Let $\tau(n)$ denote the number of divisors of $n$. OEIS sequence A309981 gives the smallest $k$ such that the tuple $(\tau(n), \tau(n+1), \ldots, \tau(n+k))$ uniquely determines $n$.

For small $n$ the value can often be verified by case analysis in residues to a suitable modulus, but $n=49$ is more resistent (and the notes in the OEIS history show that the person who posted the sequence considered it more challenging). No reference is given, and the correctness of $a(49) = 2$ is in the examples essentially as a bare assertion.

How can it be shown that $\tau(n) = 3$, $\tau(n+1) = 6$ and $\tau(n+2) = 4$ has a unique solution?

For comparison, and in case it's helpful, I give the standard case analysis.

$n$ must be a prime square $p_0^2$; it's not $2^2$ because $\tau(2^2 + 1) = 2 \neq 6$, so $n$ is odd. It's not $3^2$ because $\tau(3^2 + 1) = 4 \neq 6$. Therefore $p_0$ is coprime to $6$ and $n \equiv 1 \pmod 6$. Note also that since it's an odd square, $n \equiv 1 \pmod 8$. Combining the two, $n \equiv 1 \pmod {24}$.

$n+1$ is either $p_1^5$ or $p_1^2 q_1$. It's even; it's not $32$ because $\tau(31) = 2 \neq 3$, so it's either $4 q_1$ or $2 p_1^2$. But $4 q_1 \not \equiv 2 \pmod {24}$ so $n+1 = 2p_1^2$.

$n+2$ is either a prime cube $p_2^3$ or a semiprime $p_2 q_2$. Since $n+2 \equiv 3 \pmod {24}$ it's divisible by $3$; if it's a prime cube then it's $3^3$, but $\tau(3^3 - 1) = 4 \neq 6$, so $n+2 = 3 p_2$.

In summary \begin{array}{ccccc}n &=& p_0^2 &\equiv& 1 \pmod {24} \\ n + 1 &=& 2p_1^2 \\ n + 2 &=& 3p_3 \end{array}

posted almost 2 years ago

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2y ago

Peter Taylor‭

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