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

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 $\sum_{j=0}^\infty 1/x_j = \infty$.