OFFSET
1,2
COMMENTS
a(n) >= 2*gpf(n) for n > 1, where gpf(n) denotes the greatest prime factor of n (A006530(n)).
Conjecture: the equation a(n) = a(n+1) has no solutions. This holds up to at least n = 10^7.
LINKS
J. Sondow and E. W. Weisstein, MathWorld: Smarandache Function
FORMULA
a(p) = 2*p for prime p.
a(p_1*p_2*...*p_u) = 2*p_u, where p_i's are distinct primes and p_1 < p_2 < ... < p_u.
a(n) where n is factored as n = p_1^k_1*p_2^k_2*...*p_u^k_u is given by a(n) = max( a(p_1^k_1), a(p_2^k_2), ..., a(p_u^k_u) ), where a(p_i^k_i) = w*p_i and w is the smallest m >= 2 satisfying the inequality:
-1 + Sum_{t=1..m} floor((m*p_i)/(p_i)^t) >= k_i.
EXAMPLE
PROG
(Maxima)
f(p, k):=(z:2, for m:2 while -1+sum(floor((p*m)/(p^t)), t, 1, m)<k do (z:z+1), z*p);
a(n):=(for d:1 thru length(ifactors(n)) do (A[d]:f(ifactors(n)[d][1], ifactors(n)[d][2])), lmax(makelist(A[d], d, 1, length(ifactors(n)))));
makelist(a(n), n, 2, 200);
CROSSREFS
KEYWORD
nonn
AUTHOR
Lechoslaw Ratajczak, Mar 25 2022
STATUS
approved