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If n = Product (p_j^k_j) then a(n) = Product (2*(p_j^k_j + 1)), with a(1) = 1.
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%I #8 Apr 20 2021 10:10:31

%S 1,6,8,10,12,48,16,18,20,72,24,80,28,96,96,34,36,120,40,120,128,144,

%T 48,144,52,168,56,160,60,576,64,66,192,216,192,200,76,240,224,216,84,

%U 768,88,240,240,288,96,272,100,312,288,280,108,336,288,288,320,360,120,960,124,384,320,130

%N If n = Product (p_j^k_j) then a(n) = Product (2*(p_j^k_j + 1)), with a(1) = 1.

%F a(n) = usigma(n) * 2^omega(n).

%F a(n) = Sum_{d|n, gcd(d, n/d) = 1} usigma(d) * usigma(n/d).

%F a(n) = Sum_{d|n, gcd(d, n/d) = 1} d * 2^omega(d) * 2^omega(n/d).

%F a(n) = Sum_{d|n, gcd(d, n/d) = 1} A343525(d).

%t a[1] = 1; a[n_] := Times @@ (2 (#[[1]]^#[[2]] + 1) & /@ FactorInteger[n]); Table[a[n], {n, 64}]

%o (PARI) a(n) = my(f=factor(n)); for (k=1, #f~, f[k,1] = 2*f[k,1]^f[k,2] + 2; f[k,2] = 1); factorback(f); \\ _Michel Marcus_, Apr 20 2021

%Y Cf. A001221, A034444, A034448, A034761, A064840, A107759, A333557, A343525.

%K nonn,mult

%O 1,2

%A _Ilya Gutkovskiy_, Apr 20 2021