以下对卡迈克尔数561做重新计算
$\varphi(561)=\varphi(3*11*17)=\varphi(3)\varphi(11)\varphi(17)=2*2^3*2^4=2^8$
$\varphi(560)=\varphi(2^4*5*7)=\varphi(2^4)\varphi(5)\varphi(7)=2^4*2^2*2^2=2^8$
卡迈克尔数1105
$\varphi(1105)=\varphi(5*13*17)=\varphi(5)\varphi(13)\varphi(17)=2^2*2^3*2^4=2^9$
$\varphi(1104)=\varphi(2^4*3*23)=\varphi(2^4)\varphi(3)\varphi(23)=2^4*2*2^4=2^9$
卡迈克尔数1729
$\varphi(1729)=\varphi(7*13*19)=\varphi(7)\varphi(13)\varphi(19)=2^2*2^3*2^3=2^8$
$\varphi(1728)=\varphi(2^6*3^3)=\varphi(2^6)\varphi(3^3)=2^6*2^3=2^9$
本帖最后由 wsc810 于 2020-9-27 14:29 编辑
重算 2047
$\varphi(2047)=\varphi(23)\varphi(89)=\varphi(22)*\varphi(88)=\varphi(16)\varphi(11^2)=2^4*(2^3)^2=2^10$
$\varphi(2046)=\varphi(2)\varphi(1023)=\varphi(2)\varphi(3)\varphi(11)\varphi(31)=2*2*2^3*2^4=2^9$
4181
$\varphi(4181)=\varphi(37)\varphi(113)=\varphi(36)\varphi(112)=\varphi(36)\varphi(16)\varphi(7)=2^4*2^4*2^2=2^10$
$\varphi(4180)=\varphi(4)\varphi(5)\varphi(11)\varphi(19)=2^2*2^2*2^3*2^3=2^10$
本帖最后由 wsc810 于 2020-9-27 19:07 编辑
编写一个Mathematica函数算吧，手工计算太麻烦 第1种定义的Mathematica程序
f:=
Module[{n=n0}, f=1; f=2;
inner[{x_List, y_List}]:=Inner;
f:=Which,
OddQ@n, inner@Transpose@FactorInteger@n,
EvenQ@n, 2 inner@Transpose@FactorInteger@NestWhile[#/2&, n, EvenQ]];
f]
Table[{n, f}, {n, 25}]//Transpose//MatrixForm输出
第2种定义的Mathematica程序
ff:=
Module[{n=n0}, ff=1; ff=2;
inner[{x_List, y_List}]:=Inner;
ff:=If, inner@Transpose@FactorInteger@n];
ff]
Table[{n, ff}, {n, 25}]//Transpose//MatrixForm输出
缩进控制不准，懒得调了，将就看吧。 100//Range//{#,#//Which[#==1,0,#==2,1,PrimeQ[#],#0[#-1],True,{#0,#//FactorInteger}//Function[{RecursionFunction,nSet},RecursionFunction[#[]]#[]&/@nSet//Total]@@#&]&}&/@#&
%//ListPlot
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1
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