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A338850
Continued fraction expansion of the smallest constant 'c' such that the numbers 1+floor(c^(n^1.5)) are distinct primes for all n >= 0.
5
2, 3, 1, 2, 1, 2, 1, 1, 1, 1, 3, 1, 2, 13, 6, 1, 3, 5, 1, 5, 1, 7, 17, 1, 3, 1, 11, 18, 3, 1, 2, 1, 2, 1, 2, 17, 15, 1, 69, 3, 1, 2, 1, 1, 1, 1, 33, 1, 3, 2, 4, 17, 1, 3, 2, 2, 1, 2, 6, 1, 11, 3, 2, 1, 1, 1, 17, 1, 7, 5, 2, 2, 2, 84, 1, 8, 3, 1, 1, 22, 3698, 2, 2, 1, 1, 2, 1, 7, 2, 1, 1, 1, 1, 3, 1, 5, 15, 1, 3, 1, 2, 1, 1, 1, 1, 2, 1, 16, 1, 7, 2, 2, 3, 1, 9
OFFSET
1,1
EXAMPLE
2+1/(3+1/(1+1/(2+1/(1+1/(2+1/(1+1/(1+1/(1+1/(1+1/(3+1/(1+1/(2+1/(13+1/(6]= 590652/260429 = 2.26799626769... The constant 'c' is equal to 2.267996267706724247328553280725371774527042254400818772275…
PROG
(PARI)
c(n=40, prec=100)={
my(curprec=default(realprecision));
default(realprecision, max(prec, curprec));
my(a=List([2]), d=1.5, c=2.0, b, p, ok, smpr(b)=my(p=b); while(!isprime(p), p=nextprime(p+1)); return(p); );
for(j=1, n-1,
b=1+floor(c^(j^d));
until(ok,
ok=1;
p=smpr(b);
listput(a, p, j+1);
if(p!=b,
c=(p-1)^(j^(-d));
for(k=1, j-2,
b=1+floor(c^(k^d));
if(b!=a[k+1],
ok=0;
j=k;
break;
);
);
);
);
);
default(realprecision, curprec);
return(c);
};
contfrac(c(50, 200), 115)
\\ François Marques, Nov 17 2020
CROSSREFS
Sequence in context: A127246 A106038 A078711 * A322423 A325494 A295561
KEYWORD
nonn,cofr
AUTHOR
Bernard Montaron, Nov 12 2020
STATUS
approved