Scientia Magna Vol. 3 (2007), No. 2, 44-49 Some identities involving the near pseudo Smarandache function Yu Wang Department of Mathematics, Northwest University Xi’an, Shaanxi, P.R.China Received March 29, 2007 Abstract For any positive integer n and fixed integer t ≥ 1, we define function U (n) = t min{k : 1t+2t+···+nt+k = m, n | m, k ∈ N+, t ∈ N+}, where n ∈ N+, m ∈ N+, which is a new pseudo Smarandache function. The main purpose of this paper is using the elementary method to study the properties of U (n), and obtain some interesting identities t involving function U (n). t Keywords Some identities, reciprocal, pseudo Smarandache function. §1. Introduction and results In reference [1], A.W.Vyawahare defined the near pseudo Smarandache function K(n) as n(n+1) K(n)=m= +k,wherek isthesmallpositiveintegersuchthatndividesm. Thenhe 2 studied the elementary properties of K(n), and obtained a series interesting results for K(n). n(n+3) n(n+2) For example, he proved that K(n)= , if n is odd, and K(n)= , if n is even; 2 2 The equation K(n) = n has no positive integer solution. In reference [2], Zhang Yongfeng studied the calculating problem of an infinite series involving the near pseudo Smarandache 1 (cid:88)∞ 1 functionK(n), and proved that for anyreal numbers> , the series is convergent, 2 Ks(n) n=1 and (cid:88)∞ 1 2 5 = ln2+ , K(n) 3 6 n=1 (cid:88)∞ 1 11 22+2ln2 = π2− . K2(n) 108 27 n=1 Yang hai and Fu Ruiqin [3] studied the mean value properties of the near pseudo Smarandache function K(n), and obtained two asymptotic formula by using the analytic method. They proved that for any real number x≥1, (cid:88) (cid:88) (cid:181) n(n+1)(cid:182) 3 (cid:179) (cid:180) d(k)= d K(n)− = xlogx+Ax+O x12 log2x , 2 4 n≤x n≤x Vol. 3 SomeidentitiesofreciprocalofthenearpseudoSmarandachefunction 45 where A is a computable constant. (cid:88) (cid:181) n(n+1)(cid:182) 93 (cid:179) (cid:180) ϕ K(n)− = x2+O x32+(cid:178) , 2 28π2 n≤x where (cid:178) denotes any fixed positive number. In this paper, we define a new near Smarandache function U (n)=min{k : 1t+2t+···+ t nt+k =m, n|m, k ∈N+, t∈N+}, where n∈N+, m∈N+. Then we study its elementary properties. Aboutthisfunction,itseemsthatnonehadstudiedityet,atleastwehavenotseen such a paper before. In this paper, we using the elementary method to study the calculating problem of the infinite series (cid:88)∞ 1 , Us(n) n=1 t and give some interesting identities. That is, we shall prove the following: Theorem 1. For any real number s>1, we have the identity (cid:181) (cid:182) (cid:88)∞ 1 1 =ζ(s) 2− , Us(n) 2s n=1 1 where ζ(s) is the Riemann zeta-function. Theorem 2. For any real number s>1, we have (cid:183) (cid:181) (cid:182)(cid:181) (cid:182)(cid:184) (cid:88)∞ 1 1 1 1 1 =ζ(s) 1+ − +2 1− 1− . Us(n) 5s 6s 2s 3s n=1 2 Theorem 3. For any real number s>1, we also have (cid:34) (cid:181) (cid:182) (cid:35) (cid:88)∞ 1 1 2 =ζ(s) 1+ 1− . Us(n) 2s n=1 3 π2 π4 Taking s = 2, 4, and note that ζ(2) = , ζ(4) = , from our theorems we may 6 90 immediately deduce the following: Corollary. Let U (n) defined as the above, then we have the identities t (cid:88)∞ 1 7 (cid:88)∞ 1 2111 = π2; = π2; U2(n) 24 U2(n) 5400 n=1 1 n=1 2 (cid:88)∞ 1 25 (cid:88)∞ 1 31 = π2; = π4; U2(n) 96 U4(n) 1440 n=1 3 n=1 1 t (cid:88)∞ 1 2310671 (cid:88)∞ 1 481 = π4; = π4. U4(n) 72900000 U4(n) 23040 n=1 2 n=1 3 46 YuWang No. 2 §2. Some lemmas To complete the proof of the theorems, we need the following several lemmas. Lemma 1. For any positive integer n, we have  n  , if 2|n, U (n)= 2 1  n, if 2†n. Proof. See reference [1]. Lemma 2. For any positive integer n, we also have  5  6n, if n≡0(mod 6), n, if n≡1(mod 6) or n≡5(mod 6), U2(n)= nn2, if n≡2(mod 6) or n≡4(mod 6), , if n≡3(mod 6). 3 Proof. It is clear that U (n) = min{k :12+22+···+n2+k =m,n|m,k ∈N+} 2 n(n+1)(2n+1) = min{k : +k ≡0(mod n),k ∈N+}. 6 (1) If n≡0(mod 6), then we have n=6h (h =1,2···), 1 1 n(n+1)(2n+1) 6h (6h +1)(12h +1) = 1 1 1 6 6 = 72h3+18h2+h , 1 1 1 n(n+1)(2n+1) 5n so n| +U (n) if and only if 6h |h +U (n), then U (n)= . 6 2 1 1 2 2 6 (2) If n≡1(mod 6), then we have n=6h +1(h =0,1,2···), 2 2 n(n+1)(2n+1) (6h +1)(6h +2)(12h +3) = 2 2 2 6 6 = 12h2(6h +1)+7h (6h +1)+6h +1, 2 2 2 2 2 n(n+1)(2n+1) n(n+1)(2n+1) because n | , so n | +U (n) if and only if n | U (n), then 6 6 2 2 U (n)=n. 2 If n≡5(mod 6), then we have n=6h +5(h =0,1,2···), 2 2 n(n+1)(2n+1) (6h +5)(6h +6)(12h +11) = 2 2 2 6 6 = 12h2(6h +5)+23h (6h +5)+11(6h +5), 2 2 2 2 2 n(n+1)(2n+1) n(n+1)(2n+1) because n | , so n | +U (n) if and only if n | U (n), then 6 6 2 2 U (n)=n. 2 (3) If n≡2(mod 6), then we have n=6h +2(h =0,1,2···), 2 2 n(n+1)(2n+1) (6h +2)(6h +3)(12h +5) = 2 2 2 6 6 = 12h2(6h +2)+11h (6h +2)+2(6h +2)+3h +1, 2 2 2 2 2 2 Vol. 3 SomeidentitiesofreciprocalofthenearpseudoSmarandachefunction 47 n(n+1)(2n+1) n so n| +U (n) if and only if 6h +2|3h +1+U (n), then U (n)= . 6 2 2 2 2 2 2 If n≡4(mod 6), then we have n=6h +4(h =0,1,2···), 2 2 n(n+1)(2n+1) (6h +4)(6h +5)(12h +9) = 2 2 2 6 6 = 12h2(6h +4)+19h (6h +4)+7(6h +4)+3h +3, 2 2 2 2 2 2 n(n+1)(2n+1) n so n| +U (n) if and only if 2(3h +2)|3h +2+U (n), then U (n)= . 6 2 2 2 2 2 2 (4) If n≡3(mod 6), then we have n=6h +3(h =0,1,2···), 2 2 n(n+1)(2n+1) (6h +3)(6h +4)(12h +7) = 2 2 2 6 6 = 12h2(6h +3)+15h (6h +3)+4(6h +3)+4h +2, 2 2 2 2 2 2 n(n+1)(2n+1) n so n| +U (n) if and only if 3(2h +1)|2(2h +2)+U (n), then U (n)= . 6 2 2 2 2 2 3 Combining (1), (2), (3) and (4) we may immediately deduce Lemma 2. Lemma 3. For any positive integer n, we have  n  , if n≡2(mod 4), U (n)= 2 3  n, otherwise. Proof. From the definition of U (n) we have 3 U (n) = min{k :13+23+···+n3+k =m,n|m,k ∈N+} 3 n2(n+1)2 = min{k : +k ≡0(mod n),k ∈N+}. 4 (a) If n≡2(mod 4), then we have n=4h +2(h =0,1,2···), 1 1 n2(n+1)2 =(4h +2)3(2h +1)+(4h +2)2(2h +1)+(2h +1)2, 4 1 1 1 1 1 n2(n+1)2 n so n| if and only if 2(2h +1)|(2h +1)2+U (n), then U (n)= . 4 1 1 3 3 2 (b) If n≡0(mod 4), then we have n=4h (h =1,2···), 2 2 n2(n+1)2 =4h2(4h +1)2, 4 2 2 n2(n+1)2 so n| +U (n) if and only if n|U (n), then U (n)=n. 4 3 3 3 If n≡1(mod 4), then we have n=4h +1(h =0,1,2···), 1 1 n2(n+1)2 =(4h +1)2(2h +1)2, 4 1 1 n2(n+1)2 so n| +U (n) if and only if n|U (n), then U (n)=n. 4 3 3 3 If n≡3(mod 4), then we have n=4h +3(h =0,1,2···), 1 1 n2(n+1)2 =4(4h +3)2(h +1)2, 4 1 1 n2(n+1)2 uso n| +U (n) if and only if n|U (n), then U (n)=n. 4 3 3 3 Now Lemma 3 follows from (a) and (b). 48 YuWang No. 2 §3. Proof of the theorems Inthissection,weshallusetheelementarymethodstocompletetheproofofthetheorems. First we prove Theorem 1. For any real number s>1, from Lemma 1 we have (cid:181) (cid:182) (cid:88)∞ 1 (cid:88)∞ 1 (cid:88)∞ 1 (cid:88)∞ 1 (cid:88)∞ 1 1 = + = + =ζ(s) 2− , Us(n) (n)s ns hs (2h+1)s 2s n=1 1 h=1 2 h=0 h=1 h=0 n=2h n=2h+1 where ζ(s) is the Riemann zeta-function. This proves Theorem 1. For t=2 and real number s>1, from Lemma 2 we have (cid:88)∞ 1 (cid:88)∞ 1 (cid:88)∞ 1 (cid:88)∞ 1 (cid:88)∞ 1 (cid:88)∞ 1 = (cid:161) (cid:162) + + (cid:161) (cid:162) + (cid:161) (cid:162) + Us(n) 5n s ns n s n s ns n=1 2 h1=1 6 h2=0 h2=0 3 h2=0 2 h2=0 n=6h1 n=6h2+1 n=6h2+2 n=6h2+4 n=6h2+5 (cid:88)∞ 1 (cid:88)∞ 1 (cid:88)∞ 1 (cid:88)∞ 1 = + + + + (5h )s (6h +1)s (3h +1)s (2h +1)s 1 2 2 2 h1=1 h2=0 h2=0 h2=0 (cid:88)∞ 1 (cid:88)∞ 1 + (3h +2)s (6h +5)s 2 2 h2=0(cid:183) h2=0(cid:181) (cid:182)(cid:181) (cid:182)(cid:184) 1 1 1 1 = ζ(s) 1+ − +2 1− 1− , 5s 6s 2s 3s This completes the proof of Theorem 2. If t=3, then for any real number s>1, from Lemma 3 we have (cid:88)∞ 1 (cid:88)∞ 1 (cid:88)∞ 1 (cid:88)∞ 1 (cid:88)∞ 1 = + + (cid:161) (cid:162) + Us(n) ns ns n s ns n=1 3 h2=1 h1=0 h1=0 2 h1=0 n=4h2 n=4h1+1 n=4h1+2 n=4h1+3 (cid:88)∞ 1 (cid:88)∞ 1 (cid:88)∞ 1 (cid:88)∞ 1 = + + + (4h )s (4h +1)s (2h +1)s (4h +3)s 2 1 1 1 h2=1(cid:34) (cid:181) h1=0(cid:182) (cid:35) h1=0 h1=0 1 2 = ζ(s) 1+ 1− , 2s This completes the proof of Theorem 3. Open Problem. For any integer t > 3 and real number s > 1, whether there exists a calculating formula for the Dirichlet series (cid:88)∞ 1 ? Us(n) n=1 t This is an open problem. Vol. 3 SomeidentitiesofreciprocalofthenearpseudoSmarandachefunction 49 References [1] A. W. Vyawahare, Near pseudo Smarandache funchion, Smarandache Notion Journal, 14(2004), 42-59. [2] Yongfeng Zhang, On a near pseudo Smarandache funchion, Scientia Magna, 1(2007), No.1, 98-101. [3] Hai Yang and Ruiqin Fu, On the mean value of the Near Pseudo Smarandanche Func- tion, Scientia Magna, 2(2006), No.4, 35-39.