普洛尼克数
性質
特殊的普洛尼克數
註釋
- 若n≡0 (mod 9),則n×(n+1)≡0×1≡9 (mod 9)
- 若n≡1 (mod 9),則n×(n+1)≡1×2≡2 (mod 9)
- 若n≡2 (mod 9),則n×(n+1)≡2×3≡6 (mod 9)
- 若n≡3 (mod 9),則n×(n+1)≡3×4≡12≡3 (mod 9)
- 若n≡4 (mod 9),則n×(n+1)≡4×5≡20≡2 (mod 9)
- 若n≡5 (mod 9),則n×(n+1)≡5×6≡30≡3 (mod 9)
- 若n≡6 (mod 9),則n×(n+1)≡6×7≡42≡6 (mod 9)
- 若n≡7 (mod 9),則n×(n+1)≡7×8≡56≡2 (mod 9)
- 若n≡8 (mod 9),則n×(n+1)≡8×9≡72≡9 (mod 9)
- 若n≡0 (mod 10),則n×(n+1)≡0×1≡0 (mod 10)
- 若n≡1 (mod 10),則n×(n+1)≡1×2≡2 (mod 10)
- 若n≡2 (mod 10),則n×(n+1)≡2×3≡6 (mod 10)
- 若n≡3 (mod 10),則n×(n+1)≡3×4≡12≡2 (mod 10)
- 若n≡4 (mod 10),則n×(n+1)≡4×5≡20≡0 (mod 10)
- 若n≡5 (mod 10),則n×(n+1)≡5×6≡30≡0 (mod 10)
- 若n≡6 (mod 10),則n×(n+1)≡6×7≡42≡2 (mod 10)
- 若n≡7 (mod 10),則n×(n+1)≡7×8≡56≡6 (mod 10)
- 若n≡8 (mod 10),則n×(n+1)≡8×9≡72≡2 (mod 10)
- 若n≡9 (mod 10),則n×(n+1)≡9×10≡90≡0 (mod 10)
- 因為n與(n+1)差1,所以兩數互質,故若n×(n+1)為平方數,則n與(n+1)也皆為平方數,2個平方數差1,則必為0與1,因此唯一的普洛尼克數兼平方數為0=0×1。
参考资料
- McDaniel, Wayne L., (PDF), Fibonacci Quarterly, 1998, 36 (1): 56–59, MR 1605341
- Sloane, N.J.A. (编). . The On-Line Encyclopedia of Integer Sequences. OEIS Foundation.
- . NUMBERS APLENTY.
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