a propos divisors...

[Peter Pein]

Max Alekseyev schrieb:
> On 7/25/07, Peter Pein <petsie at dordos.net> wrote:
> 
>> a) n such there is at least one x<n such that sigma(r,x)=sigma(r,n),
>> because there exists such a series for r=1 (A069822)
>>
>> and
>>
>> b) a(n)=min({k>0: sigma(r,k)=sigma(r,n)})
> 
> [...]
> 
>> but for r = 3 the air becomes very thin. If I did not make a mistake,
>> this seq starts(!):
>>
>> 194315, 295301, 590602, 1181204, 1476505, 1886920, 2067107, 2362408,
>> 2526095, 2953010, 3248311, 3691985, 3838913, 4134214, 4469245, 4724816,
>> 5020117, 5610719, 5635135, 5906020
> 
> [...]
> 
>> And if you want to make me really happy (I have had birthday on July
>> 24th ;-) (http://www.stevesbeatles.com/songs/when_im_sixty_four.asp this
>> age will be reached in 20 years, but the symptoms... )):
> 
> My congratulations!
> Better late than never ;)

Thank you!

> 
>>  If you've got Mathematica and more RAM (4GB or so) than I do (1.5 GB),
>> could you please run this code  with, say nmax=10 or 20 million? On my
>> machine it swapped heavily with nmax=6 million and I had to kill
>> MathKernel as I tried nmax=10^7. The lines above took ~181 seconds to
>> evaluate (nmax=10^7 has been stopped by me after 15 minutes). I do not
>> expect any runtimes of more than 7 minutes. Would this be possible, please?
> 
> These are the values below 10^7 that I got with my C++ program using
> LiDIA library:
> 
> 194315 184926
> 295301 291741
> 590602 583482
> 1181204 1166964
> 1476505 1458705
> 1886920 1880574
> 2067107 2042187
> 2362408 2333928
> 2526095 2404038
> 2953010 2917410
> 3248311 3209151
> 3691985 3513594
> 3838913 3792633
> 4134214 4084374
> 4469245 4253298
> 4724816 4667856
> 5020117 4959597
> 5610719 5543079
> 5635135 5362854
> 5906020 5834820
> 6023765 5732706
> 6496622 6418302
> 6791923 6710043
> 7382525 7293525
> 7677826 7585266
> 7966915 7581966
> 8268428 8168748
> 8355545 7951818
> 8563729 8460489
> 9132805 8691522
> 9449632 9335712
> 
> Each here line contains a pair:
> n k
> such that sigma(3,k)=sigma(3,n) and k=a(n) for r=3 (following your
> notations above).
> 
> I will let my program to run for a couple more days to reach 10^8 bound.
> 
>> (AFAIK there exists no kind of "inverse
>> function" to sigma(r,n) w.r.t. n which could be calculated without this
>> brute-force method).
> 
> I disagree with this statement. There is a more or less clever way to
> reconstruct the inverse of m=sigma(r,n) w.r.t. n, using integer
> factorization of m and a kind of brute-force but of the magnitude of
> the number of divisors of m.
> 
> Regards,
> Max
> 
Dear Max,

:-) thank you very much! :-)

I just entered the extension to A13190{7|8} and of course mentioned your
name. When your program gives more results, please extend the sequences
or if you've got a lot of numbers, send Neil a b-file.

The method you mention sounds mangeable - I will look for that algorithm.

Thanks again,
Peter



Dear seqfans,
I'd appreciate information (including references) 
about conditions (necessary and/or sufficient) on 
a sequence of positive numbers a1, a2, a3, ... 
in order that their Hankel matrix be positive 
definite. 
Thanks,
Emeric



Dear seqfans,
I'd appreciate information (including references) 
about conditions (necessary and/or sufficient) on 
a sequence of positive numbers a1, a2, a3, ... 
in order that their Hankel matrix be positive 
definite. 
Thanks,
Emeric