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A Matrix-Theoretic Exact Formula for Counting Primes in Intervals Between Consecutive Odd Squares

Number Theory 2026-05-22 v1 Combinatorics

Abstract

Let Ik=[(2k1)2,(2k+1)2)I_k = [(2k-1)^2, (2k+1)^2) for k1k \geq 1. Starting from the odd-composite matrix (bij)(b_{ij}) with bij=(2i1)(2j1)b_{ij} = (2i-1)(2j-1), introduced by the author in [1], we define for each odd integer nn the \emph{matrix multiplicity} r(n)r(n), the number of times nn appears in BB. We prove the exact identity Pk=NkSk+Ek P_k = N_k - S_k + E_k where Pk=#{primes in Ik}P_k = \#\{\text{primes in } I_k\}, Nk=4kN_k = 4k counts the odd integers in IkI_k, Sk=nIk oddr(n)S_k = \sum_{n \in I_k \text{ odd}} r(n) is the total matrix multiplicity, and Ek=nIk odd(r(n)1)E_k = \sum_{n \in I_k \text{ odd}} (r(n)-1) measures the excess multiplicity of non-semiprime odd composites. All three quantities NkN_k, SkS_k, EkE_k are computable from the divisor structure of odd integers in IkI_k without primality testing. The formula yields the equivalent combinatorial condition: Pk1    EkSkNk. P_k \geq 1 \iff E_k \leq S_k - N_k. We verify Pk1P_k \geq 1 for all k108k \leq 10^8 by direct computation and establish Pk1P_k \geq 1 for all k1.37×1017k \leq 1.37 \times 10^{17} using the Baker-Harman-Pintz theorem [2]. Whether Pk1P_k \geq 1 for all kk (a weaker statement than Legendre's conjecture) remains an open problem, now equivalent to the purely combinatorial inequality EkSkNkE_k \leq S_k - N_k for all kk.

Keywords

Cite

@article{arxiv.2605.21529,
  title  = {A Matrix-Theoretic Exact Formula for Counting Primes in Intervals Between Consecutive Odd Squares},
  author = {Wujie Shi},
  journal= {arXiv preprint arXiv:2605.21529},
  year   = {2026}
}

Comments

9 pages

R2 v1 2026-07-22T07:24:37.814Z