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A304407 If n = Product (p_j^k_j) then a(n) = Product ((p_j - 1)*k_j). 5

%I #17 Nov 30 2022 08:34:58

%S 1,1,2,2,4,2,6,3,4,4,10,4,12,6,8,4,16,4,18,8,12,10,22,6,8,12,6,12,28,

%T 8,30,5,20,16,24,8,36,18,24,12,40,12,42,20,16,22,46,8,12,8,32,24,52,6,

%U 40,18,36,28,58,16,60,30,24,6,48,20,66,32,44,24,70,12,72,36,16

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

%H Andrew Howroyd, <a href="/A304407/b304407.txt">Table of n, a(n) for n = 1..1000</a>

%H Ilya Gutkovskiy, <a href="/A304407/a304407.jpg">Scatter plot of a(n) up to n=50000</a>.

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

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

%F a(n) = A005361(n)*abs(A023900(n)) = A005361(n)*A173557(n) = A005361(n)*A000010(A007947(n)).

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

%F a(n) = phi(n) if n is a squarefree (A005117), where phi() = A000010.

%F a(A002110(k)) = A005867(k).

%F Sum_{k=1..n} a(k) ~ c * n^2, where c = (Pi^4/72) * Product_{p prime} (1 - 4/p^2 + 3/p^3 + 1/p^4 - 1/p^5) = 0.2644703894... . - _Amiram Eldar_, Nov 30 2022

%e a(60) = a(2^2*3*5) = (2 - 1)*2 * (3 - 1)*1 * (5 - 1)*1 = 16.

%t a[n_] := Times @@ ((#[[1]] - 1) #[[2]] & /@ FactorInteger[n]); a[1] = 1; Table[a[n], {n, 75}]

%t Table[EulerPhi[Last[Select[Divisors[n], SquareFreeQ]]] DivisorSigma[0, n/Last[Select[Divisors[n], SquareFreeQ]]], {n, 75}]

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

%Y Cf. A000010, A000026, A000040, A002110, A003958, A005117, A005361, A005867, A006093, A007947, A023900, A048250, A059975, A068997, A081294 (numbers n such that a(n) is odd), A087656, A090624, A152649, A173557, A304117, A304408, A304409, A304411, A304412.

%K nonn,mult

%O 1,3

%A _Ilya Gutkovskiy_, May 12 2018

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