On powers of the diophantine function $\star:x\mapsto x(x+1)$

TL;DR

This paper explores the properties of iterated diophantine functions related to the sequence x(x+1), revealing infinite prime sets, sequence behaviors, and sum identities, with implications for number theory and prime distribution.

Contribution

It introduces new sequences derived from powers of the diophantine function and proves their properties, including prime divisibility and sum representations, expanding understanding of these functions.

Findings

The set of primes dividing the sequence is infinite.

Sequences derived from the function relate to known OEIS sequences.

Infinite sum identities are established for these sequences.

Abstract

We treat the functions $\star^k:{\mathbf N}\rightarrow{\mathbf N}$ where $\star:x\mapsto \star x := x(x+1)$. The set $\{\star^k x+1: \{x,k\}\subseteq{\mathbf N}\}$ is pairwise coprime; so, the set ${\mathbf P}$ of primes is infinite. Our Theorem 4 resorts to the mother sequence, M, that is obtained by factoring the infinite sequence $2,3,4,5,\ldots$ into prime powers. For each $x\ge1$ we define the gross $x$-sequence, $\gamma_\star(x) := \langle x+1; \star x+1; \star^2x+1; \star^3x+1;\ldots\rangle$, and also the star sequence, $x^\star$, obtained by factoring the terms of $\gamma_\star(x)$ into prime powers. It turns out that $\gamma_\star(1)$ is Sylvester's sequence, A00058 in the Online Encyclopedia of Integer Sequences, OEIS, and that $\gamma_\star(2)$ is the sequence A082732 in the OEIS. Theorem 3. For every integer $x\ge1$ there is a prime $p(x)$ that divides no member of $\{\star^kx+1: k\ge0\}$. Theorem 4. For each sequence $\eta$ of powers of primes there are infinitely many subsequences $c$ of M such that numerically $\eta=c_j$ but where the term-set family in M of those $c_j$ is formally. pairwise disjoint. Theorem 6. $1/x = \sum_{k=0}^{n-1} 1/(\star^kx+1) + 1/(\star^nx) = \sum_{k=0}^\infty 1/(\star^kx+1)$ for all $\{x,n\}\subseteq{\mathbf N}$. Theorem 7. For every $x\in{\mathbf N}$, when $x^\star := \langle x_j\rangle_{j=0}^\infty$ then $\sum_{j=0}^\infty 1/x_j = \infty$.