A304407 If n = Product (p_j^k_j) then a(n) = Product ((p_j - 1)*k_j).
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, 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, 40, 18, 36, 28, 58, 16, 60, 30, 24, 6, 48, 20, 66, 32, 44, 24, 70, 12, 72, 36, 16
Offset: 1
Examples
a(60) = a(2^2*3*5) = (2 - 1)*2 * (3 - 1)*1 * (5 - 1)*1 = 16.
Links
Crossrefs
Programs
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Maple
seq(mul((p-1)*padic[ordp](n, p), p in numtheory[factorset](n)), n=1..100); # Ridouane Oudra, Jun 06 2025
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Mathematica
a[n_] := Times @@ ((#[[1]] - 1) #[[2]] & /@ FactorInteger[n]); a[1] = 1; Table[a[n], {n, 75}] Table[EulerPhi[Last[Select[Divisors[n], SquareFreeQ]]] DivisorSigma[0, n/Last[Select[Divisors[n], SquareFreeQ]]], {n, 75}]
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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
Formula
a(p^k) = (p - 1)*k where p is a prime and k > 0.
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
a(n) = (-1)^A001221(n) * (Sum_{d1|n} Sum_{d2|n} mu(d1)*gcd(d1,d2)). - Ridouane Oudra, Jun 06 2025