login
A120041
Number of 10-almost primes k such that 2^n < k <= 2^(n+1).
27
0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 5, 8, 22, 47, 103, 233, 487, 1072, 2246, 4803, 10202, 21440, 45115, 94434, 197891, 412010, 858846, 1783610, 3700698, 7665755, 15853990, 32750248, 67564405, 139238488, 286625278, 589472979, 1211146741, 2486322304
OFFSET
0,12
LINKS
FORMULA
a(n) ~ 2^n log^9 n/(725760 n log 2). [Charles R Greathouse IV, Dec 28 2011]
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[10, 2^n], {n, 0, 39}]; Rest@t - Most@t
PROG
(Python)
from math import isqrt, prod
from sympy import primerange, integer_nthroot, primepi
def A120041(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, 10)+almostprimepi(m<<1, 10) # Chai Wah Wu, Aug 31 2024
KEYWORD
nonn
AUTHOR
STATUS
approved