On powers of the diophantine function $\star:x\mapsto x(x+1)$
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2024
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| _version_ | 1866910833879023616 |
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| author | Silberger, Donald |
| author_facet | Silberger, Donald |
| contents | 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, $γ_\star(x) := \langle x+1; \star x+1; \star^2x+1; \star^3x+1;\ldots\rangle$, and also the star sequence, $x^\star$, obtained by factoring the terms of $γ_\star(x)$ into prime powers. It turns out that $γ_\star(1)$ is Sylvester's sequence, A00058 in the Online Encyclopedia of Integer Sequences, OEIS, and that $γ_\star(2)$ 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 $η$ of powers of primes there are infinitely many subsequences $c$ of M such that numerically $η=c_j$ 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 $\sum_{j=0}^\infty 1/x_j = \infty$. |
| format | Preprint |
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arxiv_https___arxiv_org_abs_2412_00253 |
| institution | arXiv |
| publishDate | 2024 |
| record_format | arxiv |
| spellingShingle | On powers of the diophantine function $\star:x\mapsto x(x+1)$ Silberger, Donald Number Theory 01A55, 01A60, 11A25, 11N05, 11N25 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, $γ_\star(x) := \langle x+1; \star x+1; \star^2x+1; \star^3x+1;\ldots\rangle$, and also the star sequence, $x^\star$, obtained by factoring the terms of $γ_\star(x)$ into prime powers. It turns out that $γ_\star(1)$ is Sylvester's sequence, A00058 in the Online Encyclopedia of Integer Sequences, OEIS, and that $γ_\star(2)$ 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 $η$ of powers of primes there are infinitely many subsequences $c$ of M such that numerically $η=c_j$ 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 $\sum_{j=0}^\infty 1/x_j = \infty$. |
| title | On powers of the diophantine function $\star:x\mapsto x(x+1)$ |
| topic | Number Theory 01A55, 01A60, 11A25, 11N05, 11N25 |
| url | https://arxiv.org/abs/2412.00253 |