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a(n) = ceiling(10*(prime(n+1)-prime(n))/(2*floor(sqrt(prime(n)*log(prime(n)))))).
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%I #17 Oct 15 2020 15:13:37

%S 5,10,5,7,2,4,2,3,4,2,3,2,1,2,3,3,1,2,2,1,2,2,2,3,1,1,1,1,1,4,1,2,1,2,

%T 1,2,2,1,2,2,1,2,1,1,1,2,2,1,1,1,1,1,2,1,1,1,1,1,1,1,2,2,1,1,1,2,1,2,

%U 1,1,1,1,1,1,1,1,1,1,1,2,1,1,1,1,1,1,1,1,1,1,2,1,1,1,1,1,2,1,2,1

%N a(n) = ceiling(10*(prime(n+1)-prime(n))/(2*floor(sqrt(prime(n)*log(prime(n)))))).

%H Andrew Granville, <a href="http://www.dartmouth.edu/~chance/chance_news/for_chance_news/Riemann/cramer.pdf">Harald Cramer and the Distribution of Prime Numbers</a>

%t a = Flatten[Table[10*x /. Solve[Prime[n + 1] -Prime[n] - x*2*Floor[Sqrt[Prime[n]*Log[Prime[n]]]] == 0, x], {n, 1, 100}]]; Ceiling[a]

%o (PARI) a(n) = ceil(10*(prime(n+1)-prime(n))/ (2*floor(sqrt(prime(n)*log(prime(n)))))); \\ _Michel Marcus_, Oct 15 2020

%Y Cf. A000040.

%K nonn,less

%O 1,1

%A _Roger L. Bagula_, Apr 14 2008

%E Name edited by _Michel Marcus_, Oct 15 2020