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A120043
Number of 12-almost primes 12ap such that 2^n < 12ap <= 2^(n+1).
12
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 5, 8, 22, 47, 103, 234, 492, 1082, 2271, 4867, 10349, 21794, 45907, 96293, 202006, 421287, 879388, 1828931, 3800227, 7882784, 16325796, 33771056, 69767214, 143971956, 296771231, 611156696, 1257374970
OFFSET
0,14
COMMENTS
The partial sum equals the number of Pi_12(2^n).
EXAMPLE
(2^12, 2^13] there is one semiprime, namely 6144. 4096 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[12, 2^n], {n, 0, 30}]; Rest@t - Most@t
PROG
(Python)
from math import isqrt, prod
from sympy import primerange, integer_nthroot, primepi
def A120043(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, 12)+almostprimepi(m<<1, 12) # Chai Wah Wu, Aug 31 2024
KEYWORD
nonn
AUTHOR
STATUS
approved