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a(n) = Sum_{k=1..n} floor(p(k)/k).
1

%I #10 Nov 07 2019 14:45:52

%S 2,3,4,5,7,9,11,13,15,17,19,22,25,28,31,34,37,40,43,46,49,52,55,58,61,

%T 64,67,70,73,76,80,84,88,92,96,100,104,108,112,116,120,124,128,132,

%U 136,140,144,148,152,156,160,164,168,172,176,180,184,188,192,196,200,204,208,212,216

%N a(n) = Sum_{k=1..n} floor(p(k)/k).

%H Harvey P. Dale, <a href="/A024926/b024926.txt">Table of n, a(n) for n = 1..1000</a>

%t Table[Floor[Prime[n]/n],{n,70}]//Accumulate (* _Harvey P. Dale_, Nov 07 2019 *)

%Y Partial sums of A038605.

%K nonn

%O 1,1

%A _Clark Kimberling_