A119479(12) (was: Re: A072507)
Dean Hickerson
dean at math.ucdavis.edu
Fri Jul 28 07:18:06 CEST 2006
Mostly to Franklin T. Adams-Watters:
> The next question is, what is the length, for a given n, of the longest
> sequence of integers with exactly n divisors?
...
> 12 5 to 24
I can show that 7 <= A119479(12) <= 15. No doubt the correct value is 15,
but finding an example may be difficult.
I'll send Neil an update for this sequence in a day or two.
First, here's an example, probably not the smallest one, of 7 consecutive
numbers with 12 divisors:
907162760479789021 = 13^2 5483 978994525823
907162760479789022 = 2 11^2 3748606448263591
907162760479789023 = 3 7^2 6171175241359109
907162760479789024 = 2^5 28348836264993407
907162760479789025 = 5^2 23 1577674366051807
907162760479789026 = 2 3^2 50397931137766057
907162760479789027 = 17^2 408041 7692784523
For a number to have exactly 12 divisors, it must have one of these forms,
with primes p, q, r:
p^11
p^5 q
p^3 q^2
p^2 q r
If there were 16 consecutive numbers like this, then 2 of them, say n and
n+8, would be multiples of 8, so each must have one of these forms:
2^11
2^5 q
2^3 q^2
The first form doesn't work, since neither 2048-8 nor 2048+8 has the right
form. Exactly one of n and n+8 is divisible by 16, so either
n = 2^5 q and n+8 = 2^3 r^2
or
n = 2^3 r^2 and n+8 = 2^5 q.
In either case, r^2 = 4q +/- 1 == 3 or 5 (mod 8), which is impossible.
So A119479(12) <= 15.
If Dickson's conjecture
http://primes.utm.edu/glossary/page.php?sort=DicksonsConjecture
is true then there exist infinitely many values of n such that
n-7 = 17^2 19 p0
n-6 = 2 23^2 p1
n-5 = 3 29^2 p2
n-4 = 2^2 7 p3
n-3 = 5 31^2 p4
n-2 = 2 3^2 p5
n-1 = 11^2 43 p6
n = 2^5 p7
n+1 = 3 37^2 p8
n+2 = 2 5^2 p9
n+3 = 7^2 47 p10
n+4 = 2^2 3 p11
n+5 = 13^2 53 p12
n+6 = 2 41^2 p13
n+7 = 3^2 5 p14
with primes p0, ..., p14. For such n, the 15 values shown all have exactly
12 divisors.
Such n are congruent to 5176651285455373373146515194418848
(mod 25049078033217046360026006245313600).
I've done some searching through numbers like that, and haven't found any
in which all of p0, ..., p14 are prime. So far, the minimum number of
prime factors of p0, ..., p14 that I've found is 29, with
n=148746602012528276659207571599866575648; here the factorizations of n-7
through n+7 are:
17^2 19 3415667309629499 7930855671110333849
2 23^2 140592251429610847503976910774921149
3 29^2 775280096432899 76045088268490599659
2^2 7 484397 1482421 7398028818311104810722929
5 31^2 191 162076591260770332669620510484679
2 3^2 85138110441927629 97062291715338971443
11^2 43 397009041661387 72010003137562508527
2^5 59951450187281 77534926984596535780169
3 37^2 152898601403413 236874787514240237239
2 5^2 2974932040250565533184151431997331513
7^2 47 73 884769728659629647209462176195829
2^2 3 12395550167710689721600630966655547971
13^2 53 208684849 79578098906104400262388121
2 41^2 139 127301 2500361072448101346992875253
3^2 5 82131221 40246327821217349154310453079
Of course there are many other ways to distribute the small factors among
the 15 numbers, so it's unlikely that the smallest example of 15 will be
in the particular arithmetic progression that I looked at. Unfortunately
I can't think of an efficient way to search through a large number of such
a.p.s at once.
Dean Hickerson
dean at math.ucdavis.edu
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