The reciprocal sum of the prime-prefix-free numbers converges [pdf]
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THE RECIPROCAL SUM OF THE PRIME-PREFIX-FREE INTEGERS CONVERGES UNDER THE RIEMANN HYPOTHESIS DAN BROWN∗ Abstract. Call a positive integer prime-prefix-free if none of its proper binary prefixes ⌊𝑛/2𝑘 ⌋, 𝑘 ≥ 1, is an odd prime (OEIS A287117). Assuming the Riemann Hypothesis, we prove that the sum of 1/𝑛 over the prime-prefix-free integers converges. More precisely, the proportion of [2𝑚, 2𝑚+1) they occupy is 𝑂 𝛿 (𝑚−1/log 2+ 𝛿 ) for every 𝛿 > 0.
The heuristic exponent is 1/log 2 ≈ 1.4427, and convergence rests on the inequality 1/log 2 > 1, that is, on the base 2 being smaller than 𝑒. The proof combines Selberg’s theorem on primes in almost all short intervals with an upper-bound sieve for prime pairs, and closes with a bootstrap for a delayed recurrence.
The constants are effective: the sum is less than 𝐹72 ≈ 5 · 1014. Convergence and this bound are machine-verified in Lean 4 over Mathlib, from Mathlib’s statement of the Riemann Hypothesis. Numerically the sum exceeds 3.5276 unconditionally, and a heuristic model fitted to exact counts up to 246 puts it near 3.877. 1. Introduction Arrange the positive integers in the infinite binary tree in which the children of 𝑛 are 2𝑛 and 2𝑛 + 1. The ancestors of 𝑛 are then its proper binary prefixes ⌊𝑛/2𝑘 ⌋, 𝑘 ≥ 1. Let S be the set of 𝑛 none of whose ancestors is an odd prime. Equivalently, the odd primes are made leaves of the tree, and S is the set of nodes that remain. The prime 2 is excluded because every 𝑛 ≥ 4 has 2 or 3 as a prefix, so including it would leave a finite set. The sequence S = {1, 2, 3, 4, 5, 8, 9, 16, 17, 18, 19, 32, . . . } is A287117 in the OEIS [7], and the question whether Í𝑛∈ S 1/𝑛 converges was posed on Mathematics Stack Exchange [6]. Theorem. Assume the Riemann Hypothesis. Then Í𝑛∈ S 1/𝑛 < ∞. Moreover, for every 𝛿 > 0, #S ∩ [2𝑚, 2𝑚+1) ≪𝛿 2𝑚 𝑚−1/log 2+ 𝛿 . The implied constants are effective. Remark 7.4 makes them numeric and obtains Í𝑛∈ S 1/𝑛 < 𝐹72 ≈ 5 · 1014 under RH, while the exact count below 246 gives Í𝑛∈ S 1/𝑛 > 3.5276 unconditionally (Remark 7.3). The exponent is the one predicted by a probabilistic model. Pick 𝑥 ∈ [1, 2) at random. Its binary prefix of length 𝑚 + 1 is an integer near 2𝑚, prime with probability about 1/(𝑚 log 2), and these events are nearly independent, so the prefixes of 𝑥 avoid the odd primes up to length 𝑚 with probability aboutÎ𝑘 ≤𝑚 1 − 1/(𝑘 log 2) ≍ 𝑚−1/log 2. That survival probability is the proportion 𝑟𝑚 of the level [2𝑚, 2𝑚+1) occupied by S, and the reciprocal sum is comparable to Í𝑚 𝑟𝑚. Convergence therefore hinges on 1/log 2 ≈ 1.4427 exceeding 1. In base 𝑏 the same model gives the exponent 1/log 𝑏, below 1 for every 𝑏 ≥ 3: binary is the one base in which the analogous sum is expected to converge.
Turning the model into a proof requires lower bounds for primes in the set of survivors at every level. That set is a union of short intervals at all scales, so this is a problem about primes in short intervals, counted simultaneously along all branches of the tree. The proof uses two known results.
The first is Selberg’s theorem [8], conditional on RH, that almost all intervals [𝑥, 𝑥 + ℎ] with ℎ a small power of 𝑥 contain the expected number of primes; the exceptional set is small enough to survive a union over the polynomially many scales involved. The second is the classical upper-bound sieve for prime pairs [4]. It controls primes that have a prime ancestor only a few levels up. These pairs are beyond the reach of RH, Date: October 4, 2026. ∗Author of The Da Vinci Code.
1 2 DAN BROWN∗ and the sieve overcounts them by a constant factor, but they occupy only a small fraction of the levels, and that fraction can be made as small as we like. A telescoping identity converts the remaining long-range pairs into the same quantities 𝑟𝑚 that the argument controls, and a bootstrap for the resulting delayed recurrence finishes the proof. Each loss in the argument lowers the exponent 1/log 2, and the proof needs the exponent to stay above 1, so the losses must total less than 1/log 2 − 1 ≈ 0.44. In particular, the sieve’s constant-factor overcount can be tolerated only on a small proportion of levels.