A006558: First group of n consecutive integers with same number of divisors

[David Wasserman]

Dear fellow seqfans,
I have submitted a new upper bound: a(9) <= 2197379769820.  This is  
low enough that someone with enough computing power could check all  
smaller possibilities and conclusively determine a(9).  In fact, it's  
only about twice as big as a(8), which was found three years ago, and  
computers have improved since then.  I hope someone will take on this  
task.  At the end of this message are some ideas about how to do it.

For consecutive integers with the same prime signature, I also have a  
new upper bound for the next term: A034173(7) <= 464881210073.  This  
is a similar problem, and is probably easier.

BTW, these are probably the most interesting results from my  
arithmetic progressions project, which I posted about last April, and  
recently finished.

%I A006558 M2155
%S A006558 1,2,33,242,11605,28374,171893,1043710445721
%N A006558 Start of first run of n consecutive integers with same  
number of divisors.
%D A006558 R. K. Guy "Unsolved Problems in Number Theory", section B18.
%D A006558 David Wells, "The Penguin Dictionary of Curious and  
Interesting Numbers," Penguin Books, NY, 1986, pages 147 and 176.
%Y A006558 Cf. A000005, A005237, A005238, A006601, A049051, A019273,  
A039665.
%Y A006558 Sequence in context: A041127 A097978 A052403 this_sequence  
A002561 A030448 A093992
%Y A006558 Adjacent sequences: A006555 A006556 A006557 this_sequence  
A006559 A006560 A006561
%K A006558 nonn
%O A006558 1,2
%A A006558 njas, Robert G. Wilson v (rgwv(AT)rgwv.com)
%E A006558 Jud McCranie points out that the entry 40311 given in Guy  
and Wells is incorrect.
%E A006558 a(9) <= 17796126877482329126044.
%E A006558 One more term from Jud McCranie (j.mccranie(AT) 
adelphia.net), Jan 20 2002

%I A034173
%S A034173 1,2,33,19940,204323,380480345
%N A034173 a(n) is minimal such that prime factorizations of a 
(n), ..., a(n)+n-1 have same exponents.
%C A034173 a(7) <= 1091100709673
%e A034173 a(4)=19940 because 19940, ..., 19943 all have the form p^2  
q r.
%Y A034173 Cf. A034174.
%Y A034173 Sequence in context: A090335 A113105 A083459 this_sequence  
A003820 A109336 A098869
%Y A034173 Adjacent sequences: A034170 A034171 A034172 this_sequence  
A034174 A034175 A034176
%K A034173 hard,nonn
%O A034173 1,2
%A A034173 Dean Hickerson (dean(AT)math.ucdavis.edu)


Computing A006558(9): Note that at least one of the 9 numbers must be  
4 mod 8, so the number of divisors is divisible by 3, and each number  
is divisible by a square.  2 or 3 of them are divisible by 4, but no  
other p^2 can divide more than one of them, so at least one of them  
is divisible by p(i)^2 for some i >= 7.  So here's what I would do if  
I had a big enough computer: for each prime p from p(7) to sqrt 
(2197379769820/2), test every multiple of p^2 up to 2197379769820.   
"Test" means checking consecutive integers both above and below to  
see how many have the same number of divisors.  Many numbers will be  
tested more than once, but to eliminate the repetition would probably  
require more computation than it saves.  The total number of tests in  
this method is about 18 billion.

Maybe you can think of a more efficient way.

Thanks,
David