Divisibility of the Jordan function

Sat Dec 10 20:03:29 CET 2005

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.

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.  
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 
this note, following the notation in the paper by Vaughan and Wooley.  (The alternate lambda^*(n) in the comment for A006863 
is too awkard.)

(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 
including an edited version below.  I have also changed the initial value to 0, since by the definition K(0) is divisible by 
every integer.)

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 
many n.

A quick sketch of the proof follows.  This relies on the description of lambda^* in A006863.  The case m=2 has some 
exceptions, because 2 is the only value where K(m) is divisible by a prime raised to a power greater than m - specifically, 
2^3.

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 
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 
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) = 
(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 
J_m(n) except when n=1, n is a prime dividing K(m), or m=2 and n=4.

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 
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 
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).

Note, by the way, that the largest prime divisor of K(m) is at most m+1.

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 
and A046657).  Larger values on m are not.  In particular, J_2(n)/24 with offset 5:

1,1,2,2,3,3,5,4,7,6,8,8,12,9,15,12,16,15,22,16,25,21,27,24,
35,24,40,32,40,36,48,36,57,45,56,48,70,48,77,60,72,66,92,
64,98,75,96,84,117,81,120,96,120,105,145,96,155,120,144,128

Sum_{k=1}^n J_2(k)/24 with offset 4:

1,2,3,5,7,10,13,18,22,29,35,43,51,63,72,87,99,115,130,152,
168,193,214,241,265,300,324,364,396,436,472,520,556,613,
658,714,762,832,880,957,1017,1089,1155,1247,1311,1409

J_4(n)/240 with offset 6:

5,10,16,27,39,61,80,119,150,208,256,348,405,543,624,800,
915,1166,1280,1625,1785,2187,2400,2947,3120,3848,4096,
4880,5220,6240,6480,7809,8145,9520,9984,11774,12000

Sum_{k=1}^n J_4(k)/240 with offset 5:

4,9,19,35,62,101,162,242,361,511,719,975,1323,1728,2271,
2895,3695,4610,5776,7056,8681,10466,12653,15053,18000,
21120,24968,29064,33944,39164,45404,51884,59693,67838

One could continue, of course, but I think this is sufficient.

%I A079612
%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,
%T A079612 24,2,6960,2,171864,2,32640,2,24,2,138181680,2,24,2,1082400,2,151704,2,5520,2,
%U A079612 1128,2,4455360,2,264,2,12720,2,86184,2,13920,2,1416,2,6814407600,2,24,2
%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\
  tly the divisors of a(n).
%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 \
  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) \
  for n >= the largest exceptional value.  Frank Adams-Watters (FrankTAW(at)Netscape.com) Dec 10, 2005.
%D A079612 R. C. Vaughan and T. D. Wooley, Waring's problem: a survey, pp. 285-324 of Surveys in Number Theory (Urbana,\
   May 21, 2000), ed. M. A. Bennett et al., Peters, 2003. (The function K(n), see p. 303.)
%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\
  uns through all odd primes such that p-1 divides n and p^t exactly divides n).
%Y A079612 Cf. A006863 (bisection except for initial term); A059379 (Jordan function).
%Y A079612 Sequence in context: A089987 A054909 A100816 this_sequence A066585 A075267 A002743
%Y A079612 Adjacent sequences: A079609 A079610 A079611 this_sequence A079613 A079614 A079615
%K A079612 nonn
%O A079612 0,2
%A A079612 njas Jan 29 2003

Franklin T. Adams-Watters
16 W. Michigan Ave.
Palatine, IL 60067
847-776-7645
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