A080686 Number of 19-smooth numbers <= n.

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 22, 23, 24, 25, 26, 27, 27, 28, 28, 29, 30, 31, 32, 33, 33, 34, 35, 36, 36, 37, 37, 38, 39, 39, 39, 40, 41, 42, 43, 44, 44, 45, 46, 47, 48, 48, 48, 49, 49, 49, 50, 51, 52, 53, 53, 54, 54, 55, 55, 56

Offset: 1

Author

Cino Hilliard, Mar 02 2003

Keywords

• easy
• nonn

Comments

Range = primes 2 to 19. Input pn=19 in script below. Code below is much faster than the code for cross-reference. For input of n=200 13 times as fast and many times faster for larger input of n.

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A080682.

Number of p-smooth numbers <= n: A070939 (p=2), A071521 (p=3), A071520 (p=5), A071604 (p=7), A071523 (p=11), A080684 (p=13), A080685 (p=17), this sequence (p=19).

Programs

Mathematica

Accumulate[Table[Boole[Max[FactorInteger[n][[;; , 1]]] <= 19], {n, 100}]] (* Amiram Eldar, Apr 29 2025 *)

PARI

smoothn(n,pn) = { for(m=1,n, pr=1; forprime(p=2,pn, pr*=p; ); ct=1; for(x=1,m, f=0; forprime(y=nextprime(pn+1),floor(x), if(x%y == 0,f=1; break) ); if(gcd(x,pr)<>1,if(f==0,ct+=1; )) ); print1(ct","); ) }

Python

from sympy import integer_log, prevprime
def A080686(n):
    def g(x,m): return sum((x//3**i).bit_length() for i in range(integer_log(x,3)[0]+1)) if m==3 else sum(g(x//(m**i),prevprime(m))for i in range(integer_log(x,m)[0]+1))
    return g(n,19) # Chai Wah Wu, Sep 17 2024