Let n be a positive integer, pd(n) denotes the product of all positive divisors
of n, qd(n) denotes the product of all proper divisors of n

Fangchi, Liang

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GENERALIZATION OF THE DIVISOR PRODUCTSAND PROPER DIVISOR PRODUCTS SEQUENCESLiang FangchiSchool of Arts and Science, Air Force Engineering University, Xi’an Shaanxi, People’s Republicof ChinaAbstract Let n be a positive integer, pd(n) denotes the product of all positive divisorsof n, qd(n) denotes the product of all proper divisors of n. In this paper, westudy the properties of the sequences of {pd(n)} and {qd(n)}, and prove thatthe generalized results for the sequences {pd(n)} and {qd(n)}.Keywords: Divisor and proper divisor product; Generalization ; Sequence.§1. Introduction and resultsLet n be a positive integer, pd(n) denotes the product of all positive divisorsof n. That is, pd(n) =∏d|nd. For example, pd(1) = 1, pd(2) = 2, pd(3) = 3,pd(4) = 8, · · ·, pd(p) = p, · · ·. qd(n) denotes the product of all proper divisorsof n. That is, qd(n) =∏d|n, d<nd. For example, qd(1) = 1, qd(2) = 1, qd(3) =1, qd(4) = 2, · · ·. In problem 25 and 26 of [1], Professor F. Smarandache askedus to study the properties of the sequences {pd(n)} and {qd(n)}. About thisproblem, Liu Hongyan and Zhang Wenpeng in [2] have studied it and provedthe Makowsiki & Schinzel conjecture in [3] hold for {pd(n)} and {qd(n)}.One of them is that for any positive integer n, we have the inequality:σ(φ(pd(n))) ≥ 12pd(n), (1)where σ(n) is the divisor sum function, φ(n) is the Euler’s function.In this paper, as the generalization of [2], we will consider the properties ofthe sequences of {pd(n)} and {qd(n)} for k-th divisor sum function, and givetwo more general results. That is, we shall prove the following:30 SCIENTIA MAGNA VOL.1, NO.1Theorem 1. Let n = pα, p be a prime and α be a positive integer. Then forany fixed positive integer k, we have the inequalityσk(φ(pd(n))) ≥ 12k pkd(n),where σk(n) =∑d|ndk is the k-th divisor sum function.Theorem 2. Let n = pα, p be a prime and α be a positive integer. Then forany fixed positive integer k, we have the inequalityσk(φ(qd(n))) ≥ 12k qkd(n).§2. Proof of the theoremsIn this section, we shall complete the proof of the theorem. First we needtwo Lemmas as following:Lemma 1. For any positive integer n, then we have the identity pd(n) =nd(n)2 and qd(n) = nd(n)2−1,where d(n) =∑d|n1 is the divisor function.Proof. (See Reference [2] Lemma 1).Lemma 2. For any positive integer n, let n = pα11 pα22 · · · pαss with αi ≥ 2(1 ≤ i ≤ s), pj(1 ≤ j ≤ s) are some different primes with p1 < p2 · · · ps.Then for any fixed positive integer k, we have the estimateσk(φ(n)) ≥ φk(n) ·∏p|n(1 +1pk).Proof. From the properties of the Euler’s function we haveφ(n) = φ(pα11 )φ(pα22 ) · · ·φ(pαss ) (2)= pα1−11 pα2−12 · · · pαs−1s (p1 − 1)(p2 − 1) · · · (ps − 1).Here, let (p1 − 1)(p2 − 1) · · · (ps − 1) = pβ11 pβ22 · · · pβss qr11 qr22 · · · qrtt , whereβi ≥ 0, 1 ≤ i ≤ s; rj ≥ 1, 1 ≤ j ≤ t and q1 < q2 < · · · < qt are differentprimes. Note that σk(pα) = 1k + pk + · · ·+ pkα = pk(α+1)−1pk−1 , for any k > 0.Then for (2), we deduce thatσk(φ(n)) = σk(pα1+β1−11 pα2+β2−12 · · · pαs+βs−1s qr11 qr22 · · · qrtt ) (3)=s∏i=1pk(αi+βi)i − 1pki − 1t∏j=1qk(rj+1)j − 1qkj − 1= pk(α1+β1)1 pk(α2+β2)2 · · · pk(αs+βs)s qkr11 qkr22 · · · qkrttGeneralization of the divisor products and proper divisor products sequences 31×s∏i=11− 1pk(αi+βi)ipki − 1t∏j=11− 1qk(rj+1)j1− 1qkj= nk ·s∏i=1(1− 1pi)k s∏i=11− 1pk(αi+βi)i1− 1pxit∏j=11− 1qk(rj+1)j1− 1qkj.Becauseφ(n) = n ·∏p|n(1− 1p), (4)then from (3) and (4) we getσk(φ(n)) = nk · φk(n)nk·s∏i=11− 1pk(αi+βi)i1− 1pxit∏j=11− 1qk(rj+1)j1− 1qkj= φk(n) ·s∏i=1(1 +1pki+ · · ·+ 1pk(αi+βi)−1i)×t∏j=1(1 +1qkj+ · · ·+ 1qk(rj+1)−1j)≥ φk(n) ·∏p|n(1 +1pk).This completes the proof of Lemma 2.Now we use Lemma 1 and Lemma 2 to complete the proof of Theorem 1.Here we will debate this problem in two cases:(i) If n is a prime, then d(n) = 2. So from Lemma 1 we havePd(n) = nd(n)2 = n. (5)Noting that φ(n) = n− 1, then from (5) we immediately getσk(φ(Pd(n))) = σk(n− 1) =∑d|(n−1)dk ≥ (n− 1)k ≥ 12k· nk = 12kP kd (n).(ii) If n = pα, p be a prime and α > 1 be any positive integer. Thend(n) = α+ 1. So thatPd(n) = nd(n)2 = pα(α+1)2 . (6)Using Lemma 2 and (6), we can easily deduce thatσk(φ(Pd(n))) = σk(φ(pα(α+1)2 ))32 SCIENTIA MAGNA VOL.1, NO.1≥ φk(pα(α+1)2 )∏p1|pα(α+1)2(1 +1pk1)= pkα(α+1)2 ·(1− 1p)k·(1 +1pk)≥ p kα(α+1)2 ·(1− 1p)k≥ p kα(α+1)2 · 12k=12kP kd (n).This completes the proof of Theorem 1.Similarly, we can easily prove Theorem 2. That is,(i) If n is a prime, then d(n) = 2. So from Lemma 1 we haveqd(n) = nd(n)2−1 = 1, (7)henceσk(φ(qd(n))) = σk(1) = 1 ≥ 12k qkd(n).(ii) If n = pα, p be a prime and α > 1 be any positive integer. Thend(n) = α+ 1, so thatqd(n) = nd(n)2−1 = pα(α−1)2 . (8)Using Lemma 2 and (8), we haveσk(φ(qd(n))) = σk(φ(pα(α−1)2 )) ≥ 12kqkd(n).This completes the proof of Theorem 2.References[1] F. Smarandache, Only Problems, Not Solutions, Xiquan Publishing House,Chicago, 1993.[2] Mark Farris and Patrick Mitchell, Bounding the Smarandache function,Smarandache Notions Journal 13 (2002), 37-42.[3] Tom M. Apostol, Introduction to Analytic Number Theory, New York,Springer-Verlag, 1976.

GENERALIZATION OF THE DIVISOR PRODUCTS AND PROPER DIVISOR PRODUCTS SEQUENCES

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