A080683 23-smooth numbers: numbers whose prime divisors are all <= 23.
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 30, 32, 33, 34, 35, 36, 38, 39, 40, 42, 44, 45, 46, 48, 49, 50, 51, 52, 54, 55, 56, 57, 60, 63, 64, 65, 66, 68, 69, 70, 72, 75, 76, 77, 78, 80, 81, 84, 85, 88, 90, 91, 92, 95
Offset: 1
Links
- William A. Tedeschi, Table of n, a(n) for n = 1..10000
Crossrefs
Programs
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Magma
[n: n in [1..100] | PrimeDivisors(n) subset PrimesUpTo(23)]; // Bruno Berselli, Sep 24 2012
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Maple
select(t -> max(numtheory:-factorset(t)) <= 23, [$1..1000]); # Robert Israel, Jan 22 2016
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Mathematica
mx = 100; Sort@ Flatten@ Table[ 2^a*3^b*5^c*7^d*11^e*13^f*17^g*19^h*23^i, {a, 0, Log[2, mx]}, {b, 0, Log[3, mx/2^a]}, {c, 0, Log[5, mx/(2^a*3^b)]}, {d, 0, Log[7, mx/(2^a*3^b*5^c)]}, {e, 0, Log[11, mx/(2^a*3^b*5^c*7^d)]}, {f, 0, Log[13, mx/(2^a*3^b*5^c*7^d*11^e)]}, {g, 0, Log[17, mx/(2^a*3^b*5^c*7^d*11^e*13^f)]}, {h, 0, Log[19, mx/(2^a*3^b*5^c*7^d*11^e*13^f*17^g)]}, {i, 0, Log[23, mx/(2^a*3^b*5^c*7^d*11^e*13^f*17^g*19^h)]}] (* Robert G. Wilson v, Jan 19 2016 *)
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PARI
test(n)=m=n; forprime(p=2,23, while(m%p==0,m=m/p)); return(m==1) for(n=1,100,if(test(n),print1(n",")))
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PARI
list(lim,p=23)=if(p==2, return(powers(2, logint(lim\1,2)))); my(v=[],q=precprime(p-1),t=1); for(e=0,logint(lim\=1,p), v=concat(v, list(lim\t,q)*t); t*=p); Set(v) \\ Charles R Greathouse IV, Apr 16 2020
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Python
import heapq from itertools import islice from sympy import primerange def agen(p=23): # generate all p-smooth terms v, oldv, h, psmooth_primes, = 1, 0, [1], list(primerange(1, p+1)) while True: v = heapq.heappop(h) if v != oldv: yield v oldv = v for p in psmooth_primes: heapq.heappush(h, v*p) print(list(islice(agen(), 72))) # Michael S. Branicky, Nov 20 2022
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Python
from sympy import integer_log, prevprime def A080683(n): def bisection(f,kmin=0,kmax=1): while f(kmax) > kmax: kmax <<= 1 while kmax-kmin > 1: kmid = kmax+kmin>>1 if f(kmid) <= kmid: kmax = kmid else: kmin = kmid return kmax 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)) def f(x): return n+x-g(x,23) return bisection(f,n,n) # Chai Wah Wu, Sep 16 2024
Formula
Sum_{n>=1} 1/a(n) = Product_{primes p <= 23} p/(p-1) = (2*3*5*7*11*13*17*19*23)/(1*2*4*6*10*12*16*18*22) = 676039/110592. - Amiram Eldar, Sep 22 2020
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