

A349093


a(n) is the smallest nonprime number m (m = A018252(t)) such that n divides the product P(t) of all nonprime numbers up to and including m (P(t) = A036691(t1)).


0



1, 4, 6, 4, 10, 6, 14, 6, 9, 10, 22, 6, 26, 14, 10, 8, 34, 9, 38, 10, 14, 22, 46, 6, 15, 26, 9, 14, 58, 10, 62, 8, 22, 34, 14, 9, 74, 38, 26, 10, 82, 14, 86, 22, 10, 46, 94, 8, 21, 15, 34, 26, 106, 9, 22, 14, 38, 58
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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.
Consecutive solutions of the equation a(n) = 2*K(n) (where K(n) is the Kempner number A002034(n)) are consecutive terms of A048839.


LINKS



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

a(15) = 10 because:


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



STATUS

approved



