \(m\) and \(n\), contradicting \(\gcd(m,n)=1\). The option values contain every integer below \(k\) and exclude \(k\), so their mex is \(k\). This completes the induction.

OEIS A055396 defines \(a(1)=0\) and, for \(n>1\), \(a(n)=\pi(P(n))\), where \(P(n)\) is the least prime divisor. The theorem in this entry gives \[ g(1)=0,\qquad g(n)=a(n)-1\quad(n>1). \] The initial Grundy values are \[ 0,0,1,0,2,0,3,0,1,0,4,0,5,0,1,0,6,0,7,0,1,0,8,0,2,0,1,0,9,0. \] Iannucci and Larsson prove in their Theorem 5 that A055396 itself is the Grundy sequence of the totative game, which moves a single integer to a smaller coprime residue. The splitting game here replaces a heap by a disjunctive sum, so its recurrence is different even though the same least-prime obstruction controls the mex.

The program computes the Sprague-Grundy recurrence without using the proposed formula for its mex values. For each \(n\), it enumerates every unordered unequal split \(a+(n-a)\), tests the gcd condition, marks \(g(a)\mathbin{\mathrm{xor}}g(n-a)\), and takes the least unmarked nonnegative integer. A separate smallest-prime-factor sieve supplies the predicted value used only for comparison.

All values through 30,000 agree. The largest value in this interval is 3,244 at the prime 29,989. Compiled with Apple clang 21.0.0 using `c++ -std=c++20 -O3 -march=native -Wall -Wextra -pedantic`, the exact recurrence run took 6.57 seconds on an arm64 workstation. Its three-line output has SHA-256 digest `fbf61f7aefc06d57ab3b5adfc1bc01c915d3c1230ebae1a4ccdfae9a62bb323d`.