## phi of n factorial proof

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• Hello All: I send you a proof by Carlos Rivera of my Result Phi(n!)=Product(Ci)*Product(pi-1): The Euler Totient Function applied to factorial. (22 Feb. 2006)
Message 1 of 1 , Feb 27, 2006
Hello All:

I send you a proof by Carlos Rivera of my Result

Phi(n!)=Product(Ci)*Product(pi-1):

The Euler Totient Function applied to factorial.

(22 Feb. 2006)

I.-Theorem: Phi(n!)= P(Ci). P(pi-1)

Phi = Totient function
n! = n.(n-1).(n-2).  3.2.1
P(Ci)=C1.C2.
Ci = composite, Ci<=n
pi = prime, pi<=n

II.- Example: Phi(10!) = Phi(3628800) =
10.9.8.(7-1).6.(5-1).4.(3-1).(2-1).1 = 829440

III.- Proof:

n! =P(Ci).P(pj).P(pk)
pj = prime <= n such that gcd(P(Ci), pj)<>1
pk = prime <= n such that gcd(P(Ci), pk)=1

then gcd(P(Ci).P(pj), P(pk))=1

Phi(n!) = Phi(P(Ci).P(pj).P(pk)) =
Phi(P(Ci).P(pj)).Phi(P(pk))

Phi(P(pk)) = P(pk-1)

But P(Ci)=P(pj^aj) where aj = # times pj occurs in
P(Ci), then

P(Ci).P(pj) = P(pj^(aj+1))

Using Phi(p^b) = p^b(1-1/p) = p^(b-1).(p-1)

We obtain Phi(P(Ci).P(pj)) = Phi(P(pj^(aj+1))) =
P((pj^aj)(pj-1)) =
= P(pj^aj). P(pj-1) = P(Ci). P(pj-1)

Phi(n!) = P(Ci).P(pj-1). P(pk-1) = P(Ci).P(pi-1). End
of proof.

Corollaries:

1) Phi(n!)/Phi((n-1)!)={n, if n=composite; n-1 if
n=prime}

2) (n-1)! <=Phi(n!) <= n!

Sebastian Martin Ruiz Carlos Rivera