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If n = Product (p_j^k_j) then a(n) = Product (p_j^k_j + 3), with a(1) = 1.
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%I #13 Nov 13 2022 08:40:41

%S 1,5,6,7,8,30,10,11,12,40,14,42,16,50,48,19,20,60,22,56,60,70,26,66,

%T 28,80,30,70,32,240,34,35,84,100,80,84,40,110,96,88,44,300,46,98,96,

%U 130,50,114,52,140,120,112,56,150,112,110,132,160,62,336,64,170,120,67,128,420,70,140,156,400,74

%N If n = Product (p_j^k_j) then a(n) = Product (p_j^k_j + 3), with a(1) = 1.

%C The unitary analog of A007430.

%H Amiram Eldar, <a href="/A343570/b343570.txt">Table of n, a(n) for n = 1..10000</a>

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

%F a(n) = Sum_{d|n, gcd(d, n/d) = 1} A107759(d).

%F Sum_{k=1..n} a(k) ~ c * n^2, where c = (Pi^2/12) * Product_{p prime} (1 + 2/p^2 - 3/p^3) = 1.1848008127... . - _Amiram Eldar_, Nov 13 2022

%t a[1] = 1; a[n_] := Times @@ ((#[[1]]^#[[2]] + 3) & /@ FactorInteger[n]); Table[a[n], {n, 71}]

%o (PARI) a(n) = my(f=factor(n)); for (k=1, #f~, f[k,1] = f[k,1]^f[k,2] + 3; f[k,2] = 1); factorback(f); \\ _Michel Marcus_, Apr 20 2021

%Y Cf. A001221, A007430, A034444, A034448, A107759.

%K nonn,mult

%O 1,2

%A _Ilya Gutkovskiy_, Apr 20 2021