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Product of the divisors of n that are < sqrt(n).
1

%I #20 Jun 17 2023 11:21:35

%S 1,1,1,1,1,2,1,2,1,2,1,6,1,2,3,2,1,6,1,8,3,2,1,24,1,2,3,8,1,30,1,8,3,

%T 2,5,24,1,2,3,40,1,36,1,8,15,2,1,144,1,10,3,8,1,36,5,56,3,2,1,720,1,2,

%U 21,8,5,36,1,8,3,70,1,1152,1,2,15,8,7,36,1,320,3,2,1

%N Product of the divisors of n that are < sqrt(n).

%F a(n) = Product_{d|n, d<sqrt(n)} d.

%F a(n) = Product_{k=1..floor(sqrt(n-1))} k^c(n/k), where c(m) = 1-ceiling(m)+floor(m).

%F a(n) = A072499(n)/A000196(n)^A010052(n) for n>=1.

%e The product of divisors of 16 that are < sqrt(16) = 4 is 1*2 = 2, so a(16) = 2.

%t a[n_] := Times @@ Select[Divisors[n], #^2 < n &]; Array[a, 100]

%o (PARI) a(n) = vecprod(select(x->(x^2<n), divisors(n))); \\ _Michel Marcus_, Jun 08 2023

%Y Cf. A000196, A010052, A072499, A363521.

%Y Cf. A070039 (sum of those divisors).

%K nonn,easy

%O 1,6

%A _Wesley Ivan Hurt_, Jun 07 2023