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PreprintCC BY 4.0arXiv

On Carmichael Numbers Constructed from Consecutive Jacobsthal Numbers

  • Hayat Bensella
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Abstract

We prove that no Carmichael number has the form N=(2mJn+1)(2mJn+1+1)(2mJn+2+1), where Jn=(2n−(−1)n)/3 is the Jacobsthal sequence, m,n≥1, and all three factors are prime. Assuming that such an N is Carmichael, Korselt's criterion yields divisibility conditions relating consecutive Jacobsthal numbers to powers of two. Combining these conditions with the Binet formula and a finite case analysis for n≤4, we obtain a contradiction. Consequently, this family contains no Carmichael numbers.

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Cite

BibTeXRIS
BibTeX
@misc{bensella2026carmichael,
  title = {{On Carmichael Numbers Constructed from Consecutive Jacobsthal Numbers}},
  author = {Hayat Bensella},
  howpublished = {arXiv},
  year = {2026},
  month = oct,
  doi = {10.48550/arxiv.2610.26065},
  eprint = {2610.26065},
  archivePrefix = {arXiv},
  url = {https://arxiv.org/abs/2610.26065},
  note = {CC BY 4.0},
}
RIS
TY  - UNPB
TI  - On Carmichael Numbers Constructed from Consecutive Jacobsthal Numbers
AU  - Hayat Bensella
PY  - 2026
DA  - 2026/10/19
PB  - arXiv
DO  - 10.48550/arxiv.2610.26065
UR  - https://arxiv.org/abs/2610.26065
AB  - We prove that no Carmichael number has the form N=(2^mJ_n+1)(2^mJ_n+1+1)(2^mJ_n+2+1), where J_n=(2^n−(−1)^n)/3 is the Jacobsthal sequence, m,n≥1, and all three factors are prime. Assuming that such an N is Carmichael, Korselt's criterion yields divisibility conditions relating consecutive Jacobsthal numbers to powers of two. Combining these conditions with the Binet formula and a finite case analysis for n≤4, we obtain a contradiction. Consequently, this family contains no Carmichael numbers.
ER  -