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%I #15 Sep 08 2022 08:46:24
%S 1,2,9,6,400,12,3136,24,36,80,123904,60,692224,448,2025,120,18939904,
%T 180,94633984,240,35721,11264,2218786816,360,10000,53248,900,1344,
%U 225754218496,720,1031865892864,840,7144929,1114112,1960000,1260,94076963651584,4980736
%N a(n) = the smallest number m such that gcd(tau(m), pod(m)) = n where tau(k) = the number of the divisors of k (A000005) and pod(k) = the product of the divisors of k (A007955).
%C a(n) = the smallest number m such that A306671(m) = n.
%C If a(17) exists, it must be bigger than 10^7.
%e For n=3; a(3) = 9 because gcd(tau(9), pod(9)) = gcd (3, 27) = 3 and 9 is the smallest.
%t Array[Block[{m = 1}, While[GCD[DivisorSigma[0, m], Times @@ Divisors@ m] != #, m++]; m] &, 16] (* _Michael De Vlieger_, Mar 24 2019 *)
%o (Magma) [Min([n: n in[1..10^6] | GCD(NumberOfDivisors(n), &*[d: d in Divisors(n)]) eq k]): k in [1..16]]
%o (PARI) a(n) = {my(k=1, vk = divisors(k)); while(gcd(#vk, vecprod(vk)) != n, k++; vk = divisors(k)); k;} \\ _Michel Marcus_, Mar 06 2019
%Y Cf. A000005, A007955, A306671.
%K nonn
%O 1,2
%A _Jaroslav Krizek_, Mar 05 2019
%E a(17)-a(38) from _Jon E. Schoenfield_, Mar 07 2019