接近整数

在趣味数学中，接近整数是指很接近整数的无理数。这类数字中，有些因为其数学上的特性使其接近整数，有些还找不到其特性，看起来似乎只是巧合。

Ed Pegg jr.先生发现上图中的线段d长度为${\displaystyle {\frac {1}{2}}{\sqrt {{\frac {1}{30}}(61421-23{\sqrt {5831385}})}}}$，非常接近7（数值为7.0000000857）^[1]

有关黄金比例及其他皮索特-维贾亚拉加文数

黄金比例${\displaystyle \varphi ={\frac {1+{\sqrt {5}}}{2}}\approx 1.61803398875\,}$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${\displaystyle F_{n}}$代表费波纳契数列的第${\displaystyle n}$项

这是因为有恒等式${\displaystyle \varphi ^{n}=F_{n-1}+F_{n}\times \varphi }$^{[注 1]}，所以当${\displaystyle n}$为足够大的正整数时，

${\displaystyle \varphi ^{n}=F_{n-1}+F_{n}\times \varphi \approx F_{n-1}+F_{n}\times \left({\frac {F_{n+1}}{F_{n}}}\right)=F_{n-1}+F_{n+1}}$

这些数字接近整数的原因和黄金比例的特性有关，不是数学巧合。其原因是因为黄金比例为皮索特-维贾亚拉加文数，而皮索特-维贾亚拉加文数的高次方会是接近整数。

这些数字与费波纳契数有密切的关系，因为费波纳契数相邻两项的比值会趋近于黄金比例，而如果m整除n，则第m个费波纳契数也会整除第n个费波纳契数。

皮索特-维贾亚拉加文数是指代数数本身大于1，而且其极小多项式中另一根的绝对值小于1。像黄金比例本身大于1，${\displaystyle \varphi }$的最小多项式为 ${\displaystyle x^{2}-x-1=0}$

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

绝对值小于1，因此黄金比例为皮索特-维贾亚拉加文数，其高次方会是接近整数。

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${\displaystyle \varphi +{\overline {\varphi }}}$来表示，由于二根之和及二根之积均为整数，计算所得的结果也是一个正整数，假设为一正整数K，则${\displaystyle \varphi ^{n}}$可以用下式表示

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

由于${\displaystyle {\overline {\varphi }}}$的绝对值小于1，在n增大时，其高次方会趋于0，此时可得

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

除了黄金比例外，其他皮索特-维贾亚拉加文数的无理数也符合此一条件，例如${\displaystyle 1+{\sqrt {2}}}$。

有关黑格纳数

以下也是几个非巧合出现的接近整数，和最大三项的黑格纳数有关：

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

以上三式可以用以下的式子表示^[2]:

${\displaystyle e^{\pi {\sqrt {43}}}=12^{3}(9^{2}-1)^{3}+744-2.225\cdots \times 10^{-4}\,}$

${\displaystyle e^{\pi {\sqrt {67}}}=12^{3}(21^{2}-1)^{3}+744-1.337\cdots \times 10^{-6}\,}$

${\displaystyle e^{\pi {\sqrt {163}}}=12^{3}(231^{2}-1)^{3}+744-7.499\cdots \times 10^{-13}\,}$

其中：${\displaystyle 21=3\times 7,231=3\times 7\times 11,744=24\times 31\,}$ 由于艾森斯坦级数的关系，使得上式中出现平方项。常数${\displaystyle e^{\pi {\sqrt {163}}}\,}$有时会称为拉马努金常数。

有关π及e

许多有关π及e的常数也是接近整数，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年为止还没找到出现此特性的原因^[1]，因此只能视为一数学巧合。另一个有关格尔丰德常数的常数也是接近整数 ${\displaystyle {\frac {e^{\pi }-\pi -1}{6\pi }}=1.00793356\cdots \,}$

以下也是一些接近整数的例子

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

其他例子

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

${\displaystyle 2^{2^{2/3}}\approx 3.00507511272}$

${\displaystyle 4\ln 2.117\approx 2.99999996861}$

${\displaystyle {}_{\ln K_{0}-\ln \ln K_{0}\approx 1.0000744}}$ ，其中${\displaystyle K_{0}}$是辛钦常数

${\displaystyle {}_{{\frac {10}{81}}-\sum _{n=1}^{\infty }{\frac {\sum _{k=10^{n-1}}^{10^{n}-1}10^{-n\left[k-(10^{n-1}-1)\right]}k}{10^{\sum _{k=0}^{n-1}9\times 10^{k-1}k}}}={\frac {10}{81}}-\sum _{n=1}^{\infty }\sum _{k=10^{n-1}}^{10^{n}-1}{\frac {k}{10^{kn-9\sum _{k=0}^{n-1}10^{k}(n-k)}}}\approx 1.022344\times 10^{-9}}}$

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

${\displaystyle {}_{\qquad {\mbox{Root of }}x^{6}-615x^{5}+151290x^{4}-18608670x^{3}[phone]x^{2}-28153057165x+39605=0}\,}$

${\displaystyle {}_{{\frac {615-55{\sqrt {5}}-{\sqrt[{3}]{7451370+3332354{\sqrt {5}}+6{\sqrt {8890710030+3976046490{\sqrt {5}}}}}}-{\sqrt[{3}]{7451370+3332354{\sqrt {5}}-6{\sqrt {8890710030+3976046490{\sqrt {5}}}}}}}{6}}\approx 1.40677447684\times 10^{-6}}}$

${\displaystyle {}_{\qquad {\mbox{Root of }}312500000x^{5}-6843750000x^{4}[phone]x^{3}[phone]x^{2}-7886869750x-72099=0}\,}$

${\displaystyle {}_{\tan \left({\frac {\arctan 4}{5}}+{\frac {4\pi }{5}}\right)+{\frac {19}{50}}={\frac {219}{50}}+{\frac {-1-{\sqrt {5}}+{\sqrt {10-2{\sqrt {5}}}}{\rm {i}}}{4}}{\sqrt[{5}]{884+799{\rm {i}}}}+{\frac {-1-{\sqrt {5}}-{\sqrt {10-2{\sqrt {5}}}}{\rm {i}}}{4}}{\sqrt[{5}]{884-799{\rm {i}}}}+{\frac {-1+{\sqrt {5}}-{\sqrt {10+2{\sqrt {5}}}}{\rm {i}}}{4}}{\sqrt[{5}]{1156+289{\rm {i}}}}+{\frac {-1+{\sqrt {5}}+{\sqrt {10+2{\sqrt {5}}}}{\rm {i}}}{4}}{\sqrt[{5}]{1156-289{\rm {i}}}}\approx -9.141538637378949398666277\times 10^{-6}}}$

${\displaystyle {}_{\rm {{erfi}\left({\rm {{erfi}{\frac {\sqrt {3}}{3}}}}\right)={\frac {2}{\sqrt {\pi }}}\int _{0}^{{\frac {2}{\sqrt {\pi }}}\int _{0}^{\frac {\sqrt {3}}{3}}e^{t^{2}}{\rm {{d}t}}}e^{u^{2}}{\rm {{d}u={\frac {2}{\sqrt {\pi }}}e^{\left({\frac {2{\sqrt[{3}]{e}}}{\sqrt {\pi }}}\int _{0}^{\infty }{\frac {\sin \left({\frac {2}{3}}{\sqrt {3}}t\right)}{e^{t^{2}}}}{\rm {d}}t\right)^{2}}\int _{0}^{\infty }{\frac {\sin \left[{\frac {4u{\sqrt[{3}]{e}}}{\sqrt {\pi }}}\int _{0}^{\infty }{\frac {\sin \left({\frac {2}{3}}{\sqrt {3}}t\right)}{e^{t^{2}}}}{\rm {d}}t\right]}{e^{u^{2}}}}{\rm {d}}u={\frac {2}{\sqrt {\pi }}}\int _{0}^{{}_{{\frac {2{\sqrt[{3}]{e}}}{\sqrt {\pi }}}\int _{0}^{\infty }{\frac {\sin \left({\frac {2}{3}}{\sqrt {3}}t\right)}{e^{t^{2}}}}{\rm {d}}t}}e^{u^{2}}{\rm {d}}u={\frac {2}{\sqrt {\pi }}}e^{\left({\frac {2}{\sqrt {\pi }}}\int _{0}^{\frac {\sqrt {3}}{3}}e^{t^{2}}{\rm {{d}t}}\right)^{2}}\int _{0}^{\infty }{\frac {\sin \left({\frac {4u}{\sqrt {\pi }}}\int _{0}^{\frac {\sqrt {3}}{3}}e^{t^{2}}{\rm {{d}t}}\right)}{e^{u^{2}}}}{\rm {d}}u\approx 1.00002087363809430195879}}}}}$

${\displaystyle \sin 11=-0.999990207...}$，这是由于${\displaystyle 3.5\pi \approx 10.9956\approx 11}$的缘故，另一个类似的例子为${\displaystyle \sin 355=-0.00003014435335948844921433...}$

${\displaystyle {\sqrt {1^{2}+2^{2}+3^{2}+4^{2}+.......+552057^{2}}}\approx 236818619.0000004307}$

外部链接

• J.S. Markovitch Coincidence, data compression, and Mach's concept of economy of thought （页面存档备份，存于互联网档案馆）

注释

1. 此式可利用数学归纳法与性质${\displaystyle \varphi ^{2}=\varphi +1}$证明。

参考资料

1. Eric Weisstein, "Almost Integer" （页面存档备份，存于互联网档案馆） at MathWorld
2. 存档副本. [2011-09-17]. （原始内容存档于2009-08-11）.