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Number of 5-almost primes 5ap such that 2^n < 5ap <= 2^(n+1).
8

%I #7 Mar 30 2012 18:40:37

%S 0,0,0,0,1,1,5,8,21,41,91,199,403,873,1767,3740,7709,15910,32759,

%T 67185,138063,281566,576165,1173435,2390366,4860357,9873071,20033969,

%U 40612221,82266433,166483857,336713632,680482316,1374413154,2774347425

%N Number of 5-almost primes 5ap such that 2^n < 5ap <= 2^(n+1).

%C The partial sum equals the number of Pi_5(2^n) = 0, 0, 0, 0, 1, 2, 7, 15, 36, 77, 168, 367, 770, 1643,..

%e (2^5, 2^6] there is one semiprime, namely 48. 32 was counted in the previous entry.

%t FiveAlmostPrimePi[n_] := Sum[ PrimePi[n/(Prime@i*Prime@j*Prime@k*Prime@l)] - l + 1, {i, PrimePi[n^(1/5)]}, {j, i, PrimePi[(n/Prime@i)^(1/4)]}, {k, j, PrimePi[(n/(Prime@i*Prime@j))^(1/3)]}, {l, k, PrimePi[(n/(Prime@i*Prime@j*Prime@k))^(1/2)]}]; t = Table[ FiveAlmostPrimePi[2^n], {n, 0, 37}]; Rest@t - Most@t

%Y Cf. A014614, A114453, A036378, A120033, A120034, A120035, A120036, A120037, A120038, A120039, A120040, A120041, A120042, A120043.

%K nonn

%O 0,7

%A _Jonathan Vos Post_ and _Robert G. Wilson v_, Mar 20 2006