A120042 Number of 11-almost primes 11ap such that 2^n < 11ap <= 2^(n+1).

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 5, 8, 22, 47, 103, 234, 490, 1078, 2261, 4844, 10294, 21659, 45609, 95580, 200422, 417715, 871452, 1811412, 3761623, 7798409, 16142081, 33373093, 68906782, 142120436, 292797806, 602653984, 1239225631

Offset

0,13

Comments

The partial sum equals the number of Pi_11(2^n).

Links

Chai Wah Wu, Table of n, a(n) for n = 0..50

Example

(2^11, 2^12] there is one semiprime, namely 3072. 2048 was counted in the previous entry.

Mathematica

AlmostPrimePi[k_Integer, n_] := Module[{a, i}, a[0] = 1; If[k == 1, PrimePi[n], Sum[PrimePi[n/Times @@ Prime[Array[a, k - 1]]] - a[k - 1] + 1, Evaluate[ Sequence @@ Table[{a[i], a[i - 1], PrimePi[(n/Times @@ Prime[Array[a, i - 1]])^(1/(k - i + 1))]}, {i, k - 1}]]]]]; (* Eric W. Weisstein, Feb 07 2006 *)
t = Table[AlmostPrimePi[11, 2^n], {n, 0, 30}]; Rest@t - Most@t

Prog

(Python)
from math import isqrt, prod
from sympy import primerange, integer_nthroot, primepi
def A120042(n):
    def g(x, a, b, c, m): yield from (((d, ) for d in enumerate(primerange(b, isqrt(x//c)+1), a)) if m==2 else (((a2, b2), )+d for a2, b2 in enumerate(primerange(b, integer_nthroot(x//c, m)[0]+1), a) for d in g(x, a2, b2, c*b2, m-1)))
    def almostprimepi(n, k): return int(sum(primepi(n//prod(c[1] for c in a))-a[-1][0] for a in g(n, 0, 1, 1, k)) if k>1 else primepi(n))
    return -almostprimepi(m:=1<<n, 11)+almostprimepi(m<<1, 11) # Chai Wah Wu, Jun 17 2025

Crossrefs

Cf. A069272, A036378, A120033, A120034, A120035, A120036, A120037, A120038, A120039, A120040, A120041, A120042, A120043.

Sequence in context: A120039 A120040 A120041 * A120043 A063897 A092733

Adjacent sequences: A120039 A120040 A120041 * A120043 A120044 A120045

Keyword

nonn

Author

Jonathan Vos Post and Robert G. Wilson v, Mar 21 2006

Status

approved