This paper gives some properties of the Smarandache prime product sequence

Majumdar, A.A.K.

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A NOTE ON THE SMARANDACHE PRIME PRODUCT SEQTJENCE A.A.K.~AR Department of Mathematics, Jahangimagar University, Savar, Dhaka 1342, Bangladesh ABSTRACT This paper gives some properties of the Smarandache prime product sequence, (Pn ) , definded by where (Pn) is the sequence of primes in their natural order. k\1S (1991) Subject Classification: 11A41, llA51. 1. INTRODUCTION Let (Pn) = CPt, P2, ... ) be the (infinite) sequence of primes in their natural numbers. The first few terms of the sequence are as follows: PI =2, P2 =3,1>3 =5, P4 =7, P5 = 11, P6 = 13, P7 = 17, P8 = 19, P9 =23, PlO =29. Clearly, the sequence (Pn) is strictly increasing (in n ~ 1) with Pn > PI P2 for all n ~ 4. Furthermore, Pn > n for all n ~ 1. The Smarandache prime product sequence, (Pn), is defined by (Smarandache [5] ) (1.1 ) We note that the sequence (Pn ) is strictly increasing (in n ~ 1), satisfying the following recursion formulas: We also note that Pn is an odd (positive) integer for all n ~ 1; furthermore, Pl =3, P2 =7, P3 =31, P4 =211, Ps=2311 '37 (1.2) (1.3) are all primes, while the next five elements of the sequence (Pn ) are all composites, since Ps = 30031 = 59 x 509, P7= 510511 = 19 x 97 x 277, P8 = 9699691 = 347 x 27953, Pg = 223092871 = 317 x 703760, PlO = 6469693231 = 331 x 571 x 34231. Some of the properties of the sequence (Pn ) have been studied by Prakash [3], who conjectures that this sequence contains an infinite number of primes. This note gives some properties of the sequence (Pn ), some of which strengthens the corresponding result of Prakash [3]. This is done in §2 below, and show that for each n ~ 1, Pn is relatively prime to Pn+ 1. We conclude this paper with some remarks in the final §3. 2. MAIN RESULTS We start with the following result which has been established by Majumdar [2] by induction on n ( ~ 6), using the recurrence relationship (1.3). Lemma 2.1: Pn < (Pn+1t-2 for all n ~ 6. Exploiting Lemma 2.1, Majumdar [2] has proved the following theorem which strengthens the corresponding result of Prakash [3]. Theorem 2.1: For each n ~ 6, Pn has at most n-3 prime factors (counting multiplicities) . Another property satisfied by the sequence (Pn ) is given in Theorem 2.2. To prove the theorem, we would need the following results. Lemma 2.2: For each n ~ 1, Pn is of the form 4k+3 for some integer k ~ O. 138 Proof: Since Pn is odd for all n ~ 1, it must be of the fonn 4k+l or 4k+3 (see, for example, Shanks [4], pp. 4). But, Pn cannot be of the fonn 4k+l, othenvise, from (1.1), we would have PlP2 ... Pn = 4k, that is, 41 PlP2 ... Pn, which is absurd. Hence, Pn must be of the fonn 4k+3. 0 Lemma 2.3: (1) The product of two integers of the fonn 4k+l is an integer of the fonn 4k+l, and in general, for any integer m > 0, (4k+l)m is again of the fonn 4k+l, (2) The product of two integers of the fonn 4k+3 is an integer of the fonn 4k+ 1, and the product of two integers, one of the fonn 4k+ 1 and the other of the fonn 4k+ 3, is integer of the fonn 4k+ 3, (3) For any integer m > 0, (4k+3)m is of the fonn 4k+l or 4k+3 respectively according as m is even or odd. Proof: Part (1) has been proved by Bolker ([1], Lemma 5.2, pp. 6). The proof of the remaining parts is similar. 0 We now prove the following theorem. Theorem 2.2: For each n ~ 1, Pn is never a square or higher power of any natural number ( > 1). Proof: If possible, let Pn = N'2 for some integer N > 1. Now, since Pn is odd, N must be odd, and hence, N must be of the fonn 4k+ 1 or 4k+3 for some integer k > 0. But, in either case, by Lemma 2.3, N 2= Pn is of the fonn 4k+ 1, contradicting Lemma 2.2. Hence, Pn cannot be a square of a natural number ( > 1). To prove the remaining part, let Pn= N l for some integers N > 1, 1 ~ 3. (*) Without loss of generality, we may assume that 1 is a prime (if 1 is a composite number, let 1= rs where r is prime, and so Pn = (Nsy; setting M = NS, we may proceed with this 1\1[ in 139 place of N) .. By Theorem 2.1, 1 < n, and hence, 1 must be one of the primes Pl, P3, ... , Pn. By Fermat's Little Theorem (Bolker [1], Theorem 9.8, pp. 16), PlPl ... Pn = N 1-1 = N-l = 0 (mod 1). Thus, N = Im+ 1 for some integer m > 0, and we get PlPl ... Pn = (1m)l + (D(1m)1-l + ... + (l~l ) (1m). But the above expression shows that 121 PlPl ... Pn, which is impossible. Hence, the representation of Pn in the form (.) is not possible, which we intend to prove. 0 Some more properties related to the sequence (Pn ) are given in the following two lemmas. Lemma 2.4: For each n 2: 1, (Pn , Pn .q ) = 1. Proof: Any prime factor p of Pn+ 1 satisfies the inequality P > Pn+ 1. Now, ifplpn, then from (1.3), we see that pl(Pn+1-1), which is absurd. Hence, all the prime factors of Pn+1 are different from each of the prime factors of Pn , which proves the lemma. 0 Lemma 2.5: For each n 2: 1, Pn and Pn+2 have at most one prime factor in common. Proof: Since Pn+2-Pn = PlPl .. ·Pn (Pn-'-lPn-l-2- 1), any prime factor common to both Pn and Pn+2 must divide Pn-1Pn..;..2-1. Now, any prime factor of Pn+2 is greater than Pn+2. Hence, it follows that Pn and Pn+2 can have at most one prime factor in common, since otherwise, the product of the prime factors is greater than (Pn.,-2)2, which cannot divide Pn+1Pn+2-1 < (Pn+2)2. 0 From the proof of the above lemma we see that, if all the prime factors of Pn .... l Pn .... 2-1 are less than Pn+2, then (Pn , Pn+2) = 1. And generalizing the lemma, we have the following result: For any n 2: 1, and i 2: 1, Pn and Pn+i can have at most i-I number of prime factors mcommon. 140 3. SOME REMARKS We conclude this paper with the following remarks. (1) The sequence (Pn ) is well known, it is used in elementary texts on the Theory of Numbers (see, for example, Bolker [1 ] and Shanks [4] to prove the infinititude of the primes. Some of the properties of the sequence (Pn ) have been studied by Prakash [3]. Theorem 2.1 improves one of the results of Prakash [3], while our proof of Theorem 2.2 is much simpler than that followed by Prakash [3]. The expressions for Ps, Pr, Ps, Pg and P lO show that Theorem 2.1 is satisfied with tighter bounds, but we could not improve it further. (2) By Lemma 2.3 we see that, of all the prime factors of Pn (which is at most n-3 in number for n ~ 6, by Theorem 2.1), an odd number of these must be of the form 4k+3. In this connection, we note that, in case of Ps, one of the prime factors (namely, 59) is of the form 4k+3, while the other is of the form 4k+1; and in case of Pr , all the three prime factors are of the form 4k+ 3. (3) The Conjecture that the sequence (Pn ) contains infinitely many primes, still remains an open problem. REFERENCES [1] Bolker, E.D. (1970) Elementary Number Theory. Benjamin Inc., N.Y. [2] Majumdar, A. A. K. (1996/97) A Note on a Sequence Free From Powers. Math. Spectrum, 29 (2), pp. 41 (Letter to the Editor). [3] Prakash, K. (1990) A Sequence Free From Powers. Math. Spectrum, 22 (3), pp. 92-93. [4] Shanks, D. (1962) Solved and Unsolved Problems in Number Theorv - Volume 1. Spantan Books, Washington, D. C. [5] Smarandache, F. (1996) Collected Papers - Volume I . Tempus Publishing House, Bucharest, Romania, 141 [6] Iacobescu, F., (1997) Smarandache Partition Type Sequences. Bulletin of Pure and Applied Sciences,16E(2), pp. 237-240. 142 

A NOTE ON THE SMARANDACHE PRIME PRODUCT SEQUENCE

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