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A085818
For n > 1: a(n) = p if n = p^e with p prime and e > 1, otherwise a(n) = (n-m)-th prime, where m = number of nonprime prime powers <= n; a(1)=1.
4
1, 2, 3, 2, 5, 7, 11, 2, 3, 13, 17, 19, 23, 29, 31, 2, 37, 41, 43, 47, 53, 59, 61, 67, 5, 71, 3, 73, 79, 83, 89, 2, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 7, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241
OFFSET
1,2
COMMENTS
a(n) = A025473(n) if n = p^e with p prime and e > 1, otherwise a(n) = A008578(n-A085501(n));
n divides A085819(n) = Product_{k<=n} a(k), as by construction: a(1)=1; if n divides A085819(n-1) then a(n) = smallest prime not occurring earlier; if n does not divide A085819(n-1) then a(n) = greatest prime factor of n (A006530);
A000040 occurs infinitely many times as a subsequence.
a(A085971(n))=A000040(n) and for all k > 1: a(A000040(n)^k)=A000040(n); A085985(n)=A049084(a(n)). - Reinhard Zumkeller, Jul 06 2003
LINKS
PROG
(PARI) f(n) = 1 + sum(k=2, n, isprimepower(k) && !isprime(k)); \\ A085501
a(n) = {if (n==1, return (1)); my(p); if (isprimepower(n, &p) && !isprime(n), p, prime(n-f(n))); } \\ Michel Marcus, Jan 28 2021
(Python)
from sympy import primefactors, prime, primepi, integer_nthroot
def A085818(n): return 1 if n==1 else (f[0] if len(f:=primefactors(n))==1 and f[0]<n else prime(n-1-sum(primepi(integer_nthroot(n, k)[0]) for k in range(2, n.bit_length())))) # Chai Wah Wu, Aug 20 2024
KEYWORD
nonn
AUTHOR
Reinhard Zumkeller, Jul 04 2003
STATUS
approved