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%I #21 Sep 08 2022 08:46:25
%S 1,4,16,25,64,81,100,121,256,289,484,529,729,841,1024,1156,1296,1600,
%T 1681,1936,2116,2209,2401,2809,3025,3364,3481,4096,4624,5041,5184,
%U 6400,6724,6889,7225,7921,8464,8836,10201,11236,11449,11664,12100,12769,13225,13456
%N Numbers m such that tau(m), sigma(m) and pod(m) are pairwise relatively prime.
%C Numbers m such that A336723(m) = A000005(m) * A000203(m) * A007955(m).
%C Numbers m such that lcm(m, tau(m), sigma(m), pod(m)) = tau(m) * sigma(m) * pod(m).
%C Subsequence of numbers m such that A336722(m) = gcd(tau(m), sigma(m), pod(m)) = 1.
%C From _David A. Corneth_, Dec 11 2020: (Start)
%C a(n) is a perfect square. Proof: If a(n) is not a perfect square but is even then both tau(a(n)) and pod(a(n)) are divisible by 2. Contradiction.
%C If a(n) is not a perfect square and is odd then both tau(a(n)) and sigma(a(n)) are even. Contradiction.
%C Hence if a(n) is not a perfect square then it can be neither even nor odd. So a(n) is a perfect square. Q.E.D. (End)
%H David A. Corneth, <a href="/A338406/b338406.txt">Table of n, a(n) for n = 1..10000</a>
%e lcm(tau(4), sigma(4), pod(4)) = lcm(3, 7, 8) = tau(4) * sigma(4) * pod(4) = 3 * 7 * 8 = 168.
%t Select[Range[15000], CoprimeQ[(d = DivisorSigma[0, #]), (s = DivisorSigma[1, #])] && CoprimeQ[d, (p = #^(d/2))] && CoprimeQ[s, p] &] (* _Amiram Eldar_, Oct 25 2020 *)
%o (Magma) [m: m in [1..10^5] | LCM([#Divisors(m), &+Divisors(m), &*Divisors(m)]) eq #Divisors(m) * &+Divisors(m) * &*Divisors(m)]
%o (PARI) isok(m) = my(d=divisors(m), t=#d, s=vecsum(d), p=vecprod(d)); t*s*p == lcm([t,s,p]); \\ _Michel Marcus_, Oct 25 2020
%Y Cf. A000005 (tau), A000203 (sigma), A007955 (pod).
%Y Cf. A336722, A336723, A338395.
%K nonn
%O 1,2
%A _Jaroslav Krizek_, Oct 24 2020