login
A187737
a(n) = floor(sum_{1 < k <= n} p(k)/P(k)), where p(k) is the smallest prime factor of k and P(k) is the largest prime factor of k.
0
1, 2, 3, 4, 4, 5, 6, 7, 8, 9, 9, 10, 11, 11, 12, 13, 14, 15, 15, 16, 16, 17, 17, 18, 19, 20, 20, 21, 21, 22, 23, 24, 24, 24, 25, 26, 26, 26, 27, 28, 28, 29, 29, 30, 30, 31, 32, 33, 33, 33, 33, 34, 35, 35, 36, 36, 36, 37, 37, 38
OFFSET
2,2
COMMENTS
As p(k)/P(k)<=1, every positive integer is in this sequence. - Jon Perry, Jan 03 2013
LINKS
P. Erdős and J. H. van Lint, On the average ratio of the smallest and largest prime divisor of n, Nederl. Akad. Wetensch. Indag. Math. 44:2 (1982), pp. 127-132.
J. H. van Lint, Some sums involving the largest and smallest prime divisor of a natural number, CWI report ZW 25/74 (1974), 16 pp.
FORMULA
Erdős & Lint show that a(n) = n/log n + 3n/log^2 n + o(n/log^2 n). Lint had earlier shown that a(n) = o(n).
EXAMPLE
a(6) = floor(2/2 + 3/3 + 2/2 + 5/5 + 2/3) = floor(4 + 2/3) = 4.
PROG
(PARI) s=0.; for(n=2, 99, f=factor(n)[, 1]; print1(floor(s+=f[1]/f[#f])", "))
CROSSREFS
Sequence in context: A171974 A232748 A308936 * A109401 A307294 A280471
KEYWORD
nonn
AUTHOR
STATUS
approved