Ed Pegg jr.先生發現上圖中的線段d長度為{\displaystyle {\frac {1}{2}}{\sqrt {{\frac {1}{30}}(61421-23{\sqrt {5831385}})}}}，非常接近7（數值為7.0000000857）
在趣味數學中，**接近整数**是指很接近整數的無理數。這類數字中，有些因為其數學上的特性使其接近整数，有些還找不到其特性，看起來似乎只是巧合。

有關黃金比例及其他皮索特-维贡伊拉卡文数
--------------------

黃金比例{\displaystyle \varphi ={\frac {1+{\sqrt {5}}}{2}}\approx 1.61803398875\,}的高次方符合此特性。例如

* {\displaystyle \varphi ^{17}={\frac {3571+1597{\sqrt {5}}}{2}}\approx 3571.00028\approx F\_{16}+F\_{18}}
* {\displaystyle \varphi ^{18}=2889+1292{\sqrt {5}}\approx 5777.999827\approx F\_{17}+F\_{19}}
* {\displaystyle \varphi ^{19}={\frac {9349+4181{\sqrt {5}}}{2}}\approx 9349.000107\approx F\_{18}+F\_{20}}

其中F\_n代表費波納契數列的第n項
這是因為有恆等式{\displaystyle \varphi ^{n}=F\_{n-1}+F\_{n}\times \varphi }，所以當n為足夠大的正整數時，

{\displaystyle \varphi ^{n}=F\_{n-1}+F\_{n}\times \varphi \approx F\_{n-1}+F\_{n}\times \left({\frac {F\_{n+1}}{F\_{n}}}\right)=F\_{n-1}+F\_{n+1}}
這些數字接近整數的原因和黃金比例的特性有關，不是數學巧合。其原因是因為黃金比例為皮索特-维贡伊拉卡文数，而皮索特-维贡伊拉卡文数的高次方會是接近整數。

這些數字與費波納契數有密切的關係，因為費波納契數相鄰兩項的比值會趨近於黃金比例，而如果m整除n，則第m個費波納契數也會整除第n個費波納契數。

皮索特-维贡伊拉卡文数是指代數數本身大於1，而且其極小多項式中另一根的絕對值小於1。像黃金比例本身大於1，\varphi 的最小多項式為
{\displaystyle x^{2}-x-1=0}

另一根為
{\displaystyle {\overline {\varphi }}={\frac {1-{\sqrt {5}}}{2}}\approx -0.618\,}

絕對值小於1，因此黃金比例為皮索特-维贡伊拉卡文数，其高次方會是接近整数。

依照根和系数的关系，可得知

{\displaystyle \varphi {\overline {\varphi }}=-1}

{\displaystyle \varphi +{\overline {\varphi }}=1}

而{\displaystyle \varphi ^{n}+{\overline {\varphi }}^{n}}可以用{\displaystyle \varphi {\overline {\varphi }}} 及{\displaystyle \varphi +{\overline {\varphi }}}來表示，由於二根之和及二根之積均為整數，計算所得的結果也是一個正整數，假設為一正整數K，則{\displaystyle \varphi ^{n}}可以用下式表示

{\displaystyle \varphi ^{n}=K-{\overline {\varphi }}^{n}}

由於{\displaystyle {\overline {\varphi }}}的絕對值小於1，在n增大時，其高次方會趨於0，此時可得

{\displaystyle \varphi ^{n}\approx K}

除了黃金比例外，其他皮索特-维贡伊拉卡文数的無理數也符合此一條件，例如{\displaystyle 1+{\sqrt {2}}}。

有關黑格納數
------

以下也是幾個非巧合出現的接近整數，和最大三項的黑格納數有關：

* {\displaystyle e^{\pi {\sqrt {43}}}\approx 884736743.999777466\,}
* {\displaystyle e^{\pi {\sqrt {67}}}\approx 147197952743.999998662454\,}
* {\displaystyle e^{\pi {\sqrt {163}}}\approx 262537412640768743.99999999999925007\,}

以上三式可以用以下的式子表示:

{\displaystyle e^{\pi {\sqrt {43}}}=12^{3}(9^{2}-1)^{3}+744-2.225\cdots \times 10^{-4}\,}
{\displaystyle e^{\pi {\sqrt {67}}}=12^{3}(21^{2}-1)^{3}+744-1.337\cdots \times 10^{-6}\,}
{\displaystyle e^{\pi {\sqrt {163}}}=12^{3}(231^{2}-1)^{3}+744-7.499\cdots \times 10^{-13}\,}
其中：{\displaystyle 21=3\times 7,231=3\times 7\times 11,744=24\times 31\,}
由於艾森斯坦級數的關係，使得上式中出現平方項。常數{\displaystyle e^{\pi {\sqrt {163}}}\,}有時會稱為拉馬努金常數。

有關π及e
-----

許多有關π及e的常數也是接近整數，例如

{\displaystyle e^{\pi }-\pi =19.999099979189\cdots \,}
以及

{\displaystyle e^{5\pi }=6635623.999341134233\cdots \,}
格尔丰德常数（{\displaystyle e^{\pi }\,}）接近{\displaystyle \pi +20\,}，至2011年為止還沒找到出現此特性的原因，因此只能視為一數學巧合。另一個有關格尔丰德常数的常數也是接近整數
{\displaystyle {\frac {e^{\pi }-\pi -1}{6\pi }}=1.00793356\cdots \,}

以下也是一些接近整數的例子

* {\displaystyle 22{\pi }^{4}=2143.0000027480\cdots \,}
* {\displaystyle {\pi }^{5}=306.019684\cdots \,}
* {\displaystyle {\pi }^{3}=31.006276\cdots \,}
* {\displaystyle {\pi }^{13/2}=1704.017978\cdots \,}
* {\displaystyle {\pi }^{3}-{\frac {\pi }{500}}=30.999993494\cdots \,}
* {\displaystyle {\pi }^{2}+{\frac {\pi }{24}}=10.000504\cdots \,}
* {\displaystyle {\pi }^{5}-3{\pi }^{3}=213.0008547\cdots \,}
* {\displaystyle e^{\pi }-\pi +0.0009=19.9999999791\cdots \,}
* {\displaystyle e^{3}=20.0855369\cdots \,}
* {\displaystyle e^{9/2}=90.0171313\cdots \,}
* {\displaystyle e^{\pi {\sqrt {2}}}=85.019695\cdots \,}
* {\displaystyle e^{6}-{\pi }^{5}-{\pi }^{4}=0.00001767\cdots \,}
* {\displaystyle 9{\pi }^{5}-2e^{3}=2714.00608922\cdots \,}
* {\displaystyle e^{13/2}-\pi =662.00004039\cdots \,}
* {\displaystyle {\pi }^{13/2}-e^{9/2}=1614.00084707\cdots \,}

其他例子
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|  |  |  |  |
| --- | --- | --- | --- |
| {\displaystyle {}\_{\cos \left\{\pi \cos \left[\pi \cos \ln \left(\pi +20\right)\right]\right\}\approx -0.9999999999999999999999999999999999606783}} | {\displaystyle {}\_{\sin 2017{\sqrt[{5}]{2}}\approx -0.9999999999999999785}} | {\displaystyle {}\_{\sum \_{k=1}^{\infty }{\frac {\lfloor n\tanh \pi \rfloor }{10^{n}}}-{\frac {1}{81}}\approx 1.11\times 10^{-269}}} | {\displaystyle {}\_{{\sqrt {29}}\left(\cos {\frac {2\pi }{59}}-\cos {\frac {24\pi }{59}}\right)-{\frac {19}{5}}\approx 3.057684294154\times 10^{-6}}} |
| {\displaystyle {}\_{1+{\frac {103378831900730205293632}{e^{3\pi {\sqrt {163}}}}}-{\frac {196884}{e^{2\pi {\sqrt {163}}}}}-{\frac {262537412640768744}{e^{\pi {\sqrt {163}}}}}\approx 1.161367900476\times 10^{-59}}} | {\displaystyle {}\_{{\frac {\ln ^{2}262537412640768744}{\pi ^{2}}}-163\approx 2.32167\times 10^{-29}}} | {\displaystyle {}\_{10\tanh {\frac {28}{15}}\pi -{\frac {\pi ^{9}}{e^{8}}}\approx 3.661398\times 10^{-8}}} | {\displaystyle {}\_{{\sqrt[{4}]{\frac {91}{10}}}-{\frac {33}{19}}\approx 3.661398\times 10^{-8}}} |
| {\displaystyle {}\_{\gamma -{10 \over 81}\left(11-2{\sqrt {10}}\right)=\int \_{0}^{\infty }\left({\frac {1}{e^{x}-1}}-{\frac {1}{xe^{x}}}\right){\rm {d}}x-{10 \over 81}\left(11-2{\sqrt {10}}\right)\approx 2.72\times 10^{-7}}} | {\displaystyle {}\_{{\frac {\left(5+{\sqrt {5}}\right)\Gamma \left({3 \over 4}\right)}{e^{{\frac {5}{6}}\pi }}}\approx 1.000000000000045422}} | {\displaystyle {}\_{{1 \over 4}\left(\cos {1 \over 10}+\cosh {1 \over 10}+2\cos {{\sqrt {2}} \over 20}\cosh {{\sqrt {2}} \over 20}\right)\approx 1.000000000000248}} | {\displaystyle {}\_{e^{6}-\pi ^{5}-\pi ^{4}\approx 1.7673\times 10^{-5}}} |
| {\displaystyle {}\_{{\sqrt {29}}\left(\cos {\frac {2\pi }{59}}-\cos {\frac {24\pi }{59}}\right)\approx 3.0576842941540143382\times 10^{-6}}} | {\displaystyle {}\_{\left(3{\sqrt {5}}\right)^{\gamma }=\left(3{\sqrt {5}}\right)^{\int \_{0}^{\infty }\left({\frac {1}{e^{x}-1}}-{\frac {1}{xe^{x}}}\right){\rm {d}}x}\approx 3.000060964}} | {\displaystyle {}\_{e^{\phi \_{0}\left({\frac {2+{\sqrt {3}}}{4}}\right)}=e^{\int \_{0}^{\infty }\left({\frac {1}{te^{t}}}-{\frac {e^{{\frac {2-{\sqrt {3}}}{4}}t}}{e^{t}-1}}\right){\rm {d}}t}\approx 1.99999969}} | {\displaystyle {}\_{{\frac {\sqrt[{3}]{9}}{3\ln 2}}\approx 1.00030887}} |
| {\displaystyle {}\_{\sum \_{k=-\infty }^{\infty }10^{-{\frac {k^{2}}{10000}}}-100{\sqrt {\frac {\pi }{\ln 10}}}=\theta \_{3}\left(0,{\frac {1}{\sqrt[{10000}]{10}}}\right)-100{\sqrt {\frac {\pi }{\ln 10}}}\approx 1.3809\times 10^{-18613}}} | {\displaystyle {}\_{{\pi ^{9} \over e^{8}}\approx 9.998387}} | {\displaystyle {}\_{e^{\pi }-\pi \approx 19.999099979}} | {\displaystyle {}\_{{\frac {e^{\pi }-\ln 3}{\ln 2}}-{\frac {4}{5}}\approx 31.0000000033}} |
| {\displaystyle {}\_{{\frac {\pi ^{11}}{e^{3}}}-\Gamma \left[\Gamma \left(\pi +1\right)+1\right]={\frac {\pi ^{11}}{e^{3}}}-\int \_{0}^{\infty }{\frac {t^{\int \_{0}^{\infty }{\frac {u^{\pi }}{e^{u}}}{\rm {d}}u}}{e^{t}}}{\rm {d}}t\approx 7266.9999993632596}} | {\displaystyle {}\_{163\left(\pi -e\right)\approx 68.999664}} | {\displaystyle {}\_{\left({\frac {23}{9}}\right)^{5}={\frac {6436343}{59049}}\approx 109.00003387}} | {\displaystyle {}\_{88\ln 89\approx 395.00000053}} |
| {\displaystyle {}\_{510\lg 7\approx 431.00000040727098}} | {\displaystyle {}\_{272\log \_{\pi }97\approx 1087.000000204}} | {\displaystyle {}\_{{\frac {53453}{\ln 53453}}\approx 4910.00000122}} | {\displaystyle {}\_{{\frac {53453}{\ln 53453}}+{\frac {163}{\ln 163}}\approx 4941.99999995925082}} |
| {\displaystyle {}\_{{\sqrt[{8}]{{\frac {\sqrt {2}}{4}}\left(\pi ^{17}-4e^{2\pi }+4\pi e^{\pi }\right)}}-{\sqrt[{8}]{{\frac {\sqrt {2}}{4}}\left(\pi ^{17}-4e^{2\pi }-4\pi e^{\pi }\right)}}\approx 2.570287024592328869357\times 10^{-6}}} | {\displaystyle {}\_{10-{\sqrt[{8}]{{\frac {\sqrt {2}}{4}}\left(\pi ^{17}-4e^{2\pi }-4\pi e^{\pi }\right)}}\approx 2.57055302118\times 10^{-6}}} | {\displaystyle {}\_{10-{\sqrt[{8}]{{\frac {\sqrt {2}}{4}}\left(\pi ^{17}-4e^{2\pi }+4\pi e^{\pi }\right)}}\approx 2.65996596963\times 10^{-10}}} | {\displaystyle {}\_{{\frac {163}{\ln 163}}\approx 31.9999987343}} |

{\displaystyle 2^{2^{2/3}}\approx 3.00507511272}
{\displaystyle 4\ln 2.117\approx 2.99999996861}
{\displaystyle {}\_{\ln K\_{0}-\ln \ln K\_{0}\approx 1.0000744}} ，其中K\_0是辛钦常数
{\displaystyle {}\_{{\frac {10}{81}}-\sum \_{n=1}^{\infty }{\frac {\sum \_{k=10^{n-1}}^{10^{n}-1}10^{-n\left[k-(10^{n-1}-1)\right]}k}{10^{\sum \_{k=0}^{n-1}9\times 10^{k-1}k}}}={\frac {10}{81}}-\sum \_{n=1}^{\infty }\sum \_{k=10^{n-1}}^{10^{n}-1}{\frac {k}{10^{kn-9\sum \_{k=0}^{n-1}10^{k}(n-k)}}}\approx 1.022344\times 10^{-9}}}

  

{\displaystyle {}\_{-{\frac {1}{5}}+e^{\frac {6}{5}}{}\_{4}F\_{3}\left(-{\frac {1}{5}},{\frac {1}{20}},{\frac {3}{10}},{\frac {11}{20}};{\frac {1}{5}},{\frac {2}{5}},{\frac {3}{5}};{\frac {256}{3125e^{6}}}\right)+{\frac {2}{25e^{\frac {6}{5}}}}{}\_{4}F\_{3}\left({\frac {1}{5}},{\frac {9}{20}},{\frac {7}{10}},{\frac {19}{20}};{\frac {3}{5}},{\frac {4}{5}},{\frac {7}{5}};{\frac {256}{3125e^{6}}}\right)-{\frac {4}{125e^{\frac {12}{5}}}}{}\_{4}F\_{3}\left({\frac {2}{5}},{\frac {13}{20}},{\frac {9}{10}},{\frac {23}{20}};{\frac {4}{5}},{\frac {6}{5}},{\frac {8}{5}};{\frac {256}{3125e^{6}}}\right)+{\frac {7}{625e^{\frac {18}{5}}}}{}\_{4}F\_{3}\left({\frac {3}{5}},{\frac {17}{20}},{\frac {11}{10}},{\frac {27}{20}};{\frac {6}{5}},{\frac {7}{5}},{\frac {9}{5}};{\frac {256}{3125e^{6}}}\right)-\pi \approx 2.89221114964408683\times 10^{-8}}}

  

{\displaystyle {}\_{\qquad {\mbox{Root of }}x^{6}-615x^{5}+151290x^{4}-18608670x^{3}+1144433205x^{2}-28153057165x+39605=0}\,}
{\displaystyle {}\_{{\frac {615-55{\sqrt {5}}-{\sqrt[{3}]{7451370+3332354{\sqrt {5}}+6{\sqrt {8890710030+3976046490{\sqrt {5}}}}}}-{\sqrt[{3}]{7451370+3332354{\sqrt {5}}-6{\sqrt {8890710030+3976046490{\sqrt {5}}}}}}}{6}}\approx 1.40677447684\times 10^{-6}}}

  

{\displaystyle {}\_{\qquad {\mbox{Root of }}312500000x^{5}-6843750000x^{4}+6826250000x^{3}+10476025000x^{2}-7886869750x-72099=0}\,}
{\displaystyle {}\_{\tan \left({\frac {\arctan 4}{5}}+{\frac {4\pi }{5}}\right)+{\frac {19}{50}}={\frac {219}{50}}+{\frac {-1-{\sqrt {5}}+{\sqrt {10-2{\sqrt {5}}}}{\rm {i}}}{4}}{\sqrt[{5}]{884+799{\rm {i}}}}+{\frac {-1-{\sqrt {5}}-{\sqrt {10-2{\sqrt {5}}}}{\rm {i}}}{4}}{\sqrt[{5}]{884-799{\rm {i}}}}+{\frac {-1+{\sqrt {5}}-{\sqrt {10+2{\sqrt {5}}}}{\rm {i}}}{4}}{\sqrt[{5}]{1156+289{\rm {i}}}}+{\frac {-1+{\sqrt {5}}+{\sqrt {10+2{\sqrt {5}}}}{\rm {i}}}{4}}{\sqrt[{5}]{1156-289{\rm {i}}}}\approx -9.141538637378949398666277\times 10^{-6}}}

  

{\displaystyle {}\_{\rm {{erfi}\left({\rm {{erfi}{\frac {\sqrt {3}}{3}}}}\right)={\frac {2}{\sqrt {\pi }}}\int \_{0}^{{\frac {2}{\sqrt {\pi }}}\int \_{0}^{\frac {\sqrt {3}}{3}}e^{t^{2}}{\rm {{d}t}}}e^{u^{2}}{\rm {{d}u={\frac {2}{\sqrt {\pi }}}e^{\left({\frac {2{\sqrt[{3}]{e}}}{\sqrt {\pi }}}\int \_{0}^{\infty }{\frac {\sin \left({\frac {2}{3}}{\sqrt {3}}t\right)}{e^{t^{2}}}}{\rm {d}}t\right)^{2}}\int \_{0}^{\infty }{\frac {\sin \left[{\frac {4u{\sqrt[{3}]{e}}}{\sqrt {\pi }}}\int \_{0}^{\infty }{\frac {\sin \left({\frac {2}{3}}{\sqrt {3}}t\right)}{e^{t^{2}}}}{\rm {d}}t\right]}{e^{u^{2}}}}{\rm {d}}u={\frac {2}{\sqrt {\pi }}}\int \_{0}^{{}\_{{\frac {2{\sqrt[{3}]{e}}}{\sqrt {\pi }}}\int \_{0}^{\infty }{\frac {\sin \left({\frac {2}{3}}{\sqrt {3}}t\right)}{e^{t^{2}}}}{\rm {d}}t}}e^{u^{2}}{\rm {d}}u={\frac {2}{\sqrt {\pi }}}e^{\left({\frac {2}{\sqrt {\pi }}}\int \_{0}^{\frac {\sqrt {3}}{3}}e^{t^{2}}{\rm {{d}t}}\right)^{2}}\int \_{0}^{\infty }{\frac {\sin \left({\frac {4u}{\sqrt {\pi }}}\int \_{0}^{\frac {\sqrt {3}}{3}}e^{t^{2}}{\rm {{d}t}}\right)}{e^{u^{2}}}}{\rm {d}}u\approx 1.00002087363809430195879}}}}}
{\displaystyle \sin 11=-0.999990207...}，這是由於{\displaystyle 3.5\pi \approx 10.9956\approx 11}的緣故，另一個類似的例子為{\displaystyle \sin 355=-0.00003014435335948844921433...}
{\displaystyle {\sqrt {1^{2}+2^{2}+3^{2}+4^{2}+.......+552057^{2}}}\approx 236818619.0000004307}

外部連結
----

* J.S. Markovitch Coincidence, data compression, and Mach's concept of economy of thought （页面存档备份，存于）

註釋
--

1. ↑  此式可利用數學歸納法與性質{\displaystyle \varphi ^{2}=\varphi +1}證明。

參考資料
----

1. 1 2  Eric Weisstein, "Almost Integer" （页面存档备份，存于） at MathWorld
2. ↑ .  [2011-09-17]. （原始内容存档于2009-08-11）.