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Let b = A360519; let Lg = gcd(b(n-1),b(n)), Rg = gcd(b(n),b(n+1)); let L(n) = prod_{primes p|Lg} p-part of b(n), R(n) = prod_{primes p|Rg} p-part of b(n), M(n) = b(n)/(L(n)*R(n)); sequence gives M(n).
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%I #17 Mar 09 2023 23:09:42

%S 3,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,3,1,1,1,1,1,1,1,7,1,1,1,1,

%T 1,1,1,1,1,1,1,1,1,1,1,7,1,1,1,1,1,5,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,

%U 1,1,11,1,1,11,1,1,1,1,13,1,1,1,1,1,1,9,1,1,1,1,1,1,5,17,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,3,1

%N Let b = A360519; let Lg = gcd(b(n-1),b(n)), Rg = gcd(b(n),b(n+1)); let L(n) = prod_{primes p|Lg} p-part of b(n), R(n) = prod_{primes p|Rg} p-part of b(n), M(n) = b(n)/(L(n)*R(n)); sequence gives M(n).

%C The p-part of a number k is the highest power of p that divides k. For example, the 2-part of 24 is 8, the 3-part is 3.

%C Since so many of the initial terms are 1, we show more than the usual number of terms in the DATA section.

%C Conjecture: All terms are odd, and every odd number eventually appears.

%H N. J. A. Sloane, <a href="/A361129/b361129.txt">Table of n, a(n) for n = 2..20000</a>

%Y Cf. A360519, A361118, A361128, A361130.

%K nonn

%O 2,1

%A _Scott R. Shannon_, _Rémy Sigrist_, and _N. J. A. Sloane_, Mar 09 2023