The Binary Totient Series

Iterated prime dilations of divisor counts.

Multiplying an argument by a new prime adds exactly one new family of divisors. For the divisor count $d_A$ above and $n\ge 1$, put ${\mathrm{\Delta }}_{p}f(n)=f(pn)-f(n)$. If $p_1,\dots ,p_r$ are distinct primes, $P=p_1\cdots p_r$ and $gcd(n,P)=1$, then

For one prime, the divisors of $pn$ not dividing $n$ are exactly $pd$ with $d\mid n$, because $p\nmid n$. Iteration gives the displayed identity: distinctness ensures that each new prime is coprime to the arguments produced at the earlier steps. Equivalently, inclusion–exclusion cancels every divisor not containing all the new prime factors.

Let $\mathrm{\Omega }(a)$ count the prime factors of $a\ge 1$ with multiplicity, and let $K\ge 0$ be an integer. If $\mathrm{\Omega }(a)\le K$ for every positive $a\in A$ and $r>K$, the set on the right is empty, since $\mathrm{\Omega }(Pd)=r+\mathrm{\Omega }(d)>K$. This proves the stated vanishing with all its hypotheses. Erdos249257.SupportDilationDifferences.iteratedDilationDifference_eq_zero_of_cardFactors_le The source also proves vanishing of the right-hand divisor count for a list with repeated primes. That is a statement about the condition $Pd\in A$, not an identity with repeated dilation differences.

For example, with $A=\{2\}$ and $n=1$,

Thus the bound $\mathrm{\Omega }(a)\le 1$ and two prime factors do not by themselves make the repeated difference vanish. The strict inequality $r>K$ is also material: ${\mathrm{\Delta }}_2{d}_{\{2\}}(1)=1$ when $r=K=1$. Finally, at $n=2$ the one-prime identity without the coprimality hypothesis would give ${\mathrm{\Delta }}_2{d}_{\{2\}}(2)=0$ on the left and $\mathrm{\#}\{d\mid 2:2d\in \{2\}\}=1$ on the right. This last example concerns the identity, not the necessity of coprimality for every possible vanishing theorem. These divisor counts belong to the unweighted support problem; they are not the totient coefficients.