接近整数

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。 黃金比例 φ = 1 + 5 2 ≈ 1.61803398875 {\displaystyle \varphi ={\frac {1+{\sqrt {5}}}{2}}\approx 1.61803398875\,} 的高次方符合此特性。例如 φ 17 = 3571 + 1597 5 2 ≈ 3571.00028 ≈ F 16 + F 18 {\displaystyle \varphi ^{17}={\frac {3571+1597{\sqrt {5}}}{2}}\approx 3571.00028\approx F_{16}+F_{18}} φ 18 = 2889 + 1292 5 ≈ 5777.999827 ≈ F 17 + F 19 {\displaystyle \varphi ^{18}=2889+1292{\sqrt {5}}\approx 5777.999827\approx F_{17}+F_{19}} φ 19 = 9349 + 4181 5 2 ≈ 9349.000107 ≈ F 18 + F 20 {\displaystyle \varphi ^{19}={\frac {9349+4181{\sqrt {5}}}{2}}\approx 9349.000107\approx F_{18}+F_{20}} 其中 F n {\displaystyle F_{n}} 代表費波納契數列的第 n {\displaystyle n} 項 這是因為有恆等式 φ n = F n − 1 + F n × φ {\displaystyle \varphi ^{n}=F_{n-1}+F_{n}\times \varphi } ，所以當 n {\displaystyle n} 為足夠大的正整數時， φ n = F n − 1 + F n × φ ≈ F n − 1 + F n × ( F n + 1 F n ) = F n − 1 + F n + 1 {\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-维贾亚拉加文数，而皮索特-维贾亚拉加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-维贾亚拉加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， φ {\displaystyle \varphi } 的最小多項式為 x 2 − x − 1 = 0 {\displaystyle x^{2}-x-1=0} 另一根為 φ ¯ = 1 − 5 2 ≈ − 0.618 {\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-维贾亚拉加文数，其高次方會是接近整数。 依照根和系数的关系，可得知 φ φ ¯ = − 1 {\displaystyle \varphi {\overline {\varphi }}=-1} φ + φ ¯ = 1 {\displaystyle \varphi +{\overline {\varphi }}=1} 而 φ n + φ ¯ n {\displaystyle \varphi ^{n}+{\overline {\varphi }}^{n}} 可以用 φ φ ¯ {\displaystyle \varphi {\overline {\varphi }}} 及 φ + φ ¯ {\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 φ n {\displaystyle \varphi ^{n}} 可以用下式表示 φ n = K − φ ¯ n {\displaystyle \varphi ^{n}=K-{\overline {\varphi }}^{n}} 由於 φ ¯ {\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 φ n ≈ K {\displaystyle \varphi ^{n}\approx K} 除了黃金比例外，其他皮索特-维贾亚拉加文数的無理數也符合此一條件，例如 1 + 2 {\displaystyle 1+{\sqrt {2}}} 。 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： e π 43 ≈ 884736743.999777466 {\displaystyle e^{\pi {\sqrt {43}}}\approx 884736743.999777466\,} e π 67 ≈ 147197952743.999998662454 {\displaystyle e^{\pi {\sqrt {67}}}\approx 147197952743.999998662454\,} e π 163 ≈ 262537412640768743.99999999999925007 {\displaystyle e^{\pi {\sqrt {163}}}\approx 262537412640768743.99999999999925007\,} 以上三式可以用以下的式子表示: e π 43 = 12 3 ( 9 2 − 1 ) 3 + 744 − 2.225 ⋯ × 10 − 4 {\displaystyle e^{\pi {\sqrt {43}}}=12^{3}(9^{2}-1)^{3}+744-2.225\cdots \times 10^{-4}\,} e π 67 = 12 3 ( 21 2 − 1 ) 3 + 744 − 1.337 ⋯ × 10 − 6 {\displaystyle e^{\pi {\sqrt {67}}}=12^{3}(21^{2}-1)^{3}+744-1.337\cdots \times 10^{-6}\,} e π 163 = 12 3 ( 231 2 − 1 ) 3 + 744 − 7.499 ⋯ × 10 − 13 {\displaystyle e^{\pi {\sqrt {163}}}=12^{3}(231^{2}-1)^{3}+744-7.499\cdots \times 10^{-13}\,} 其中： 21 = 3 × 7 , 231 = 3 × 7 × 11 , 744 = 24 × 31 {\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 e π 163 {\displaystyle e^{\pi {\sqrt {163}}}\,} 有時會稱為拉馬努金常數。 許多有關π及e的常數也是接近整數，例如 e π − π = 19.999099979189 ⋯ {\displaystyle e^{\pi }-\pi =19.999099979189\cdots \,} 以及 e 5 π = 6635623.999341134233 ⋯ {\displaystyle e^{5\pi }=6635623.999341134233\cdots \,} 格尔丰德常数（ e π {\displaystyle e^{\pi }\,} ）接近 π + 20 {\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 e π − π − 1 6 π = 1.00793356 ⋯ {\displaystyle {\frac {e^{\pi }-\pi -1}{6\pi }}=1.00793356\cdots \,} 以下也是一些接近整數的例子 22 π 4 = 2143.0000027480 ⋯ {\displaystyle 22{\pi }^{4}=2143.0000027480\cdots \,} π 5 = 306.019684 ⋯ {\displaystyle {\pi }^{5}=306.019684\cdots \,} π 3 = 31.006276 ⋯ {\displaystyle {\pi }^{3}=31.006276\cdots \,} π 13 / 2 = 1704.017978 ⋯ {\displaystyle {\pi }^{13/2}=1704.017978\cdots \,} π 3 − π 500 = 30.999993494 ⋯ {\displaystyle {\pi }^{3}-{\frac {\pi }{500}}=30.999993494\cdots \,} π 2 + π 24 = 10.000504 ⋯ {\displaystyle {\pi }^{2}+{\frac {\pi }{24}}=10.000504\cdots \,} π 5 − 3 π 3 = 213.0008547 ⋯ {\displaystyle {\pi }^{5}-3{\pi }^{3}=213.0008547\cdots \,} e π − π + 0.0009 = 19.9999999791 ⋯ {\displaystyle e^{\pi }-\pi +0.0009=19.9999999791\cdots \,} e 3 = 20.0855369 ⋯ {\displaystyle e^{3}=20.0855369\cdots \,} e 9 / 2 = 90.0171313 ⋯ {\displaystyle e^{9/2}=90.0171313\cdots \,} e π 2 = 85.019695 ⋯ {\displaystyle e^{\pi {\sqrt {2}}}=85.019695\cdots \,} e 6 − π 5 − π 4 = 0.00001767 ⋯ {\displaystyle e^{6}-{\pi }^{5}-{\pi }^{4}=0.00001767\cdots \,} 9 π 5 − 2 e 3 = 2714.00608922 ⋯ {\displaystyle 9{\pi }^{5}-2e^{3}=2714.00608922\cdots \,} e 13 / 2 − π = 662.00004039 ⋯ {\displaystyle e^{13/2}-\pi =662.00004039\cdots \,} π 13 / 2 − e 9 / 2 = 1614.00084707 ⋯ {\displaystyle {\pi }^{13/2}-e^{9/2}=1614.00084707\cdots \,} 2 2 2 / 3 ≈ 3.00507511272 {\displaystyle 2^{2^{2/3}}\approx 3.00507511272} 4 ln ⁡ 2.117 ≈ 2.99999996861 {\displaystyle 4\ln 2.117\approx 2.99999996861} ln ⁡ K 0 − ln ⁡ ln ⁡ K 0 ≈ 1.0000744 {\displaystyle {}_{\ln K_{0}-\ln \ln K_{0}\approx 1.0000744}} ，其中 K 0 {\displaystyle K_{0}} 是辛钦常数 10 81 − ∑ n = 1 ∞ ∑ k = 10 n − 1 10 n − 1 10 − n [ k − ( 10 n − 1 − 1 ) ] k 10 ∑ k = 0 n − 1 9 × 10 k − 1 k = 10 81 − ∑ n = 1 ∞ ∑ k = 10 n − 1 10 n − 1 k 10 k n − 9 ∑ k = 0 n − 1 10 k ( n − k ) ≈ 1.022344 × 10 − 9 {\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}}} − 1 5 + e 6 5 4 F 3 ( − 1 5 , 1 20 , 3 10 , 11 20 ; 1 5 , 2 5 , 3 5 ; 256 3125 e 6 ) + 2 25 e 6 5 4 F 3 ( 1 5 , 9 20 , 7 10 , 19 20 ; 3 5 , 4 5 , 7 5 ; 256 3125 e 6 ) − 4 125 e 12 5 4 F 3 ( 2 5 , 13 20 , 9 10 , 23 20 ; 4 5 , 6 5 , 8 5 ; 256 3125 e 6 ) + 7 625 e 18 5 4 F 3 ( 3 5 , 17 20 , 11 10 , 27 20 ; 6 5 , 7 5 , 9 5 ; 256 3125 e 6 ) − π ≈ 2.89221114964408683 × 10 − 8 {\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}}} Root of  x 6 − 615 x 5 + 151290 x 4 − 18608670 x 3 + 1144433205 x 2 − 28153057165 x + 39605 = 0 {\displaystyle {}_{\qquad {\mbox{Root of }}x^{6}-615x^{5}+151290x^{4}-18608670x^{3}+1144433205x^{2}-28153057165x+39605=0}\,} 615 − 55 5 − 7451370 + 3332354 5 + 6 8890710030 + 3976046490 5 3 − 7451370 + 3332354 5 − 6 8890710030 + 3976046490 5 3 6 ≈ 1.40677447684 × 10 − 6 {\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}}} Root of  312500000 x 5 − 6843750000 x 4 + 6826250000 x 3 + 10476025000 x 2 − 7886869750 x − 72099 = 0 {\displaystyle {}_{\qquad {\mbox{Root of }}312500000x^{5}-6843750000x^{4}+6826250000x^{3}+10476025000x^{2}-7886869750x-72099=0}\,} tan ⁡ ( arctan ⁡ 4 5 + 4 π 5 ) + 19 50 = 219 50 + − 1 − 5 + 10 − 2 5 i 4 884 + 799 i 5 + − 1 − 5 − 10 − 2 5 i 4 884 − 799 i 5 + − 1 + 5 − 10 + 2 5 i 4 1156 + 289 i 5 + − 1 + 5 + 10 + 2 5 i 4 1156 − 289 i 5 ≈ − 9.141538637378949398666277 × 10 − 6 {\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}}} e r f i ( e r f i 3 3 ) = 2 π ∫ 0 2 π ∫ 0 3 3 e t 2 d t e u 2 d u = 2 π e ( 2 e 3 π ∫ 0 ∞ sin ⁡ ( 2 3 3 t ) e t 2 d t ) 2 ∫ 0 ∞ sin ⁡ [ 4 u e 3 π ∫ 0 ∞ sin ⁡ ( 2 3 3 t ) e t 2 d t ] e u 2 d u = 2 π ∫ 0 2 e 3 π ∫ 0 ∞ sin ⁡ ( 2 3 3 t ) e t 2 d t e u 2 d u = 2 π e ( 2 π ∫ 0 3 3 e t 2 d t ) 2 ∫ 0 ∞ sin ⁡ ( 4 u π ∫ 0 3 3 e t 2 d t ) e u 2 d u ≈ 1.00002087363809430195879 {\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}}}}} sin ⁡ 11 = − 0.999990207... {\displaystyle \sin 11=-0.999990207...} ，這是由於 3.5 π ≈ 10.9956 ≈ 11 {\displaystyle 3.5\pi \approx 10.9956\approx 11} 的緣故，另一個類似的例子為 sin ⁡ 355 = − 0.00003014435335948844921433... {\displaystyle \sin 355=-0.00003014435335948844921433...} 1 2 + 2 2 + 3 2 + 4 2 + . . . . . . . + 552057 2 ≈ 236818619.0000004307 {\displaystyle {\sqrt {1^{2}+2^{2}+3^{2}+4^{2}+.......+552057^{2}}}\approx 236818619.0000004307} 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） 此式可利用數學歸納法與性質 φ 2 = φ + 1 {\displaystyle \varphi ^{2}=\varphi +1} 證明。 Eric Weisstein, "Almost Integer" （页面存档备份，存于互联网档案馆） at MathWorld 存档副本. [2011-09-17]. （原始内容存档于2009-08-11）.