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费马数是表格的数字
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F类n个 := 2 2 n个+ 1, n个≥ 0. |
A000215号费马数:.-
{3, 5, 17, 257, 65537, 4294967297, 18446744073709551617, 340282366920938463463374607431768211457, 115792089237316195423570985008687907853269984665640564039457584007913129639937, ...}
A080176号费马数在基地2代表。(请参见A080176号,注释部分。)-
{11、101、10001、100000001、10000000000000001、10000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000
公式
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定期
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F类0 = 三;F类n个 = (F类n个 − 1− 1) 2+ 1, n个≥ 1. |
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F类n个 = F类k个+2的情况下, n个≥ 0, |
在哪里我们有空产品(给出乘法恒等式,即。1)+ 2,给予,如预期。属性
费马数的序列是互质序列,自
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F类n个 = F类k个+2的情况下, n个≥ 0, |
在哪里我们有空产品(给出乘法恒等式,即。1)+ 2,给予.由于存在无穷多个费马数,所有这些费马数都是互质的,这意味着存在无穷多质数。
正在生成函数
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远期差额
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F类n个 + 1−F类n个 = (F类n个 ) 2− 3 F类n个+ 2 = (F类n个− 1) (F类n个− 2), n个≥ 0. |
部分金额
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F类n个 = (2 2 n个+1) = 米+ 1 + 2 2 n个 = ?. |
部分倒数和
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倒数总和
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费马数的素因式分解
这个基本因子费马数的形式[1]
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F类n个 = 2 2 n个+ 1 = (k个 1·2 n个+ 1) (k个 2·2 n个+ 1)⋯, k个 我∈ ℕ +,n个≥ 0. |
此外,费马数的素因子的形式如下[2]
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F类n个 = 2 2 n个+ 1 = ( ·2 n个 + 2+ 1) ( ·2 n个 + 2+ 1)⋯, ∈ ℕ +,n个≥ 2. |
费马数的素因式分解
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主要因素 |
0
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3 = 2 1+ 1 |
3 = (1 / 2)·2 2+ 1 |
1
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5 = 2 2+ 1 |
5 = (1 / 2)·2 三+1个 |
2
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17 = 2 4+ 1 |
17 = 1·2 4+ 1 |
三
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257 = 2 8+ 1 |
257 = 8·2 5+ 1 |
4
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65537 = 2 16+ 1 |
65537 = 1024·2 6+ 1 |
5
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4294967297 = 2 32+ 1 |
641·6700417 = (5·2 7+1)(52347·2 7+ 1) |
6
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18446744073709551617 = 2 64+ 1 |
274177·67280421310721 = (1071·2 8+ 1) (262814145745·2 8+ 1) |
7
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340282366920938463463374607431768211457 = 2 128+ 1 |
59649589127497217·5704689200685129054721 = (116503103764643·2 9+ 1) (11141971095088142685·2 9+ 1) |
F类8 = 2 2 8+1个 | = 115792089237316195423570985008687907853269984665640564039457584007913129639937 |
| = 1238926361552897·93461639715357977769163558199606896584051237541638188580280321 |
| = (1209889024954·2 10+ 1) (91271132534529275165198787304303609945362536661756043535430·2 10+ 1) |
F类9 = 2 2 9+ 1 | = 13407807929942597099574024998205846127479365820592393377723561443721764030073546976801874298166903427690031858186486050853753882811946569946433649006084097 |
| = 2424833·7455602825647884208337395736200454918783366342657·741640062627530801524787141901937474059940781097519023905821316144415759504705008092818711693940737 |
| =(1184·2 11+ 1) (3640431067210880961102244011816628378312190597·2 11+ 1) (362128936829849024182024971631805407255830459520272960891514314523640507570656742232821636569307·2 11+ 1) |
F类10 = 2 2 10+ 1 | = 179769...137217 (309十进制数字) |
| = 45592577·6487031809·4659775785220018543264560743076778192897·P(P)252* |
| = (11131·2 12+ 1) (1583748·2 12+ 1) (1137640572563481089664199400165229051·2 12+ 1)P(P)252*
|
F类11 = 2 2 11+ 1 | = 323170...230657 (617十进制数字) |
| = 319489·974849·167988556341760475137·3560841906445833920513·P(P)564* |
| = (39·2 13+ 1) (119·2 13+ 1) (20506415569062558·2 13+ 1) (434673084282938711·2 13+ 1)P(P)564*
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*下标表示素因子的小数位数。
A093179号最小因子-第th个费马数.-
{3, 5, 17, 257, 65537, 641, 274177, 59649589127497217, 1238926361552897, 2424833, 45592577, 319489, 114689, 2710954639361, 116928085873074369829035993834596371340386703423373313, 1214251009, ...}
A070592号的最大素因子-第th个费马数.-
{3, 5, 17, 257, 65537, 6700417, 67280421310721, 5704689200685129054721, 93461639715357977769163558199606896584051237541638188580280321, ...}
费马素数
据推测,第一个5这个序列中的数字是素数(费马素数).
A019434年的列表费马素数:素数形式的,对于一些.-
{3, 5, 17, 257, 65537, ?}
不同Fermat素数的乘积
既然有5已知的费马素数,{F类0 ,F类1 ,F类2 ,F类三 ,F类4} = {3, 5, 17, 257, 65537} |
,那么就有 ()+()+()+()+()+() = 1+5+10+10+5+1 = 32 = 2 5 |
不同已知产品费马素数. The31不同已知的非空产品费马素数给出的边数可构造的奇边多边形(因为多边形至少有三侧面)。
另请参见
注意事项