A356683 a(n) is the smallest positive k such that the count of squarefree numbers <= k that have n prime factors is equal to the count of squarefree numbers <= k that have n-1 prime factors (and the count is positive).
2, 39, 1279786, 8377774397163159586
Offset: 1
Examples
The first two squarefree numbers are 1 and 2; 1 has 0 prime factors and 2 has 1 prime factor, so a(1)=2. At k=39, in the interval [1..k], there are 12 squarefree numbers with 1 prime factor (i.e., 12 primes: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37), and 12 squarefree numbers with 2 prime factors (i.e., 6, 10, 14, 15, 21, 22, 26, 33, 34, 35, 38, 39). k=39 is the smallest such positive number for which these two counts are the same (and are positive), so a(2)=39. At k=1279786, the interval [1..k] includes 265549 squarefree numbers with 2 prime factors and the same number of squarefree numbers with 3 prime factors, and there is no smaller positive number k that has this property (where the counts are positive), so a(3)=1279786. There are 75 numbers with this property, the last one being 1281378. At k=8377774397163159586, the interval [1..k] includes 1356557942402075858 squarefree numbers with 3 prime factors and the same number of squarefree numbers with 4 prime factors, and there is no smaller positive number k that has this property (where the counts are positive), so a(4)=8377774397163159586. There are 14 numbers with this property, the last one being 8377774397163162544. - _Henri Lifchitz_, Jan 31 2025
Crossrefs
Programs
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PARI
a(n) = my(nbm = 0, nbn = 0); for (k=1, oo, if (issquarefree(k), my(o=omega(k)); if (o==n, nbn++); if (o==n-1, nbm++); if (nbm && (nbn==nbm), return(k)))); \\ Michel Marcus, Nov 25 2022
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Python
from itertools import count from math import prod, isqrt from sympy import primerange, integer_nthroot, primepi def A356683(n): def g(x,a,b,c,m): yield from (((d,) for d in enumerate(primerange(b+1,isqrt(x//c)+1),a+1)) if m==2 else (((a2,b2),)+d for a2,b2 in enumerate(primerange(b+1,integer_nthroot(x//c,m)[0]+1),a+1) for d in g(x,a2,b2,c*b2,m-1))) def f(k,n): return sum(primepi(k//prod(c[1] for c in a))-a[-1][0] for a in g(k,0,1,1,n)) if n>1 else primepi(k) return 2 if n==1 else next(k for k in count(1) if (x:=f(k,n-1))>0 and x==f(k,n)) # Chai Wah Wu, Aug 31 2024
Extensions
a(4) from Henri Lifchitz, Jan 31 2025