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Does the digit sum of triangular numbers yield infinitely many distinct primes?

数论 Math StackExchange -2 票 1 回答 40 浏览 提问者: Knut Sylvén 2026-08-11 19:33
number-theory

问题内容

Let $T_n = \frac{n(n+1)}{2}$ denote the $n$-th triangular number, and let $S(m)$ denote the sum of the digits of $m$ in base 10.

I am investigating a strong version of a digit-sum conjecture: Does $S(T_n)$ yield infinitely many distinct prime numbers?

Modular and Growth Behavior

We know that $T_n \pmod 9$ repeats the periodic sequence $1, 3, 6, 1, 6, 3, 1, 9, 9$. Since $S(m) \equiv m \pmod 9$, the only small prime solution available when $S(T_n) \equiv 0 \pmod 3$ is the number $3$ itself (which appears early on for $n = 2, 6, 15, 20, 24$).

However, as $n$ grows, the number of digits in $T_n$ approaches infinity, and the average value of $S(T_n)$ increases logarithmically. For the conjecture to hold with distinct primes, $S(T_n)$ must continually hit larger prime targets of the form $9k+1$ instead of just cycling through a finite set of small primes.

Computational Evidence up to $n = 50,000,000$

A large-scale computation up to $n = 50,000,000$ yields 1,832,941 total prime matches. More importantly, the data confirms that the primes discovered are scaling upward with the number of digits:

  • For small $n$, the primes are tiny (e.g., $3, 19, 37$).
  • Near $n = 50,000,000$, where $T_n$ becomes a 15-digit number, the algorithm routinely discovers much larger distinct prime digit sums like 73, 109, and 127.
  • The cumulative count of these prime hits grows smoothly and continuously outpaces a scaled square root curve ($c \cdot \sqrt{n}$), showing no signs of stagnation or structural boundaries.

Question

Proving that a polynomial sequence generates infinitely many distinct primes in its digit sum seems notoriously difficult due to the mixing of base-dependent operations and pure algebra.

Is it currently proven (or conditionally known under conjectures like Bateman-Horn or Hardy-Littlewood) whether $S(T_n)$ contains infinitely many distinct primes? Are there any analogous results for quadratic forms where the infinitude of distinct digit-sum primes has been established?

回答 (1)

French Man 0 票 2026-08-12 00:02 原文

Yes, the conjecture's true. Take $$T(2\cdot 10^k-1)=(2\cdot 10^k-1)10^k=1\underbrace{99\dots9}_{k \text{ copies}}\,\underbrace{00\dots0}_{k \text{ copies}}$$ $$\implies S(T(2\cdot 10^k-1))=1+9k$$ Of which there's infinitely many primes due to Dirichlet's theorem.