|
%I #13 Dec 18 2014 02:07:37
%S 2,2,2,2,2,2,2,2,2,6,2,2,2,8,3,2,2,2,2,6,3,12,2,2,2,14,2,8,2,6,2,2,12,
%T 18,2,2,2,20,14,6,2,8,2,12,3,24,2,2,2,6,18,14,2,2,4,8,20,30,2,6,2,32,
%U 3,2,4,12,2,18,24,8,2,2,2,38,3,20,4,14,2,6,2,42,2,8,5,44,30,12,2,6,4,24,32
%N a(n) = smallest r such that r^k <= spf(n) and gpf(n) < r^(k+1), for some k >= 0, where spf and gpf (smallest and greatest prime factor of n) are given by A020639(n) and A006530(n).
%H Antti Karttunen, <a href="/A252375/b252375.txt">Table of n, a(n) for n = 1..10000</a>
%F If A251725(n) = 1, a(n) = 2, otherwise a(n) = A251725(n).
%F Other identities. For all n >= 1:
%F a(n) = a(A066048(n)). [The result depends only on the smallest and the largest prime factor of n.]
%o (Scheme)
%o (define (A252375 n) (let ((spf (A020639 n)) (gpf (A006530 n))) (let outerloop ((r 2)) (let innerloop ((rx 1)) (cond ((and (<= rx spf) (< gpf (* r rx))) r) ((<= rx spf) (innerloop (* r rx))) (else (outerloop (+ 1 r))))))))
%o (define (A252375 n) (let ((x (A251725 n))) (if (= 1 x) 2 x))) ;; Alternatively, using the implementation of A251725.
%Y A252374 gives the corresponding exponents.
%Y Cf. A251726 (those n for which a(n) <= A006530(n)).
%Y Cf. A251727 (those n > 1 for which a(n) = A006530(n)+1).
%Y Cf. A006530, A020639, A066048, A138510, A251725.
%K nonn
%O 1,1
%A _Antti Karttunen_, Dec 17 2014
|
