Index                                                                                                                                                                                               Modular Sieves

Three Theorems on Modular Sieves that suggest the Prime Difference is


Bhupinder Singh Anand[1]

Although this 1964 paper is not, strictly, a foundational paper, it developed as an off-shoot to the foundational query: Do we discover the natural numbers (Platonically), or do we construct them linguistically? The paper also assumes computational significance in the light of Agrawal, Kayal and Saxena's August 2002 paper, “PRIMES is in P”, since both the Trim and Compact Number Generating algorithms - each of which generates all the primes - are deterministic algorithms that run in polynomial time, and suggest that the Prime Difference, dP(n), is


1.   A Prime Number Generating Theorem

2.   Trim Numbers

2.1  A Trim Number Theorem …

2.2  … and two conjectures

2.3  Is the Theorem significant?

3.   Compact Numbers

3.1  Two Compact Number Theorems …

3.2  … and two more conjectures



We address the question:


Do we necessarily discover the primes, as in the sieve of Eratosthenes, or can we also generate them sequentially?


In other words, given the first k primes, can we generate the prime p(k+1) algorithmically, without testing any number larger than p(k) for primality?


We give an affirmative answer, defining two algorithms based on the following theorem (where i, j, k, … represent natural numbers):


Theorem 1.1: For any given natural number k, and all i < k, let p(i) denote the i'th prime, and define the sequence {aP(i): i = 1 to k} such that aP(i) < p(i), and:


p(k) + aP(i) ≡ 0 (mod p(i))


Let dP(k) be such that, for all l < dP(k), there is some i such that:


laP(i) (mod p(i)).


Then dP(k) is the Prime Difference, defined by:


p(k+1) - p(k) = dP(k).


Proof: For all l < dP(k), there is some i such that:


p(k) + l ≡ 0 (mod p(i)).


Since, by Bertrand’s Postulate ([HW60], p343), p(k+1) < 2p(k) for all natural numbers k, it follows that:


dP(k) < p(k)


Hence, p(k) + dP(k) < p(k)2, and so it is the prime p(k+1)

2: TRIM NUMBERS (1964)      


The significance of the Prime Number Generating Theorem is seen in the following algorithm.


We define Trim numbers recursively by t(1) = 2, and t(n+1) = t(n) + dT(n), where p(i) is the i'th prime and:


(i)      dT(1) = 1, and aT(2, 1) = 1:


(ii)     dT(n) is the smallest even integer that does not occur in the n'th sequence:


{ aT(n, 1), …, aT(n, n-1)};


(iii)    ji ≥ 0 is selected so that, for all 0 < i ≤ (n-1):


0 ≤ aT(n+1, i) = {( aT(n, i) - dT(n) + ji*p(i)} < p(i)


It follows that the Trim number t(n+1) is, thus, a prime unless all its prime divisors are less than dT(n).


(In [An64] I illustrate how Trim numbers are generated sequentially as a 2-dimensional modular sieve.)




Theorem 2.1: For all n > 1, t(n) < n2.


Proof:  t(n) = {t(n-1) + dT(n-1)}


dT(n-1) < 2(n-1)


t(n) < {t(n-1) + 2(n-1)}


t(n) < [{t(n-2) + 2(n-2)} + 2(n-1)]


t(n) < {2 + 2(n-1) + 2(n-2) + … + 2(1)}


t(n) < [2 + 2{(n-1)*n/2}]


t(n) < (n2 - n + 2)


Hence, for all n > 1, t(n) < n2


2.2: ... AND TWO CONJECTURES (1964)


Conjecture 1.1: For all k, t(k) ≤ p(n) ≤ t(k+1) for some n.


Conjecture 1.2: If p(n+1) = {p(n) + dP(n)}, then dP(n) = O(n).




The significance of the above theorem lies in the following observation [Ca05]:


“Is there always a prime between n2 and (n+1)2? In 1882 Opperman stated π(n2+n) > π(n2) > π(n2-n), (n>1), which also seems very likely, but remains unproven ([Ri95], p248). 


Both of these conjectures would follow if we could prove the conjecture that the prime gap following a prime p is bounded by a constant times (log p)2.”




Compact numbers are defined recursively by c(1) = 2, and c(n+1) = c(n) + dC(n),

where p(i) is the i'th prime and:


(i)      dC(1) = 1, and aC(2, 1) = 1;


(ii)     dC(n) is the smallest even integer that does not occur in the n’th sequence:


{aC(n, 1), …, aC(n, k)};


(iii)    ji ≥ 0 is selected so that, for all 0 < ik:


0 ≤ aC(n+1, i) = { aC(n, i) - dC(n) + ji*p(i)} < p(i);


(iv)    k is selected so that:


p(k)2 < c(n) ≤ p(k+1)2;


(v)     if c(n) = p(k+1)2, then:


aC(n, k+1) = 0.


It follows that the compact number c(n+1) is either a prime, or a prime square, unless all, except a maximum of 1, prime divisors of the number are less than dC(n).


(In [An64] I illustrate how Compact numbers are generated sequentially as a significantly more compact, 2-dimensional, modular sieve.)




Theorem 3.1: There is always a compact number c(m) such that n2 < c(m) ≤ (n+1)2.


Proof: If p(k) ≤ n < p(k+1), then  (n+1) ≤ p(k+1).


Now {(n+1)2n2} = {2n + 1} > 2p(k) > 2k.


Hence (n+1)2 > {n2 + 2k}.


Since there is always a compact number in any interval larger than 2k between p(k)2 and p(k+1)2, the theorem follows


Theorem 3.2: For sufficiently large n, dC(n) < constant*{c(n)/log c(n)}1/2


Proof: By definition, dC(n) ≤ 2k, where p(k)2 < c(n) ≤ p(k+1)2.


The theorem follows from the Prime Number Theorem, since k is the number of primes less than c(n)1/2




Conjecture 3.1: If p(k)2 < p(n) < p(k+1)2, then:


{p(n+1) - p(n)} = O(k)

= O().


Conjecture 3.2:


{Number of primes < n}/{Number of compacts < n} = 1/2.


[An64]    Anand, B. S. 1964. Two Series containing the Primes and a Conjecture concerning the Prime Difference Transcribed notes.




[Ca05]    Caldwell, Chris. K. 2005. Prime Conjectures and Open Questions. Web resource.




[HW60]  Hardy, G. H. and Wright, E. M. 1960. An Introduction to the Theory of Numbers. 4th edition. Clarendon Press, Oxford.


[Ri95]     Ribenboim, P. 1995. The new book of prime number records. 3rd edition, Springer-Verlag, New York, NY, pp. xxiv+541, ISBN 0-387-94457-5. 1995.


(Transcribed from original notes: Thursday 8th December 2005 2:17:06 AM IST by Updated: Saturday June 09, 2007 08:14:21 PM IST by

[1] The author is an independent scholar. E-mail:; Postal address: 32, Agarwal House, D Road, Churchgate, Mumbai - 400 020, INDIA. Tel: +91 (22) 2281 3353. Fax: +91 (22) 2209 5091.