#A8 INTEGERS 5 (205) ON SPARSEL SCHEMMEL TOTIENT NUMBERS Colin Defant Depatment of Mathematics, Univesity of Floida, Gainesville, Floida cdefant@ufl.edu Received: 7/30/4, Revised: 2/23/4, Accepted: 4/26/5, Published: 5/8/5 Abstact Fo each positive intege, let S denote the th Schemmel totient function, a multiplicative aithmetic function defined by ( S (p 0, if p apple ; ) = p (p ), if p > fo all pimes p and positive integes. The function S is simply Eule s totient function. Masse and Shiu have established seveal fascinating esults concening spasely totient numbes, positive integes n satisfying (n) < (m) fo all integes m > n. We define a spasely Schemmel totient numbe of ode to be a positive intege n such that S (n) > 0 and S (n) < S (m) fo all m > n with S (m) > 0. We then genealize some of the esults of Masse and Shiu.. Intoduction Thoughout this pape, we will let N and P denote the set of positive integes and the set of pime numbes, espectively. Fo any pime p and positive intege n, we will let!(n) denote the numbe of distinct pime factos of n, and we will let p(n) denote the exponent of p in the pime factoization of n. Futhemoe, we will let n# denote the poduct of all the pime numbes less than o equal to n (with the convention # = ), and we will let p i denote the i th pime numbe. The Eule totient function (n) counts the numbe of positive integes less than o equal to n that ae elatively pime to n. In 869, V. Schemmel intoduced a class of functions S, now known as Schemmel totient functions, that genealize Eule s totient function. S (n) counts the numbe of positive integes k apple n such that gcd(k + j, n) = fo all j 2 {0,,..., }. Clealy, S =. It has been This wok was suppoted by National Science Foundation gant no. 262930.
INTEGERS: 5 (205) 2 shown [6] that S is a multiplicative function that satisfies ( S (p 0, if p apple ) = p (p ), if p > fo all pimes p and positive integes. It follows easily fom this fomula that, fo any positive integes, u, v, we have S (u)s (v) apple S (uv) apple min(us (v), vs (u)). () Fo any positive intege, we will let B denote the set of positive integes whose smallest pime facto is geate than, and we will convene to let 2 B. Equivalently, B = {n 2 N: S (n) > 0}. Masse and Shiu have studied the set F of positive integes n that satisfy (n) < (m) fo all m > n [5]. These integes ae known as spasely totient numbes, and they motivate the following definition. Definition.. Let be a positive intege. A positive intege n is a spasely Schemmel totient numbe of ode if n 2 B and S (n) < S (m) fo all m 2 B with m > n. We will let F be the set of all spasely Schemmel totient numbes of ode. Remak.. Lee-Wah ip has shown that if is a positive intege, then thee c ()n exists a positive constant c () such that S (n) (log log 3n) fo all n 2 B [8]. In fact, this esult follows quite easily fom Metens s estimates. Theefoe, each set F is infinite. The aim of this pape is to modify some of the poofs that Masse and Shiu used to establish esults concening spasely totient numbes in ode to illustate how those esults genealize to esults concening spasely Schemmel totient numbes. 2. A Fundamental Constuction The fundamental esult in Masse and Shiu s pape, upon which all subsequent theoems ely, is a constuction of a cetain subset of F, so we will give a simila constuction of subsets of the sets F. Lemma. Fix some positive intege, and suppose x, x 2,..., x s, y, y 2,..., y s, X, ae eal numbes such that < x i apple y i fo all i 2 {, 2,..., s}. If
INTEGERS: 5 (205) 3 s s max(x, x 2,..., x s ) and X x i < y i, then i= i= s s (X ) (x i ) < ( ) (y i ). i= i= Poof. The poof is by induction on s, so we will assume that s lemma is tue if we eplace s with s. Note that 2 and that the s s (x i ) apple ( ) (y i ), i= so the poof is simple if X < y s. Theefoe, we will assume that X y s. If we s s wite the inequality X x i < y i as Xx s s s x i < y i, then the induction y i= i= s i= i= hypothesis tells us that i= Xxs y s s s (x i ) < ( ) (y i ). (2) i= i= Multiplying each side of (2) by y s, we see that it su ces to show, in ode to complete the induction step, that Xxs (y s ) (X )(x s ). (3) y s Xxs We may ewite (3) as X x s y s + y s apple (X + x s ), o, equivalently, y s x s apple y s. This inequality holds because X y s, so we have completed the induction step of the poof. Fo the case s =, we note again that the poof istivial if X < y, so we will assume that X y. This implies that y x apple X, which we may ewite as y + Xx apple X + x. Multiplying this last inequality by and adding y Xx + 2 to each side, we get Xx y + Xx + 2 Xx (X + x ) + 2, y Xx so (y ) y that (y )( ) > (x )(X ). x y (x )(X ). As x X < y by hypothesis, we find
INTEGERS: 5 (205) 4 In what follows, we will let () denote the numbe of pimes less than o equal to. Theoem. Let be a positive intege, and let ` and k be nonnegative integes such that k () + 2. Suppose d is an element of B such that d < p k+ and k d(p k+` ) < (d + )(p k ). If we set n = dp k+` p i, then n 2 F. i= ()+ k Poof. Fist, note that n 2 B. Setting u = d and v = p k+` S (n) apple d(p k+` ) k i= ()+ Using the hypothesis d(p k+` ) < (d + )(p k ), we get S (n) < (d + ) k i= ()+ i= ()+ p i in () yields (p i ). (4) (p i ), (5) fom which the hypothesis d < p k+ yields S (n) < k+ i= ()+ (p i ). (6) Now, choose some abitay m 2 B with m > n. We will show that S (m) > S (n). Thee is a unique intege t > () such that Clealy,!(m) apple t t i= ()+ t i= ()+ (), so S (m) m p i apple m < t i= ()+ t+ i= ()+ p i p i.. This implies that S (m) (p i ). If t k +, then we may use (6) to conclude that S (n) < S (m). Theefoe, let us assume that t apple k. Then!(m) apple k (). Suppose!(m) apple k () so that S k (m). Fom (4), we m p i i= ()+ have S k (n) <, so S (n) < S (m). Because m > n, we see that n n m S (n) < S (m). i= ()+ p i
INTEGERS: 5 (205) 5 Now, assume!(m) = k (). Then we may wite m = µ k () i= q i, whee µ is a positive intege whose pime factos ae all in the set {q, q 2,..., q k () } and, fo all i, j 2 {, 2,..., k ()} with i < j, q i is a pime and p ()+i apple q i < q j. This means that S (m) = µ S (n) < µ k i= ()+ k () i= (p i ) = µ (q i ). If µ d +, then we may use (5) to find that k () i= that µ apple d. Because m > n, we have k () i= (p ()+i ) apple S (m). Hence, we may assume q i > d µ p k+` k i= ()+ p i. (7) Fo each i 2 {, 2,..., k ()}, let x i = p ()+i, and let y i = q i. If we set s = k (), X = d µ p k+`, and = q k (), then we may use Lemma and (7) to conclude that k () i= (q i ) > Thus, because µ apple d, we have S (m) = µ k () i= d µ p k+` (p i ). i= ()+ (q i ) > (dp k+` µ) d(p k+` ) k i= ()+ (p i ). k i= ()+ Recalling (4), we have S (m) > S (n), so the poof is complete. (p i ) 3. Pime Divisos of Spasely Schemmel Totient Numbes In thei pape, Masse and Shiu casually mention that 2 is the only spasely totient pime powe [5], but thei bief poof utilizes the fact that, fo =, + is pime. We will see that if + is pime, then + is indeed the only spasely Schemmel totient numbe of ode that is a pime powe. Howeve, if + is composite, thee could easily be multiple spasely Schemmel totient numbes of ode that ae pime powes. The following esults will povide an uppe bound (in tems of ) fo the values of spasely Shemmel totient pime powes of ode.
INTEGERS: 5 (205) 6 Spasely Schemmel Totient Numbes of Ode 2 3, 5*, 2*, 45*, 05*, 65*, 95, 35*, 345, 525*, 585, 735*, 55*, 365*, 785*, 995*, 245, 245, 2625, 3465*, 4095*, 4305, 455, 5775*, 5985, 6825, 8085, 8925, 9555, 0395*, 505*, 9635*, 2945*, 23205, 25935, 26565*, 28875, 3395, 33495, 3395, 35805, 45045*, 47355, 49665, 50505, 58905, 65835, 75075*, 77805, 79695, 82005, 8435, 9875 3, 5, 7, 35*, 55, 65, 85, 95, 75*, 245*, 385*, 455*, 595*, 665, 805, 875, 05, 085, 295, 435, 925*, 2275, 2695*, 385, 5005*, 6545*, 735*, 7735, 8855, 0465, 65, 935, 395, 4245, 5785, 6555, 6835, 8095, 8655, 8865, 25025*, 25795, 27335, 35035*, 36575, 38675, 4585, 55055*, 65065*, 85085*, 95095* 4 5, 35*, 55, 65, 85, 75*, 385*, 455*, 595*, 665, 805, 05, 085, 295, 925*, 2275, 2695*, 5005*, 6545*, 735*, 7735, 8855, 0465, 65, 935, 395, 4245, 5785, 6555, 6835, 8095, 25025*, 35035*, 36575, 38675, 4585, 55055*, 85085*, 95095* 5, 7,, 3, 77*, 9*, 9, 33, 6, 203, 27, 259, 287, 30, 539*, 00*, 309*, 463*, 547, 77, 2233, 2387, 2849, 357, 33, 369, 408, 4543, 7007*, 963, 0*, 707*, 909*, 23023*, 2487, 29029, 303, 33649, 37037, 404, 43043, 47047, 53053, 59059, 606, 67067, 707, 73073, 79079, 83083 6 7, 77*, 9*, 9, 33, 6, 203, 27, 259, 00*, 309*, 463*, 547, 77, 2233, 2387, 2849, 357, 33, 369, 408, 7007*, 707*, 909*, 23023*, 2487, 29029, 303, 37037, 404, 43043, 47047, 53053, 59059, 606, 67067, 707, 73073, 79079 Table : This table lists all Spasely Schemmel totient numbes of ode that ae less than 0 5 fo each 2 {2, 3, 4, 5}. A table of Spasely totient numbes ( = ) is given in [5]. Those numbes constucted by Theoem ae maked with asteisks. Lemma 2. If j 2 N\{, 2, 4}, then p j+ p j apple 7 5. Poof. Piee Dusat apple[2] has shown that, fo x 396 738, thee must be at least one x pime in the inteval x, x + 25 log 2 x. Theefoe, wheneve p j > 396 738, we may p j + set x = p j + to get p j+ apple (p j + ) + 25 log 2 (p j + ) < 7 5 p j. Using Mathematica 9.0 [7], we may quickly seach though all the pimes less than 396 738 to conclude the desied esult. Lemma 3. Let p be a pime, and let,, and > and p -. If p 2 F, then p 2 F. be positive integes such that Poof. Suppose, fo the sake of finding a contadiction, that p 62 F and p 2
INTEGERS: 5 (205) 7 F. Because p 2 F B, we know that p 2 B. Then, because p 62 F, thee must exist some m 2 B such that m > p and S (m) apple S (p ) = p 2 (p )S ( ). Howeve, this implies that pm > p and S (pm) apple ps (m) apple p (p )S ( ) = S (p ), which contadicts the fact that p 2 F. Theoem 2. If p is a pime and is a positive intege, then p 2 F if and only if < p < (p ()+ )(p ()+2 ) +. Poof. Fist, suppose < p < (p ()+ )(p ()+2 ) +, and let m be an abitay element of B that is geate than p. We will show that S (m) > p.!(m) If!(m) 2, then S (m) (p ()+i ) (p ()+ )(p ()+2 ) > p i=. Theefoe, we may assume that!(m) = so that we may wite m = q fo some pime q > and positive intege. Futhemoe, we may assume > because if =, then S (m) = q = m > p. If 62 {, 2, 3, 5}, then it is easy to see, with the help of Lemma 2, that p ()+2 < 2. Thus, if 62 {, 2, 3, 5}, then we have S (m) = q (q ) q(q ) p ()+ (p ()+ ) > (p ()+ ) > (p ()+ )(p ()+2 ) > p. If =, then the inequality p < (p ()+ )(p ()+2 ) + foces p = 2, so S (m) = q (q ) > = p. If = 2, then the inequality < p < (p ()+ )(p ()+2 ) + foces p = 3, so S (m) = q (q ) > = p. If = 3, then q 5 and eithe p = 5 o p = 7. Theefoe, S (m) = q (q ) 5(5 3) > p. Finally, if = 5, then p 2 {7,, 3} and q 7. Thus, S (m) = q (q ) 7(7 5) > p. To pove the convese, suppose p (p ()+ )(p ()+2 ) +. We wish to find some m 2 B such that m > p and S (m) apple p. We may assume that p > p ()+ p ()+2 because, othewise, we may simply set m = p ()+ p ()+2. We know that thee exists a unique intege t ()+2 such that p ()+ p t < p < p ()+ p t+. Suppose > 3 so that, with the help of Lemma 2 and some vey shot casewok, we may conclude that p ()+ apple 7 and p t+ apple 7 p t. Then, setting m = p ()+ p t+, we have S (m) = (p ()+ )(p t+ ) apple 4 7 7 p t < 44 49 p 4 ()+p t 7 2 < p ()+ p t < p. We now handle the cases in which apple 3. If =, then p is odd, so we may set m = 2p to get S (m) = S (2)S (p) = p = p. If = 2, then 3 - p, so we may set m = 3p to find S 2 (m) = S 2 (3)S 2 (p) = p 2 = p. Finally, if = 3, then we have 5p t < p < 5p t+. Set m = 5p t+. As p t+ < 5 2 p t, we have 2p t+ S 3 (m) = 2(p t+ 3) < p 3 = p. 3 < p, so
INTEGERS: 5 (205) 8 Theoem 3. Let be a positive intege, and let p be a pime. Then p 2 ()+ 62 F and p 3 62 F. Poof. Suppose p 2 ()+ 2 F. Then, as p ()+ p ()+2 > p 2 ()+, we must have (p ()+ )(p ()+2 ) > p ()+ (p ()+ ). Theefoe, < p ()+2 p ()+. It is easy to see that this inequality fails to hold fo all apple 0. Fo, we may use Lemma 2 to wite p ()+ < p 2 and p ()+2 < p 2p ()+. Hence, p ()+2 p ()+ < ( p p 2 )p ()+ < (2 2) <, which is a contadiction. Now, suppose p 3 2 F. Then, by Lemma 3, we know that p 2 2 F, so p > p ()+. Let t be the unique intege such that p ()+ p t < p 2 < p ()+ p t+. Then p 3 < p ()+ p t+ p and p ()+ < p < p t+. Theefoe, as p 3 2 F, we see that S (p 3 ) = p 2 (p ) < (p ()+ )(p t+ )(p ), implying that p 2 < (p t+ )(p ()+ ) < p t+ (p ()+ ). Using Betand s Postulate, we see that p t+ < 2p t and p ()+ apple 2. Theefoe, p ()+ p t < p 2 < p t+ (p ()+ ) < 2p t (p ()+ ), so 2 < p ()+. This is ou desied contadiction. Combining Lemma 3, Theoem 2, and Theoem 3, we see that any n 2 F satisfying n ((p ()+ )(p ()+2 ) + ) 2 must have at least two pime factos. Futhemoe, we ecod the following conjectue about the nonexistence of spasely Schemmel totient numbes that ae squaes of pimes. This conjectue has been checked fo all apple 6 and p apple 37. Conjectue. Fo any pime p and positive intege, p 2 62 F. We now poceed to establish asymptotic esults concening the pimes that divide and do not divide spasely Schemmel totient numbes. Fo a given 2 N and n 2 F, we will define P k (n) to be the k th lagest pime diviso of n (povided!(n) k), and we will let Q k (n) denote the k th smallest pime that is lage than and does not divide n (the functions Q k depends on, but this should not lead to confusion because we will wok with fixed values of ). We will let R(n) = n p. We will also make use of the Jacobsthal function J. Fo a positive p2p p n intege n, J(n) is defined to be the smallest positive intege a such that evey set of a consecutive integes contains an element that is elatively pime to n [4]. In paticula, fo any positive intege, J(#) is the lagest possible di eence between consecutive elements of B. This means that, fo any positive eal x, the smallest element of B that is geate than x is at most x + J(#). Fo convenience, we will wite J = J(#). The fist sixteen values of J (stating with = ) ae, 2, 4, 6, 0, 4, 22, 26, 34, 40, 46, 58, 66, 74, 90, 00. Finally, we will let k() be the unique positive eal oot of the polynomial J xk + kx (k ).
INTEGERS: 5 (205) 9 Lemma 4. If, n, and k ae positive integes such that k!(n) k, then Q k (n) > k ()(P k (n) ). 2, n 2 F, and Poof. Wite M = k P i (n) and N = i= k i= Q i (n). Let µ be the smallest element of B that is geate than M N. As noted above, we must have µ < M N + J. Let us put m = µn M n so that m 2 B and < m n < + J N M < + J Q k (n) k P k (n) k. n Since, is an intege that is elatively pime to N, we may use the multiplicativity M of S along with () to wite S (m) = S µn n apple µs N n n = µs (N)S apple µ S (N)S (n). M M M S (M) This implies that S (m) m apple S (N) N M S (M) S (n) n < Q k (n) Theefoe, Q k (n) S (m) < k + J P k (n) k so the fact that n 2 F implies that Q k (n) k + J P k (n) k = k i= Q k Q k Q i (n) (n) j= S (n) P k (n) n. (n) S (n) P j (n) n S (n), P k (n) >. (8) P k (n) Q k (n) k Wite x = J P k (n) k, x 2 = Q k (n), and x 3 = so that (8) becomes P k (n) (+x )( x 2 ) k ( x 3 ) k >. Because x and x 2 ae positive and 0 < x 3 <, we may invoke the inequalities +x < e x, x 2 < e x2, and ( x 3 ) x3/( < e x3) to wite e x (k )x2+kx3/( x3) >. (9) Afte a little algebaic manipulation, (9) becomes J Qk (n) P k (n) + k Q k (n) P k (n) (k ) > 0.
INTEGERS: 5 (205) 0 Thus, if we wite A(x) = J Qk (n) xk + kx (k ), then A P k (n) means that Q k (n) P k (n) > k(), so we ae done. Lemma 5. Fo any positive integes and n with n 2 F, P (n) < Q (n) J + J Q (n). > 0. This Poof. Fix and n, and wite P = P (n) and Q = Q (n). Suppose, fo the sake of finding a contadiction, that P Q J + J Q. Let µ be the smallest element of B that is geate than P Q. Then µ < P Q + J. Wite m = Qµ P n so that m 2 B and < m n < + J Q P apple + J Because m J + J Q. is divisible by Q and all the pime divisos of n except possibly P, we have S (m) m apple S (n) Q P n. Theefoe, apple = S (m) < Q Q Q J + J Q Q + P J J + J Q! S (n)!! J J + J Q + J + J Q S (n) + J Q Q S (n) = S (n). This is ou desied contadiction, so the poof is complete. Fo the following lemma, ecall that we defined R(n) = n p2p p n Lemma 6. Let be a positive intege, and let n 2 F. Then p. R(n) < J Q (n)(q (n) ).
INTEGERS: 5 (205) Poof. Fix, and n, and wite Q = Q (n) and R = R(n). Suppose R J Q(Q ). Let µ be the smallest element of B geate than R Q. Then µ < R Q + J. If we put m = Qµ R n, then m 2 B and < m n < + J Q R apple + Q. Because m is divisible by Q and all the pime divisos of n, we have S (m) m apple S (n) Q n. This implies that S (m) < + Q Q S (n) = S (n), which is a contadiction. Coollay. Let and be positive integes, and let V ( ) = J p ()+ (p ()+ ). If n 2 F and n V ( ) (V ( ) + p ()+ )#, then!(n). In paticula, fo p () # n 2 F, lim!(n) =. n! Poof. We pove the contapositive. Suppose n 2 F and!(n) <. Then Q (n) apple p ()+, so P (n) < p ()+ J + J p ()+ = V ( ) + p ()+ by Lemma 5. This implies that R(n) = n p p () # n. Lemma 6 tells us (V ( ) + p p2p ()+ )# p n that R(n) < V ( ), so n apple R(n) (V ( ) + p ()+ )# p () # < V ( ) (V ( ) + p ()+ )#. p () # Coollay 2. Let be a positive intege. Fo su ciently lage n 2 F, P (n) 4 - n.
INTEGERS: 5 (205) 2 Poof. Fo any intege n >, wite fo any n 2 F satisfying n >, P(n)(n) = (n). Using Lemma 6, we see that, P (n) (n) apple R(n) < J Q (n)(q (n) Because Q (n) is at most the smallest pime exceeding P (n), we may use Betand s Postulate to wite ). J Q (n)(q (n) ) apple 2 J P (n)(2p (n) ) < 4 J P (n) 2. If P (n) 4 n, then (n) 3, so P (n) < 4 J. By Coollay, we see that this is impossible fo su ciently lage n. Masse and Shiu show that P (n) 3 - n fo all spasely totient numbes n, but thei methods ae not obviously genealizable [5]. Thus, we make the following conjectue, which has been checked fo all apple 6 and n apple 00000. Conjectue 2. Fo any positive integes and n with n 2 F and n >, P (n) 3 - n. Fo small values of, we may e otlessly make small amounts of pogess towad Conjectue 2. Fo example, it is easy to use Lemma 6 to show that P (n) 4 - n fo all n 2 F 2. Indeed, if P (n) 4 n fo some n 2 F 2, then P (n) < 4 J2 = 4. This foces 2 n to be a powe of 3, but Theoem 3 tells us that thee ae no powes of 3 in F 2 except 3 itself. We ae finally eady to establish ou pomised asymptotic esults. Theoem 4. Let, K, and L be positive integes with K (a) lim sup n! (b) lim sup n! (c) lim sup n! (d) lim sup n! P (n) log n 2, Q L (n) log n =, P K (n) log n apple K(), P (n) log 2 n apple J. 2. Fo n 2 F, we have Poof. To pove (a), let us begin by choosing some intege k ()+2. Let `(k) be k the lagest intege such that p k+`(k) < 2p k. Setting n(k) = p k+`(k) p i, we i= ()+
INTEGERS: 5 (205) 3 see, by Theoem, that n(k) 2 F. Futhemoe, as k i= ()+ p i apple n(k) < k+ i= ()+ the Pime Numbe Theoem tells us that p k log n(k) as k!. Thus, as k!, P (n(k)) = p k+`(k) 2p k 2 log n(k). To pove (b), choose any n 2 F with n >, and let k(n) be the unique intege satisfying p i apple n < p i. Using the Pime Numbe Theoem again, k(n) k(n)+ we i= ()+ i= ()+ have Q L (n) apple p k(n)+l log n as n!. In addition, fo those n 2 F (guaanteed by Theoem ) of the fom n = k(n) i= ()+ p i, we see that Q L (n) = p k(n)+l log n. Coollay guaantees that the limit in (c) is well-defined. To pove the limit, we use Lemma 4 to find that if n 2 F and!(n) K, then P K (n) log n < K() Q K (n) + log n log n. Then the desied esult follows fom setting L = K in (b). Finally, (d) follows immediately fom Lemma 5 and fom setting L = in (b). p i, Acknowledgments The autho would like to thank the efeee fo many valuable suggestions. Refeences [] Bake, R. C. and Haman, G., Spasely totient numbes, Annales de la Faculté des Sciences de Toulouse, 5, 996, 8390. [2] Dusat, Piee. Estimates of some functions ove pimes without R.H., axiv:002.0442 (200). [3] Haman G., On spasely totient numbes, Glasgow Math. J., 33, 99, 349 358. [4] Jacobsthal, E., Übe sequenzen ganze Zahlen von denen keine zu n teilefemd ist, I III Noske Vidensk. Selsk. Foh. Tondheim 33 (960), 7 39. [5] Masse, D. W. and Shiu, P., On spasely totient numbes. Pacific J. Math. 2 (986), no. 2, 407 426. [6] Schemmel, V. Übe elative Pimzahlen, Jounal fü die eine und angewandte Mathematik, Band 70 (869), S. 9 92 [7] Wolfam Reseach, Inc., Mathematica, Vesion 9.0, Champaign, IL (202). [8] ip, Lee-Wah. On Camichael type poblems fo the Schemmel totients and some elated questions. Thesis (Ph.D.)Univesity of Albeta (Canada). PoQuest LLC, Ann Abo, MI, 989, pp 2.