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
Keywords
Examples
(2^11, 2^12] there is one semiprime, namely 3072. 2048 was counted in the previous entry.
Links
- Chai Wah Wu, Table of n, a(n) for n = 0..50
Crossrefs
Programs
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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
-
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<
Chai Wah Wu, Jun 17 2025
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