[seqfan] Re: Giuga numbers
mathar at strw.leidenuniv.nl
Mon Dec 7 12:53:03 CET 2009
I have part of half of the answer to the question in
> From seqfan-bounces at list.seqfan.eu Fri Nov 13 10:37:28 2009
> Date: Fri, 13 Nov 2009 10:28:03 +0100
> From: Paolo Lava <paoloplava at gmail.com>
> Subject: [seqfan] Giuga numbers
> Is it correct to state that Giuga numbers
> A007850 <http://www.research.att.com/~njas/sequences/A007850>
> are all and only numbers n whose arithmetic derivative n' is equal to n+1 ?
Concerning the question: Is the arithmetic derivative g' of Giuga numbers g,
g'=A003414(g), equal to g+1?
I can show that g'-1 =g times a factor relative prime to g.
Details: First, all Giuga numbers g are square-free.
[This is known, see Luca, Pomerance and Shparlinski in http://math.dartmouth.edu/~carlp/giugafinal.pdf ,
Int J. Mod Math 4 (1) (2009) 13-18, MR2508938,
or Borwein^3, and Girgensohn in Am Math Monthly 103 (1996) 40-50,
or the Borwein, Bailey book "Experimentation in Mathematics: Comput. Paths
Proof by contradiction: if they were not square-free, they could be
written g=p^2*r with p some prime divisor and r some integer.
Then g/p-1 = p*r-1 would by definition of the Giuga numbers have
to equal alpha*p, with alpha some integer. Now p*r-1=alpha*p means
p*(r-alpha)=1 and this cannot happen since a prime multiplied by an integer
cannot generate the 1.]
Second, we have by definition of the Giuga numbers g/p-1=alpha*p
for all primes that divide g (alpha an integer depending on p).
and by the multiplicative rule of the arithmetic derivative
g' = (alpha*p+1)+ p*(alpha*p+1)'
g' = alpha*p+1+ p*(alpha*p+1)'
g' = 1+ p* ( (alpha*p+1)'+alpha)
This must be valid for all pairs (p_i,alpha_i) of primes p_i dividing g',
and since g is one unique number with a single, unique value of g', this
yields the pile of equations
1+ p_1* ( (alpha_1*p_1+1)'+alpha_1)
=1+ p_2* ( (alpha_2*p_2+1)'+alpha_2)
=1+ p_3* ( (alpha_3*p_3+1)'+alpha_3)
or, subtracting 1,
= p_1* ( (alpha_1*p_1+1)'+alpha_1)
= p_2* ( (alpha_2*p_2+1)'+alpha_2)
= p_3* ( (alpha_3*p_3+1)'+alpha_3)
We see that this constant g'-1 has each prime p_i that divides g as a
factor, and since g is square-free (see the Lemma above), this number g'-1
is g potentially multiplied by a number which is relative prime to g.
It remains to show that this factor is 1, to complete the answer to
half of the question, which I have not managed to do.
[Perhaps one can put a strict upper bound on this factor along the
following lines of thought :
= p_i* (g/p_i)'+ p_i*alpha_i
= p_i* (g/p_i)'+ (g/p_i) -1
= p_i* ( product of dist. prime factors of g excluding p_1)'+ (g/p_1) -1
= p_i* (sum of products of dist prime factors excluding p_i and a p_j)+ (g/p_1) -1
= sum of products of dist prime factors, one omitted, p_i included + product of dist prime factors, p_i excluded -1
= sum of products of dist prime factors, one omitted -1 (trivial by
definition of the arith. deriv.
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