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A321096型 常数z的十进制展开式满足:CF(5*z,n)=CF(z,n。 8
8, 8, 0, 5, 4, 0, 1, 7, 5, 7, 6, 8, 8, 6, 6, 3, 3, 7, 3, 4, 7, 6, 9, 3, 9, 9, 9, 5, 4, 3, 7, 6, 3, 9, 8, 7, 6, 4, 1, 1, 5, 6, 2, 8, 7, 1, 4, 8, 4, 5, 3, 8, 1, 4, 6, 8, 5, 1, 6, 9, 4, 3, 7, 5, 8, 9, 0, 0, 1, 7, 7, 4, 6, 6, 9, 6, 5, 1, 6, 8, 3, 8, 3, 7, 9, 0, 1, 2, 1, 1, 0, 8, 8, 4, 4, 5, 0, 5, 7, 5, 7, 1, 7, 1, 8, 7, 9, 9, 9, 4, 5, 4, 4, 0, 7, 3, 4, 8, 1, 0, 6, 4, 9, 6, 6, 3, 6, 1, 5, 9, 4, 2, 7, 2, 3, 4, 2, 7, 4, 7, 0, 9, 6, 6, 5, 0, 6, 3, 5, 3, 0, 5, 0, 9, 3, 5, 7, 7, 7, 2, 0, 5, 7, 6, 7, 8, 5, 2 (列表;常数;图表;参考;;历史;文本;内部格式)
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例子
此常数z的十进制展开式开始于:
z=8.80540175768866337347693999543763987641156287148453。。。
z的简单连分式展开开始:
z=[8;1、4、7、4、1、7、4,1、7、1、4,7,1、4,7。。。,A321095型(n) ,…];
这样,5*z的简单连分式展开就开始了:
5*z=[44;37、40、43、40、37、43、四十、37、四十三、三十七。。。,A321095型(n) +36,…]。
扩展条款。
z的十进制展开式中的前1000位是
z=8.80540175768866337347693999543763987641156287148453\
81468516943758900177466965168383790121108844505757\
17187999454407348106496636159427234274709665063530\
5093577720576785270678192087967312074186445658376\
11976420857048526287399074761091708055469240672330\
4870032646205845266355562520962867414041721519998\
23160265836475138977655743106219535120079474624210\
99859736180993210725127756524124066610420161356552\
06399576189405297872838825683944974598313320897473\
82753910120167655227857213444857908106596044697304\
03731161925824151533224356077992140610806046874738\
76664918873754737673637688749508970659125480247854\
23148383885317039738394794978662476138806965606055\
89270492560115659978879195004407252000769277753763\
08004863113948791353859078952125599689193183276991\
65636567095719857927702210317282103234986565300353\
38502840972838489688713570280513772584527070132759\
37605094965136681249812588515170458336825974959207\
93310692922707508597239859725358109680410147516880\
36970398236866976270861217620395537245456726267033...
...
z的连分数的最初1020项是
z=[8;1,4,7,4,1,7,4-1,7,1,4,7,14,71,4,4,4,17,4,1,4,7,
4,1,7,4,1,7,1,4,7,4,1,7,4,1,7,1,4,7,1,4,7,4,1,7,1,4,7,4,1,7,
4,1,7,1,4,7,4,1,7,4,1,7,1,4,7,1,4,7,4,1,7,1,4,7,4,1,7,4,1,7,
1,4,7,1,4,7,4,1,7,1,4,7,1,4,7,4,1,7,1,4,7,4,1,7,4,1,7,1,4,7,
1,4,7,4,1,7,1,4,7,1,4,7,4,1,7,1,4,7,4,1,7,4,1,7,1,4,7,4,1,7,
4,1,7,1,4,7,1,4,7,4,1,7,1,4,7,4,1,7,4,1,7,1,4,7,1,4,7,4,1,7,
1,4,7,1,4,7,4,1,7,1,4,7,4,1,7,4,1,7,1,4,7,1,4,7,4,1,7,1,4,7,
1,4,7,4,1,7,1,4,7,4,1,7,4,1,7,1,4,7,4,1,7,4,1,7,1,4,7,1,4,7,
4,1,7,1,4,7,4,1,7,4,1,7,1,4,7,1,4,7,4,1,7,1,4,7,1,4,7,4,1,7,
1,4,7,4,1,7,4,1,7,1,4,7,4,1,7,4,1,7,1,4,7,1,4,7,4,1,7,1,4,7,
4,1,7,4,1,7,1,4,7,4,1,7,4,1,7,1,4,7,1,4,7,4,1,7,1,4,7,4,1,7,
4,1,7,1,4,7,1,4,7,4,1,7,1,4,7,1,4,7,4,1,7,1,4,7,4,1,7,4,1,7,
1,4,7,4,1,7,4,1,7,1,4,7,1,4,7,4,1,7,1,4,7,4,1,7,4,1,7,1,4,7,
4,1,7,4,1,7,1,4,7,1,4,7,4,1,7,1,4,7,4,1,7,4,1,7,1,4,7,1,4,7,
4,1,7,1,4,7,1,4,7,4,1,7,1,4,7,4,1,7,4,1,7,1,4,7,1,4,7,4,1,7,
1,4,7,1,4,7,4,1,7,1,4,7,4,1,7,4,1,7,1,4,7,4,1,7,4,1,7,1,4,7,
1,4,7,4,1,7,1,4,7,4,1,7,4,1,7,1,4,7,1,4,7,4,1,7,1,4,7,1,4,7,
4,1,7,1,4,7,4,1,7,4,1,7,1,4,7,4,1,7,4,1,7,1,4,7,1,4,7,4,1,7,
1,4,7,4,1,7,4,1,7,1,4,7,4,1,7,4,1,7,1,4,7,1,4,7,4,1,7,1,4,7,
4,1,7,4,1,7,1,4,7,1,4,7,4,1,7,1,4,7,1,4,7,4,1,7,1,4,7,4,1,7,
4,1,7,1,4,74,1,7,4,1,7,1,4,4,7,1,4,7,14,7,4,1,7,1,4,7,74,1,7,4,1,7,4,7,4,4,7,4,1,7,
1,4,7,4,1,7,4,1,7,1,4,7,14,7,1,4,74,1,7,1,4,74,1,7,4,7,4,4,4,4,4,1,7,4,1,7,1,4,4,7,
1,4,7,4,1,7,1,4,7,4,1,7,1,4,7,4,1,7,1,7,1,4,7,1,4,7,
4,1,7,1,4,7,1,4,7,4,1,7,1,4,7,4,1,7,4,1,7,1,4,7,4,1,7,4,1,7,
1,4,7,1,4,7,4,1,7,1,4,7,4,1,7,4,1,7,1,4,7,1,4,7,4,1,7,1,4,7,
1,4,7,4,1,7,1,4,7,4,1,7,4,1,7,1,4,7,4,1,7,4,1,7,1,4,7,1,4,7,
4,1,7,1,4,7,4,1,7,4,1,7,1,4,7,4,1,7,4,1,7,1,4,7,1,4,7,4,1,7,
1,4,7,4,1,7,4,1,7,1,4,7,1,4,7,4,1,7,1,4,7,1,4,7,4,1,7,1,4,7,
4,1,7,4,1,7,1,4,7,1,4,7,4,1,7,1,4,7,1,4,7,4,1,7,1,4,7,4,1,7,
4,1,7,1,4,7,4,1,7,4,1,7,1,4,7,1,4,7,4,1,7,1,4,7,4,1,7,4,1,7,
1,4,7,1,4,7,4,1,7,1,4,7,1,4,7,4,1,7,1,4,7,4,1,7,4,1,7,1,4,7,
1,4,7,4,1,7,1,4,7,1,4,7,4,1,7,1,4,7,4,1,7,4,1,7,1,4,7,4,1,7,
4,1,7,1,4,7,1,4,7,4,1,7,1,4,7,4,1,7,4,1,7,1,4,7,1,4,7,4,1,7,
1,4,7,1,4,7,4,1,7,1,4,7,4,1,7,4,1,7,1,4,7,4,1,7,4,1,7,1,4,7, ...].
...
生成方法。
从CF=[8]开始,然后重复(PARI代码):
{M=contfracpnqn(CF+向量(#CF,i,36));
z=(1/5)*M[1,1]/M[2,1];CF=控制(z)}
该方法可以如下所示。
z0=[8]=8;
z1=(1/5)*[44]=[8;1,4]=44/5;
z2=(1/5)*[44;37,40]=[8;1,4,7,4,1,7,5]=65204/7405;
z3=(1/5)*[44;37,40,43,40,37,43,41]=[8;1,4,7,4,1,7,1,4、7,1、4、7、1、4,7、4、1、7、4,8]=1467584898352/166668703909;
z4=(1/5)*[44;37,40,43,40,37,43,40,37,43第4、7、1、4、7和4、1、7和1、4和7、4和1、7、4、1和8] = 38573876143771692243421005778572392/4380705980858842541559248009960149 ; ...
其中,常数z等于上述过程的迭代极限。
黄体脂酮素
(PARI)/*生成超过5300位数字*/
{CF=[8];对于(i=1,8,M=contfracpnqn(CF+向量(#CF,i,36));z=(1/5)*M[1,1]/M[2,1];CF=contfrac(z))}
对于(n=1200,打印1(楼层(10^(n-1)*z)%10,“,”)
交叉参考
关键字
非n,欺骗
作者
保罗·D·汉纳2018年10月28日
状态
经核准的

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