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%I #16 May 25 2022 11:23:37
%S 1,2,12,4,34,1,22,1,4,1,1,2,1,1,17,16,3,1,1,2,1,1,1,5,1,1,3,3,14,2,
%T 107,1,1,8,5,4,7,1,4,1,6,3,19,3,1,1,2,3,5,76,1,1,2,1,1,90,2,2,48717,1,
%U 1,1,3,1,1,2,4,1,1,1,14,2,1,1,2,4,28,2,3,46,1,1,3,1,1,1,2,1,5,12,1,1,3,3,1,2,3,1,78,1,1,1,3,2,4,1,6,1,1,1048,1,3,1,1,2,3,4,1,2,4,3,8,1,12,5,1,1,7,1,11,11,1,118,6,1,2,1,5,3,1,1,1,2,3,1,2,1,1,2,2,3,5,4,1,12,147838832589501802758390,1,10,1,1,1,2,4,6,10,2,8,1,2,1,1,7,1,1,1,3,9,1,1,1,55,1,1,1,1,2
%N The leading digits in the terms in A067103 converge; dividing by a suitable power of 10 they converge to the number shown below; sequence gives continued fraction for this number.
%e 1.480389426511475059423875475946678140937510326102334419703757169...
%o (PARI) {A067103(n)= c=0; d=0; for(i=1,n, c=c*10^(1+floor(3*log(i)/log(10)))+i^3; d=d*10^(1+floor(log(i)/log(10)))+i; ); floor(c/d) }
%o (PARI) c1(n) = my(s=""); for(k=1, n, s=Str(s, k)); eval(s); \\ A007908
%o c3(n) = my(s=""); for(k=1, n, s=Str(s, k^3)); eval(s); \\ A019522
%o lista() = my(nn=1000); default(realprecision, 1000); my(x=c3(nn)\c1(nn)); x = x/10.^(#Str(x)-1); contfrac(x); \\ _Michel Marcus_, May 25 2022
%Y Cf. A007908, A019522, A067103.
%Y For a more dramatic continued fraction see A030167.
%K easy,cofr,nonn,base
%O 1,2
%A _Randall L Rathbun_, Jan 12 2002
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