来自在线整数百科全书的问候语!051599,1A-1,1,A065,15%,1,4,15,16,85,1085,645 8605888,166644,63555,252564,9,%,T A065 159 467845 1520838,191515315369336361056810986182565,% U U A065 159 11838 4196182120353553541 31 1024338 25348 84149151553500,http://oei.org/y*搜索:ID:A:1217189πn A065 159二进制串自替换:A(n)是通过n为n的每一个比特的二进制扩展而获得的,n为%%C A065 159A(0)=0。A(2 ^ n)=4 ^ n(4n+2)=(4n+1)*(1 +a(4n+1)/(4n+1)).{%cA065 159a(n)=a065 157(n,n)=a065 158(n,n)*n=a065 160(n)*n %h h a065 159 Paul Tek,n,a(n)n=0…10000的表%F A065159 a(n)=z(n, n) with z(u, v) = if u=0 then 0 else if u mod 2 = 0 then z(u/2, v)*2 else z([u/2], v)*A062383(v)+v. - _Reinhard Zumkeller_, Feb 15 2004 %e A065159 a(5): 5=101->(101)0(101)=1010101=85. %t A065159 bss[n_]:=Module[{idn2=IntegerDigits[n,2]},FromDigits[Flatten[idn2/.{1-> idn2}],2]]; Array[bss,40,0] (* _Harvey P. Dale_, Aug 15 2017 *)%0YA065 159 CF.A065 157,A065 158,A065 160.0%K A065 159基座,易,非N%%O A065 159 0,3‰A065 159 MARC勒布吕,10月18日2001‰内容在OEIS最终用户许可协议下可用:HTTP:/OEIS.Org/许可证