如何判断[根号(2^n)]的奇偶性

王守恩 · 发表于 10 小时前

马上注册，结交更多好友，享用更多功能，让你轻松玩转社区。

您需要登录才可以下载或查看，没有账号？欢迎注册

×

如何判断[根号(2^n)]的奇偶性？

Table[Mod[Floor[Sqrt[2^n]], 2], {n, 98}]——标准。
{1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0}
Table[Mod[Floor[1.5^n], 2], {n, 3}]——跟标准比,相同数位=2, a(1) = 2。
{1, 0, 1}
Table[Mod[Floor[1.42^n], 2], {n, 9}]——跟标准比,相同数位=8, a(2) = 8。
{1, 0, 0, 0, 1, 0, 1, 0, 1}
Table[Mod[Floor[1.415^n], 2], {n, 13}]——跟标准比,相同数位=12, a(3) = 12。
{1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1}
Table[Mod[Floor[1.4143^n], 2], {n, 20}]——跟标准比,相同数位=19, a(4) = 19。
{1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 1}
Table[Mod[Floor[1.41422^n], 2], {n, 25}]——跟标准比,相同数位=24, a(5) = 24。
{1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1}
Table[Mod[Floor[1.414214^n], 2], {n, 31}]——跟标准比,相同数位=30, a(6) = 30。
{1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1}
Table[Mod[Floor[1.4142136^n], 2], {n, 40}]——跟标准比,相同数位=39, a(7) = 39。
{1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 1}
Table[Mod[Floor[1.41421357^n], 2], {n, 43}]——跟标准比,相同数位=42, a(8) = 42。
{1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 0, 1}
Table[Mod[Floor[1.414213563^n], 2], {n, 51}]——跟标准比,相同数位=50, a(9) = 50。
{1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1}
Table[Mod[Floor[1.4142135624^n], 2], {n, 59}]——跟标准比,相同数位=58, a(10) = 58。
{1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1}
Table[Mod[Floor[1.41421356238^n], 2], {n, 59}]——跟标准比,相同数位=58, a(11) = 58。
{1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1}
Table[Mod[Floor[1.414213562374^n], 2], {n, 59}]——跟标准比,相同数位=58, a(12) = 58。
{1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1}
Table[Mod[Floor[1.4142135623731^n], 2], {n, 77}]——跟标准比,相同数位=76, a(13) = 76。
{1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 0, 1}
Table[Mod[Floor[1.41421356237310^n], 2], {n, 77}]——跟标准比,相同数位=76, a(14) = 76。
{1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 0, 1}
Table[Mod[Floor[1.414213562373096^n], 2], {n, 85}]——跟标准比,相同数位=84, a(15) = 84。
{1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1}

得到一串数——2, 8, 12, 19, 24, 30, 39, 42, 50, 58, 58, 58, 76, 76, 84, 88, 104, 110, 113, 120, 129, 138, 144, 144, 144, 156, 166, 170, 186, 186,——有规律吗？
Sqrt(2) = 1.4, 1, 4, 2, 1, 3, 5, 6, 2, 3, 7, 3, 0, 9, 5, 0, 4, 8, 8, 0, 1, 6, 8, 8, 7, 2, 4, 2, 0, 9, 69807856967187537694807317667973799073......

mathe · 发表于 6 小时前

https://oeis.org/A004539

王守恩 · 发表于 5 小时前

mathe 发表于 2025-10-15 14:42
https://oeis.org/A004539

A004539——Expansion of sqrt(2) in base 2.——可能不是这串数。

{1, 0, 1, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 1, 1, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 0, 1, 1, 1, 1, 0, 0, 1, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 1, 0, 0, 1, 0, 1, 1, 1, 1, 1, 0, 1, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 1, 1, 0, 0,

Mod[Floor[Sqrt[2^n]], 2],——我们这串数有个显著特征——不会有2个“1”连在一起。

{1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0}

账号		自动登录	找回密码
密码			欢迎注册

[讨论] 如何判断[根号(2^n)]的奇偶性

马上注册，结交更多好友，享用更多功能，让你轻松玩转社区。

评分

点评