A result obtained using Smarandache Function

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A result obtained using Smarandache Function Sebastian Martin Ruiz 21 November 1998 Smarandache Function is defined as followed:

SCm) = The smallest positive integer so that SCm)! is divisible by m. [1] Let's see the value which such function takes for m = ppn with n integer, n and p prime number. To do so a Lemma is required.

2:

2

Lemma 1 V m, n E N m, n 2: 2

Where E( x) gives the greatest integer less than or equal to x.

Demonstration: Let's see in the first place the value taken by EUogm (mn+l - mn + m)]. If n 2: 2: mm+l - mn + m < m n+1 and therefore logm(mn+l - mn + m) < logm mn+l = n + 1. As a result EUogm (mn+l - mn + m)] < n + l. And if m 2: 2: mmn 2: 2m n => mn+l 2: 2m n => m n + 1 + m 2: 2m n => n 1 m + _ mn + m > > 100"Om mn = n => _ mn => log m (mn+l - mn + m) _ n 1 EUogm(m + - mn + m)] 2: n As a result: n ~ EUogm(m n + 1 - mn + m)] < n + 1 therefore:

Now let's see the value which it takes for 1 S; k S; n: E [mn+l;;;::n+m]

E

[

, ] = E [ mn+1-" _ mn-k m n+l ' - mn ,m m" n

If k = 1: E [m +

1

n

-;:"m m] = ...

mn - m n-

If 1 < k S; n: E rmn+';;;::n+m] = m n + 1 -

123

1

+1

k -

mn -

k

1 ] + _._

m",-l


Let's see what is the value of the sum: _m n -

k=l k=2 k=3

1

mn - 1

.. , +1 _m n -

2

mn-

2

_m n -

3

?

m-

k=n-1 k=n

-m

m

-1

Therefore:

m,n

Proposition 1 V p prime number 'lin S(V

n

~

2

~ 2 :

= pn+l _ pn + p

)

Demonstration: Having ep (k) = exponent of the prime number p in the prime numbers descomposition of k. We get:

k p

k p

k p

ep (k!)=E(-)+E(-2)+E(-3)+路路路+E(

k pE(l og.".禄)

And using the Lemma we have:

e [( p

p

n+l_

n+ )!J p p

= E [pn+l -

pn P

+ p] +E [pn+l

- pn p2

+ p] +- ..+E [

pn+l_ pn

Therefore:

And:

References: [lJ C. Dumitrescu and R. }liiller: To Enjoy is a Permanent Component of Mathematics. SYIARA.NDACHE NOTIONS JOURNAL Vol. 9, ':;-0. 1-2.(1998) pp 21-26.

Author: Sebastian ylartin Ruiz. Avda. de RegIa 43. SPAIN.

124

CHIPIO~A

+p

,.]

mE[Iog.lpnTl_pn+PJJ

(CADIZ) 11550

=

n P


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