A229990 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.
1, 3, 4, 8, 12, 19, 20, 24, 28, 31, 40, 44, 48, 52, 55, 64, 67, 68, 71, 72, 84, 91, 92, 99, 100, 104, 108, 111, 115, 120, 127, 128, 131, 132, 140, 148, 151, 152, 155, 160, 171, 175, 180, 184, 187, 188, 204, 208, 211, 220, 224, 231, 232, 235, 239, 244, 248, 252
Offset: 1
Keywords
Examples
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.
Links
- Nathaniel Johnston, Table of n, a(n) for n = 1..10000
Programs
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Maple
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
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Mathematica
z = 1000; c[n_] := PrimePi[Floor[3 n/2]] - PrimePi[Floor[n/2]-1]; t = Table[c[n], {n, 1, z}]; (* A229989 *) Flatten[Position[Differences[t], -1]] (* A076274? *) Flatten[Position[Differences[t], 1]] (* A229990 *)