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If n = Product (p_j^k_j) then a(n) = Product ((p_j - 1)*(k_j + 1)).
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%I #13 Jan 05 2021 22:29:51

%S 1,2,4,3,8,8,12,4,6,16,20,12,24,24,32,5,32,12,36,24,48,40,44,16,12,48,

%T 8,36,56,64,60,6,80,64,96,18,72,72,96,32,80,96,84,60,48,88,92,20,18,

%U 24,128,72,104,16,160,48,144,112,116,96,120,120,72,7,192,160,132,96,176,192

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

%H Andrew Howroyd, <a href="/A304408/b304408.txt">Table of n, a(n) for n = 1..1000</a>

%H Ilya Gutkovskiy, <a href="/A304408/a304408.jpg">Scatter plot of a(n) up to n=50000</a>

%H <a href="/index/Pri#prime_indices">Index entries for sequences computed from indices in prime factorization</a>

%H <a href="/index/Eu#epf">Index entries for sequences computed from exponents in factorization of n</a>

%F a(n) = A000005(n)*abs(A023900(n)) = A000005(n)*A173557(n) = A000005(n)*A000010(A007947(n)).

%F a(p^k) = (p - 1)*(k + 1) where p is a prime and k > 0.

%F a(n) = 2^omega(n)*phi(n) if n is a squarefree (A005117), where omega() = A001221 and phi() = A000010.

%e a(20) = a(2^2*5) = (2 - 1)*(2 + 1) * (5 - 1)*(1 + 1) = 24.

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

%p seq(a(n), n=1..80); # _Alois P. Heinz_, Jan 05 2021

%t a[n_] := Times @@ ((#[[1]] - 1) (#[[2]] + 1) & /@ FactorInteger[n]); a[1] = 1; Table[a[n], {n, 70}]

%t Table[DivisorSigma[0, n] EulerPhi[Last[Select[Divisors[n], SquareFreeQ]]], {n, 70}]

%o (PARI) a(n)={my(f=factor(n)); prod(i=1, #f~, my(p=f[i,1], e=f[i,2]); (p-1)*(e+1))} \\ _Andrew Howroyd_, Jul 24 2018

%Y Cf. A000005, A000010, A000026, A000040, A000302 (numbers n such that a(n) is odd), A001221, A006093, A007947, A023900, A034444, A059975, A062355, A173557, A304407, A304409, A304411, A304412.

%K nonn,mult

%O 1,2

%A _Ilya Gutkovskiy_, May 12 2018