infinitary harmonic numbers needed
Meeussen Wouter (bkarnd)
wouter.meeussen at vandemoortele.com
Thu Aug 30 17:55:14 CEST 2001
the infinitary divisors of k are produced by: (cfr
http://xraysgi.ims.uconn.edu/amicable.html )
Table[ (Apply[Times, (First[it]^(# /.
z->List))]&/@Flatten[Outer[z,Sequence@@( bitty /@ Last[
it=Transpose[FactorInteger[k]] ] ),1]]) ,{k,2,75}]
with helper function:
bitty[k_]:=Union[Flatten[Outer[Plus,Sequence@@
({0,#}&/@Union[(2^Range[0,Floor[Log[2,k]]]) Reverse[IntegerDigits[k,2]]])
]]]
their harmonic mean is integer for k= :
1+Flatten[Position[Table[(Length[#]/(Plus@@(1/#))&)@ (Apply[Times,
(First[it]^(# /. z->List))]&/@Flatten[Outer[z,Sequence@@( bitty /@ Last[
it=Transpose[FactorInteger[k]] ] ),1]]) ,{k,2,2^16+1}] ,_Integer]] //Timing
{238.9*Second,
{6, 45, 60, 90, 270, 420, 630, 2970, 5460, 8190, 9100, 15925, 27300, 36720,
40950, 46494, 54600}}
***************************************************************
Sadly,
I have to warn that the above procedures do not reproduce A037445, but they
do fit with A049417.
disagreement with A037445 is found for {36, 47, 48, 51, 56, 58, 61, 70, 74,
75}, where
A037445 gives {8, 4, 8, 2, 4, 8, 4, 4, 8, 8},
but the infinitary divisors are calculated as:
{1, 9, 4, 36}
{1, 47}
{1, 3, 16, 48}
{1, 17, 3, 51}
{1, 7, 2, 14, 4, 28, 8, 56}
{1, 29, 2, 58}
{1, 61}
{1, 7, 5, 35, 2, 14, 10, 70}
{1, 37, 2, 74}
{1, 25, 3, 75}
I don't have the nerve to "correct" A037445,
but could someone please check this out?
I also propose to replace its comments
Comments: A divisor of n is called infinitary if it is a product of
divisors of the
form p^{y_a 2^a}, where p^y is an exact-power divisor of n and
sum_a y_a
2^a is the binary representation of y.
with that from the website mentioned at the top (not equivalent!) :
if the prime factorization of n is p1a1 p2a2 ... pmam, then
d divides n just
when d can be written in the form p1c1 p2c2 ... pmcm, where
c1 is
between 0 and a1, inclusive, c2 is between 0 and a2,
inclusive, and so on.
d is called an infinitary divisor of n if each ci has a zero
bit in its binary
expansion everywhere that the corresponding ai does
Wouter Meeussen
tel +32 (0)51 33 21 24
fax +32 (0)51 33 21 75
wouter.meeussen at vandemoortele.com
> -----Original Message-----
> From: N. J. A. Sloane [SMTP:njas at research.att.com]
> Sent: Wednesday, August 29, 2001 3:45 AM
> To: seqfan at ext.jussieu.fr
> Subject: infinitary harmonic numbers needed
>
> Dear Seqfans
> there's a sequence that maybe someone could compute
>
> To start with, recall that there are numbers that are
> called the infinitary divisors of n
>
> Their sum gives A049417,
> the number of infinitary divisors gives A037445.
> They are defined there and (maybe clearer) in A006407
>
> ok, so take n, make a list of its infinitary divisors,
> (call them d_1, d_2, ... d_k)
> compute their harmonic mean, which is
>
> 1 / ( (1/k) * (Sum 1/d_i ) )
>
> The sequence i'm interested in is the set of n for which
> this harmonic mean is an integer.
>
> There's a paper, Infinitary harmonic numbers,
> by Hagis and Cohen, Math. Rev. 91d:11001,
> Bull. Australian math. Soc., 41 (1990), 151-158
> that discusses their their asymptotics.
> If you have access to math sci net, see
> http://www.ams.org/msnmain?co3=AND&co4=AND&dr=all&fmt=doc&fn=105&id=91d_11
> 001&l=100&pg3=ICN&pg4=TI&r=2&s3=Cohen&s4=infinitary
>
> The sequence may be in the EIS already under a different name.
>
> NJAS
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