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Smallest exponent of 2 when the number of primes in the neighborhood of center=2^n and radius=ceiling(log(2^n)) equals n.
2

%I #9 Aug 30 2019 15:18:41

%S 25,9,1,2,20,18,127,844,573

%N Smallest exponent of 2 when the number of primes in the neighborhood of center=2^n and radius=ceiling(log(2^n)) equals n.

%e First in the suitable neighborhood of 2^25 no primes occur: a[0]=25, while corresponding around 2^127 6 primes arise: a[6]=127.

%t t=Table[Count[Table[PrimeQ[2^n+i], {i, -Ceiling[Log[2^n]//N], Ceiling[Log[2^n]//N]}], True], {n, 1, 256}] Table[Min[Flatten[Position[t, j]]], {j, 0, 10}]

%Y Cf. A096509-A096523.

%K nonn,more

%O 0,1

%A _Labos Elemer_, Jul 12 2004