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A348582 a(n) is the greatest factor among all the products A307720(k) * A307720(k+1) equal to n.

%I A348582 #8 Oct 25 2021 11:06:55
%S A348582 1,2,3,2,5,3,7,4,3,5,11,4,13,7,5,8,17,6,19,5,7,11,23,8,5,13,9,7,29,6,
%T A348582 31,8,11,17,7,9,37,19,13,8,41,7,43,11,9,23,47,8,7,10,17,13,53,9,11,8,
%U A348582 19,29,59,10,61,31,9,8,13,11,67,17,23,10,71,9,73,37
%N A348582 a(n) is the greatest factor among all the products A307720(k) * A307720(k+1) equal to n.
%C A348582 We know there are n ways to get n as a product of terms A307720(k)*A307720(k+1) for various k's. Look at these 2*n numbers from A307720. Then a(n) is the largest of them.
%H A348582 Rémy Sigrist, <a href="/A348582/b348582.txt">Table of n, a(n) for n = 1..10000</a>
%H A348582 Rémy Sigrist, <a href="/A348582/a348582.txt">C program for A348582</a>
%F A348582 a(p) = p for any prime number p.
%F A348582 a(n) * A348581(n) = n.
%e A348582 For n = 6:
%e A348582 - we have the following products equal to 6:
%e A348582     A307720(7)  * A307720(8)  = 3 * 2 = 6
%e A348582     A307720(12) * A307720(13) = 2 * 3 = 6
%e A348582     A307720(13) * A307720(14) = 3 * 2 = 6
%e A348582     A307720(14) * A307720(15) = 2 * 3 = 6
%e A348582     A307720(15) * A307720(16) = 3 * 2 = 6
%e A348582     A307720(16) * A307720(17) = 2 * 3 = 6
%e A348582 - the corresponding distinct factors are 2 and 3,
%e A348582 - hence a(6) = 3.
%o A348582 (C) See Links section.
%Y A348582 Cf. A307720, A307730, A348581.
%K A348582 nonn
%O A348582 1,2
%A A348582 _Rémy Sigrist_ and _N. J. A. Sloane_, Oct 24 2021