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A157754
a(1) = 0, a(n) = lcm(A051904(n), A051903(n)) for n >= 2.
4
0, 1, 1, 2, 1, 1, 1, 3, 2, 1, 1, 2, 1, 1, 1, 4, 1, 2, 1, 2, 1, 1, 1, 3, 2, 1, 3, 2, 1, 1, 1, 5, 1, 1, 1, 2, 1, 1, 1, 3, 1, 1, 1, 2, 2, 1, 1, 4, 2, 2, 1, 2, 1, 3, 1, 3, 1, 1, 1, 2, 1, 1, 2, 6, 1, 1, 1, 2, 1, 1, 1, 6, 1, 1, 2, 2, 1, 1, 1, 4, 4, 1, 1, 2, 1, 1, 1, 3, 1, 2, 1, 2, 1, 1, 1, 5, 1, 2, 2, 2
OFFSET
1,4
COMMENTS
a(n) for n >= 2 equals LCM of minimum and maximum exponents in the prime factorization of n.
a(n) for n >= 2 deviates from A072411, first different term is a(360), a(360) = 3, A072411(360) = 6.
FORMULA
a(1) = 0, a(p) = 1, a(pq) = 1, a(pq...z) = 1, a(p^k) = k, for p = primes (A000040), pq = product of two distinct primes (A006881), pq...z = product of k (k > 2) distinct primes p, q, ..., z (A120944), p^k = prime powers (A000961(n) for n > 1) k = natural numbers (A000027).
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = A033150. - Amiram Eldar, Sep 11 2024
EXAMPLE
For n = 12 = 2^2 * 3^1 we have a(12) = lcm(2,1) = 2.
For n = 144 = 2^4 * 3^2 we have a(144) = lcm(4,2) = 4.
MATHEMATICA
Table[LCM @@ {Min@ #, Max@ #} - Boole[n == 1] &@ FactorInteger[n][[All, -1]], {n, 100}] (* Michael De Vlieger, Jul 12 2017 *)
PROG
(PARI) a(n) = if(n == 1, 0, my(e = factor(n)[, 2]); lcm(vecmin(e), vecmax(e))); \\ Amiram Eldar, Sep 11 2024
KEYWORD
nonn
AUTHOR
Jaroslav Krizek, Mar 05 2009
STATUS
approved