A recurrence formula for prime numbers using the Smarandache or Totient functions

A recurrence formula for prime numbers using the Smarandache or Totient functions Felice Russo Via A. Infante 67051 Avezzano (Aq) Italy felice.russo@katamail.com

Abstract In this paper we report a recurrence formula to obtain the n-th prime in terms of (n-l)th prime and as a function of Smarandache or Totient function.

In [1] the Srnarandache Prime Function is defined as follows:

P: N

o

if n is prime

1

ifn is composite

~(O,l)

where: P(n)=

This function can be used to determine the number of primes 7r(N) less or equal to some integer

N and to determine a recurrence formula to obtain the n-th prime starting from the (n-l )th one.

In fact: N

7r(N)= L(1-P(i)

;=1 and 2Pn

Pn+l =1+ Pn +

L

where Pn is the n-th prime.

193

j

[lPU)

The first equation is obvious because (1- P(i)) is equal to 1 every time i is a prime. Let's prove the second one.

Since Pn+l < 2Pn [2] where Pn+l and Pn are the (n+I)th and n-th prime respectively, the following equality holds [3]:

j

j

Pn+!-l

I1P(i)=

I1

L

j= Pn+!

Pn+!-I

P(i)

=

j

L I1

PC;)

i= Pn +1

j

I1 j=Pn+!

P(i)

=0

being P(Pn+l) = 0 by definition.

i=Pn +1

As: j

I1

Pn+!-I

P(i)

i=Pn +1

=

L

1 = (Pn+1 -I)-(Pn +1)+1

= Pn+l -

Pn- 1

j=Pn +1

we get: j

Pn+l =1+ Pn

+

I1P(i)

q.e.d

j=Pn +1 i=p" +1

According to this result we can obtain Pn+1 once we know Pn and P(i).

We report now two expressions for P(i) using the Smarandache function Sen) [4] and the well known Totient function rp(n) [5].

194

â&#x20AC;˘

P(i) =

â&#x20AC;˘

P(i)

=

l-l S~i)J l-l~(i)J 1-1

for; > 4

for;>1

LJ

where n is the floor function [6]. Let's prove now the first equality.

By definition ofSmarandache function SU) numbers [6]. Then

lS~i) J

=i

for;

E

Pu {1,4} where P is the set of prime

is equal to 1 if ; is a prime number and equal to zero for all

composite> 4 being S(i) ~ i [4].

About the second equality we can notice that by definition rp(n) < n for n > 1 and q:(n)=n-l if and only if n is a prime number [5]. So rp(n) ~ n -1 for n > 1 and this implies that

l~(i)J 1-1

is equal to 1 if ; is a prime number and equal to zero otherwise ..

Then:

and

According to the result obtained in [7] for the Smarandache function:

195

S(i)=i+l-

l

~ -(i.sin(k!~))2 J

L.,j

I

k=1

and in [3] for the following function:

1

if k divide i

o

if k doesn't divide i

l~J-li~lJ= the previous recurrence fonnulas can be further semplified as follows:

.( .

l~ -(i.sin(k!.~ Âť2 JJ

1+1- ~l

1-

I

k=1

for n >2

and

I(l-(liJ-l ~ JJ) i

Pn+l =1+ Pn

+

1- hI

i-I

196

I

for

n~l

References.

[1] http://wwwogallup.unmedu/-smarandache/primfuct.txt [2] P. Ribenboim, The book ofprime numbers records, Second edition, New York, SpringerVerlag, 1989 [3] S. M. Ruiz, Afunctional recurrence to obtain the prime numbers using the Smarandache primefunction, SNJ VoL 11 N. 1-2-3, Spring 2000 [4] C. Ashbacher, An introduction to the Smarandachefunction, Erhus Univ. Press, 1995.

[5] E. Weisstein, CRC Concise Encyclopedia ofMathematics, CRe Press, 1999 [6] E. Burton, S-prima/ity degree ofa number and S-prime numbers, SNJ VoL 11 N. 1-2-3, Spring 2000 [7] M. Le, A formula ofthe Smarandache Function, SNJ Vol. 10 1999

197