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Numbers k such that the interval [floor((k+1)/2), floor(3*(k+1)/2)] contains more primes than the interval [floor(k/2), floor(3*k/2)] does.
2

%I #13 Apr 21 2021 03:50:41

%S 1,3,4,8,12,19,20,24,28,31,40,44,48,52,55,64,67,68,71,72,84,91,92,99,

%T 100,104,108,111,115,120,127,128,131,132,140,148,151,152,155,160,171,

%U 175,180,184,187,188,204,208,211,220,224,231,232,235,239,244,248,252

%N Numbers k such that the interval [floor((k+1)/2), floor(3*(k+1)/2)] contains more primes than the interval [floor(k/2), floor(3*k/2)] does.

%H Nathaniel Johnston, <a href="/A229990/b229990.txt">Table of n, a(n) for n = 1..10000</a>

%e 4 is in this sequence because [[5/2], [15/2]] contains the primes 2,3,5,7, while [[4/2], [12/2]] contains the primes 2,3,5.

%p with(numtheory): isA229990 := proc(n) return pi(floor(3*(n+1)/2))-pi(floor((n+1)/2)-1) > pi(floor(3*n/2))-pi(floor(n/2)-1): end proc: seq(`if`(isA229990(n),n,NULL), n=1..252); # _Nathaniel Johnston_, Oct 11 2013

%t z = 1000; c[n_] := PrimePi[Floor[3 n/2]] - PrimePi[Floor[n/2]-1];

%t t = Table[c[n], {n, 1, z}]; (* A229989 *)

%t Flatten[Position[Differences[t], -1]] (* A076274? *)

%t Flatten[Position[Differences[t], 1]] (* A229990 *)

%Y Cf. A076274, A229989, A056172.

%K nonn

%O 1,2

%A _Clark Kimberling_, Oct 09 2013