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<DIV>The Jordan function(s) J_m(n) can be defined as multiplicative with J_m(p^e) = (p^m-1)*p^(m*(e-1)). Cf A059379.</DIV>
<DIV> </DIV>
<DIV>Looking at the sequences J_m(n) for fixed m, one is struck by the fact that all but a few early terms have a common factor. </DIV>
<DIV>In fact, this sequence of common factors is in the OEIS: A079612. I will refer to this sequence as K(n) for the remainder of </DIV>
<DIV>this note, following the notation in the paper by Vaughan and Wooley. (The alternate lambda^*(n) in the comment for A006863 </DIV>
<DIV>is too awkard.)</DIV>
<DIV> </DIV>
<DIV>(Incidently, the name for A079612 is wrong. According to it, K(n) for odd n should be 4, but 2 is the correct value. I am </DIV>
<DIV>including an edited version below. I have also changed the initial value to 0, since by the definition K(0) is divisible by </DIV>
<DIV>every integer.)</DIV>
<DIV> </DIV>
<DIV>In fact, K(m) not only divides J_m(n) for all but finitely many n; it also divides Sum_{k=1}^n J_m(k) for all but finitely </DIV>
<DIV>many n.</DIV>
<DIV> </DIV>
<DIV>A quick sketch of the proof follows. This relies on the description of lambda^* in A006863. The case m=2 has some </DIV>
<DIV>exceptions, because 2 is the only value where K(m) is divisible by a prime raised to a power greater than m - specifically, </DIV>
<DIV>2^3.</DIV>
<DIV> </DIV>
<DIV>First, note that if a prime p does not divide K(m), p^m = 1 (mod K(m)), and so K(m) divides J_m(p^e). For a prime p dividing </DIV>
<DIV>K(m), if q^f is another prime power factor of K(m), p^m = 1 (mod q^f). This shows that J_m(n) is divisible by K(m) whenever </DIV>
<DIV>n is the product of two or more distinct primes. Furthermore, if e is the multiplicity of p as a divisor of K(m), J_m(p^2) = </DIV>
<DIV>(p^m-1)*p^m = 0 (mod p^e) except when m=2 and p=2, and for all higher powers of p. Thus we have proved that K(m) divides </DIV>
<DIV>J_m(n) except when n=1, n is a prime dividing K(m), or m=2 and n=4.</DIV>
<DIV> </DIV>
<DIV>Looking at the sum, J_m(p) = -1 (mod p^e) (except m=2, p=2, where we have to use J_2(2)+J_2(4)). J_m(1) = 1, so these two </DIV>
<DIV>terms cancel out mod p^e; and all other J_m(n) are divisible by p^e. So the Sum_{k=1}^n J_m(k) = 0 (mod p^e) for n >= p (4 </DIV>
<DIV>for m=2, p=2), and thus this sum = 0 (mod K(m)) from the largest prime divisor of K(m) on (or 4 for m=2).</DIV>
<DIV> </DIV>
<DIV>Note, by the way, that the largest prime divisor of K(m) is at most m+1.</DIV>
<DIV> </DIV>
<DIV>This does suggest a few new sequences. J_1(n) = phi(n), and both phi(n)/2 and Sum_{k=1}^n phi(n)/2 are in the OEIS (A023022 </DIV>
<DIV>and A046657). Larger values on m are not. In particular, J_2(n)/24 with offset 5:</DIV>
<DIV> </DIV>
<DIV>1,1,2,2,3,3,5,4,7,6,8,8,12,9,15,12,16,15,22,16,25,21,27,24,<BR>35,24,40,32,40,36,48,36,57,45,56,48,70,48,77,60,72,66,92,<BR>64,98,75,96,84,117,81,120,96,120,105,145,96,155,120,144,128</DIV>
<DIV> </DIV>
<DIV>Sum_{k=1}^n J_2(k)/24 with offset 4:</DIV>
<DIV> </DIV>
<DIV>1,2,3,5,7,10,13,18,22,29,35,43,51,63,72,87,99,115,130,152,<BR>168,193,214,241,265,300,324,364,396,436,472,520,556,613,<BR>658,714,762,832,880,957,1017,1089,1155,1247,1311,1409</DIV>
<DIV> </DIV>
<DIV>J_4(n)/240 with offset 6:</DIV>
<DIV> </DIV>
<DIV>5,10,16,27,39,61,80,119,150,208,256,348,405,543,624,800,<BR>915,1166,1280,1625,1785,2187,2400,2947,3120,3848,4096,<BR>4880,5220,6240,6480,7809,8145,9520,9984,11774,12000</DIV>
<DIV> </DIV>
<DIV>Sum_{k=1}^n J_4(k)/240 with offset 5:</DIV>
<DIV> </DIV>
<DIV>4,9,19,35,62,101,162,242,361,511,719,975,1323,1728,2271,<BR>2895,3695,4610,5776,7056,8681,10466,12653,15053,18000,<BR>21120,24968,29064,33944,39164,45404,51884,59693,67838</DIV>
<DIV> </DIV>
<DIV>One could continue, of course, but I think this is sufficient.</DIV>
<DIV> </DIV>
<DIV>%I A079612<BR>%S A079612 0,2,24,2,240,2,504,2,480,2,264,2,65520,2,24,2,16320,2,28728,2,13200,2,552,2,131040,2,<BR>%T A079612 24,2,6960,2,171864,2,32640,2,24,2,138181680,2,24,2,1082400,2,151704,2,5520,2,<BR>%U A079612 1128,2,4455360,2,264,2,12720,2,86184,2,13920,2,1416,2,6814407600,2,24,2<BR>%N A079612 Largest number m such that a^n = 1 (mod m) whenever a is prime to m; the numbers with this property are exac\<BR> tly the divisors of a(n).<BR>%C A079612 a(m) divides the Jordan function J_m(n) for all n except when n is a prime dividing a(m) or m=2, n=4; it is \<BR> the largest number dividing all but finitely many values of J_m(n). For m > 0, a(m) also divides Sum_{k=1}^n J_m(k) \<BR> for n >= the largest exceptional value. Frank Adams-Watters (FrankTAW(at)Netscape.com) Dec 10, 2005.<BR>%D A079612 R. C. Vaughan and T. D. Wooley, Waring's problem: a survey, pp. 285-324 of Surveys in Number Theory (Urbana,\<BR> May 21, 2000),!
ed. M. A. Bennett et al., Peters, 2003. (The function K(n), see p. 303.)<BR>%F A079612 a(n)=2 for n odd; for n even, a(n) = product of 2^{t+2} (where 2^t exactly divides n) and p^{t+1} (where p r\<BR> uns through all odd primes such that p-1 divides n and p^t exactly divides n).<BR>%Y A079612 Cf. A006863 (bisection except for initial term); A059379 (Jordan function).<BR>%Y A079612 Sequence in context: A089987 A054909 A100816 this_sequence A066585 A075267 A002743<BR>%Y A079612 Adjacent sequences: A079609 A079610 A079611 this_sequence A079613 A079614 A079615<BR>%K A079612 nonn<BR>%O A079612 0,2<BR>%A A079612 njas Jan 29 2003</DIV>
<DIV> </DIV>
<DIV>Franklin T. Adams-Watters<BR>16 W. Michigan Ave.<BR>Palatine, IL 60067<BR>847-776-7645</DIV></DIV></DIV>
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