BULLETIN OF THE
GREEK MATHEMATICAL SOCIETY
Volume 53, 2007 (147–149)
Formulae for the kth Prime Number
Panayiotis G. Tsangaris
Received 28/03/2006
Accepted 10/06/2006
Introduction
In
which
prime
which
this note we first prove an explicit formula for the kth prime (Theorem 1),
improves a previous result of S. Regimbal [2]. Then we determine the smallest
which is greater than a given natural number (Theorem 2), using a method
extends a previous idea of R. Ernvall [1].
Theorem 1. Let k be a natural number. Then










1

 m,
¯
¯




pk =

¯
¯

¯
¯
m=2 
m

P

¯ 
¯

1
 1 + ¯k −  [√m] 1

√
¯



[P
n]
P [m/d]
¯
¯
[ m/d ] n=2
[ [n/d]
¯
¯
n/d ]
d=1
d=1
k2
X
where [
] is the integral part function.
Proof. The proof of Theorem has its origin to Regimbal’s proof. We have:
·
¸ ½
[n/d]
1 if d|n
=
0 if d - n.
n/d
Let
√
[ n] ·
X [n/d] ¸
f (n) =
.
n/d
d=1
Mathematics subject classification: 11A51
147
148
Panayiotis G. Tsangaris
A natural number n > 1 is composite if and only if it has a divisor d such that
√
1 < d 5 [ n]. Therefore
½
1 if n is a prime
f (n) =
0 if n is
composite.
We also define the function h by h(0) = h(1) = 0 and h(n) = [1/f (n)] for n > 1.
Then h is the characteristic function for the set of all prime numbers.
Let π(m) be the number of primes 5 m. Then
π(m) =
m
X
h(n).
n=2
Also
½
h(m)π(m) =
π(m),
0,
if n is a
if m is
If k is a natural number, then
¸ ½
·
1
1,
=
0,
1+ | k − h(m)π(m) |
prime
composite.
if k = h(m)π(m)
if k =
6 h(m)π(m).
Therefore
·
¸
½
1
m, if m is the kth prime
m=
0, otherwise.
1+ | k − h(m)π(m) |
J. Rosser and L. Schoenfeld [2] have proved that pk < k(log + log log k) for k > 5.
Hence, pk < k 2 for k = 3. Consequently
pk =
k2 ·
X
m=2
¸
1
m,
1+ | k − h(m)π(m) |
which proves the Theorem.
Theorem 2. (The Smallest Prime Greater than a Given Positive Integer)
Let m be a natural number with m = 2. Let p1 , p2 , . . . , pk be the primes not exceeding
m. µ ¶
Let
∞
P
2m
ak
a1
s
s
c =
/p1 · · · pk , where ai =
([2m/pi ] − 2[m/pi ]) for each i = 1, 2, . . . , k,
m
s=1
P
(or let c = (2m)!/pb11 · · · pbkk , where bi =
[2m/psi ] for each i =
15s5[(log 2m)/(log pi )]
1, 2, . . . , k). Let t = cc /(cc , c!). Let r be the height of c in t, i.e. cr |t but cr+1 - t.
Then, the smallest prime greater than m is equal to the natural number
c/(t/cr , c).
149
Formulae for the kth Prime Number
Consequently, if p1 , p2 , . . . , pk , . . . is the sequence of all primes, then
pk+1 = c/(t/cr , c),
where c, t, r etc are defined as above, by putting m = pk .
Proof. Let q1 , . . . , qn be all possible prime numbers in increasing order such that
m < qi < 2m for any i = 1, 2, . . . , n. It is easily seen that c = q1 q2 · · · qn . Let ai be
the height of qi in c!. Then
ai =
∞
X
[c/qis ]
s=1
<
∞
X
c/qis < c.
(1)
s=1
Also, the following hold true:
[c/qi+1 ] = c/qi+1 < c/qi = [c/qi ]
s
[c/qi+1
] 5 [c/qis ]
for
for
15i5n−1
s = 2, 3, . . . .
(2)
(3)
From (1), (2) and (3) we obtain:
c > a1 > a2 > · · · > an > 0.
Hence (cc , c!) = q1a1 · · · qnan . Thus t = q1e1 · · · qnen , where ei = c − ai and consequently:
0 < e1 < · · · < en . Moreover, cr kt implies r = e1 . Therefore t/cr = q2e2 −e1 · · · qnen −e1
and (t/cr , c) = q2 . . . qn , where c/(t/cr , c) = q1 . Finally,
pk+1 = c/(t/cr , c),
which ends the proof.
References
[1] R. Ernvall, A formula for the least Prime greater than a given integer, Elem.,
Math. 30(1975), pp. 13–14.
[2] S. Regimbal, An Explicit Formula for the kth Prime Number, Math. Mag.
48(1975), pp. 230–232.
[3] J.B. Rosser and L. Schoenfeld, Approximate formulas for some functions of prime
numbers, Illinois J. Math. 6(1972), pp. 64–94.
¦ Panayiotis G. Tsangaris
Department of Mathematics,
Athens University,
15784 Panepistimiopolis,
Athens, GREECE
ptsangaris@math.uoa.gr