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A080189
a(n) = k such that f^k(prime(n)) = 2, where f is the mapping of primes > 2 to primes defined by A052248.
2
0, 1, 2, 3, 2, 4, 2, 3, 5, 3, 3, 4, 4, 6, 3, 4, 3, 4, 6, 2, 5, 5, 7, 4, 3, 3, 5, 2, 5, 5, 7, 7, 6, 6, 3, 4, 6, 8, 7, 5, 3, 5, 2, 4, 3, 6, 6, 6, 4, 4, 6, 3, 8, 8, 8, 5, 3, 7, 7, 4, 4, 5, 6, 5, 7, 9, 8, 6, 4, 5, 4, 6, 5, 4, 3, 5, 4, 7, 4, 7, 4, 7, 2, 6, 3, 7, 5, 5, 3, 7, 4, 9, 9, 8, 9, 8, 9, 4, 4, 8, 8, 5, 5, 4, 3
OFFSET
1,3
COMMENTS
Since the largest of all prime factors of the numbers between prime p and the next prime is smaller than p, we have p > f(p) > f^2(p) > ... > 2, so a(n) is finite for all n.
EXAMPLE
prime(6) = 13, f(13) = 7, f(7) = 5, f(5) = 3, f(3) = 2, so f^4(13) = 2 and a(6) = 4.
PROG
(PARI) {forprime(k=2, 580, c=0; p=k; while(p>2, q=nextprime(p+1); m=0; for(j=p+1, q-1, f=factor(j); a=f[matsize(f)[1], 1]; if(m<a, m=a)); p=m; c++); print1(c, ", "))}
CROSSREFS
Cf. A052248.
Sequence in context: A368710 A080771 A025477 * A076399 A360729 A370328
KEYWORD
nonn
AUTHOR
_Klaus Brockhaus_, Feb 10 2003
STATUS
approved