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%I #10 Jun 12 2022 07:22:36
%S 1,1,1,1,3,32,2,3,3,1,3,2,3,6,1,3,1,3,6,8,1,4,17,1,2,3,3,1,2,2,3,5,15,
%T 1,1,5,1,1,3,2,1,1,1,3,1,2,2,2,1,1,1,25,1,8,10,1,1,11,2,2,25,2,16,1,2,
%U 1,2,1,1,1,2,11,4,35,3,1,1,9,59,1,2,2,1
%N Continued fraction of A137245 = sum( 1/(p log p), p prime ) = 1.63661632335126...
%o (PARI) s=1.6366163233512608685696580039218636711815970761312; e=.1^default(realprecision); a=[contfrac(s+e),contfrac(s-e)]; for(n=1,min(#a[1],#a[2]), a[1][n] == a[2][n] & print1(a[1][n]","))
%K nonn,cofr
%O 0,5
%A _M. F. Hasler_, Mar 27 2012
%E More terms from _Vaclav Kotesovec_, Jun 12 2022
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