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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 #16 Jul 24 2024 09:20:48

%S 1,5,7,9,11,35,15,17,19,55,23,63,27,75,77,33,35,95,39,99,105,115,47,

%T 119,51,135,55,135,59,385,63,65,161,175,165,171,75,195,189,187,83,525,

%U 87,207,209,235,95,231,99,255,245,243,107,275,253,255,273,295,119,693,123,315,285,129,297,805

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

%C The unitary analog of A060640.

%H Alois P. Heinz, <a href="/A343525/b343525.txt">Table of n, a(n) for n = 1..20000</a>

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

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

%F Dirichlet g.f.: zeta(s-1) * zeta(s) * Product_{p prime} (1 + 1/p^(s-1) - 2/p^(2*s-1)). - _Amiram Eldar_, Jul 24 2024

%p a:= n-> mul(2*i[1]^i[2]+1, i=ifactors(n)[2]):

%p seq(a(n), n=1..80); # _Alois P. Heinz_, Apr 18 2021

%t a[1] = 1; a[n_] := Times @@ ((2 #[[1]]^#[[2]] + 1) & /@ FactorInteger[n]); Table[a[n], {n, 66}]

%o (PARI) a(n) = my(f=factor(n)); for (k=1, #f~, f[k,1] = 2*f[k,1]^f[k,2]+1; f[k,2]=1); factorback(f); \\ _Michel Marcus_, Apr 18 2021

%Y Cf. A001221, A034444, A034448, A060640, A107759, A145388, A298473.

%K nonn,look,mult

%O 1,2

%A _Ilya Gutkovskiy_, Apr 18 2021