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%I #11 Jan 11 2021 23:35:42
%S 1,1,1,63,7,3,2,2,1,1,1,250,2,1,2,1,2,3,1,4,1,1,3,1,1,2,1,1,1,2,1,2,2,
%T 1,1,6,7,1,1,1,6,1,1,9,9,2,1,6,2,5,1,25,1,1,1,2,18,1,3,5,1,1,5,1,3,1,
%U 1,4,1,1,3,2,2,3,40,2,3,8,2,2,25,1,5,2,1,1,3,2,2,1,10,1,1,2,1,2,1,1,2,1,3,2,420,2,2,1
%N Continued fraction expansion of the smallest constant d such that the numbers floor(2^(n^d)) are distinct primes for all n >= 1.
%e 1+1/(1+1/(1+1/(63+1/(7+1/(3+1/(2+1/(2+1/(1+1/(1+1/(1+1/(250] = 22739482/15120055 = 1.503928524069522...
%e The constant is equal to d=1.503928524069520633527689067897583199190738849581138429002999...
%o (PARI) A339458(n=63, prec=200)={
%o my(curprec=default(realprecision));
%o default(realprecision, max(prec,curprec));
%o my(a=List([2]), d=1.0, c=2.0, b, p, ok, smpr(b)=my(p=b); while(!isprime(p), p=nextprime(p+1)); return(p); );
%o for(j=1, n-1,
%o b=floor(c^(j^d));
%o until(ok,
%o p=smpr(b);
%o ok = 1;
%o listput(a,p,j);
%o if(p!=b,
%o d=log(log(p)/log(c))/log(j);
%o for(k=1,j-2,
%o b=floor(c^(k^d));
%o if(b!=a[k],
%o ok=0;
%o j=k;
%o break;
%o );
%o );
%o );
%o );
%o );
%o default(realprecision, curprec);
%o return(contfrac(d));
%o } \\ _François Marques_, Dec 08 2020
%Y Cf. A339459, A339457, A338613, A338837, A338850.
%K nonn,cofr
%O 1,4
%A _Bernard Montaron_, Dec 06 2020
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