Browse Groups

• ## Another large pair of consecutive smooth numbers

(2)
• NextPrevious
• Let f(x) = x*(x-11)*(x-24)*(x-65)*(x-90)*(x-129)*(x-173)* (x-212)*(x-237)*(x-278)*(x-291)*(x-302)/67440294559676054016000 then the smooth pair is Q and Q+1
Message 1 of 2 , Mar 20, 2012
View Source
Let
f(x) = x*(x-11)*(x-24)*(x-65)*(x-90)*(x-129)*(x-173)*
(x-212)*(x-237)*(x-278)*(x-291)*(x-302)/67440294559676054016000
then the smooth pair is
Q and Q+1 where

Q=f(2037335926565)
=
7582846503419980985727018264244718835660110144701285792357\
0157185560815861224078102966540735794780952727297976931615\
819922375
=
5^3*11*13*23*29*37*53*59*73*101^2*109*151*193*277*
293*563*617*743*821*947*1459*2097803*1220327*718139*
97523*54581*28807*207293*1610639*835633*80777*322939*11681*
1847*4231*1871*2791
=
7.58 * 10^124 roughly

and

Q+1=
2^3*3^5*7*41*43*47*67*89*113^2*223*431*433*461*491*577*1163*
1249*1303*1543*337013*2573827*422069*123427*
303587*36683*755173*3167*1682999*1914811*3593*17807*61031*
2741*225149*5953

max prime involved in either is 2573827=prime[188076].

This one should be fairly difficult to beat, I think, even though I estimate
that with this same prime bound an example with Q over 2000 digits long
should exist (which I am not smart enough to find), albeit I think no Q over
3000 digits long should exist.

Although this might be beatable using the same idea and a larger search (and that would be interesting), it would be more interesting to employ a different idea.

Another fairly strong example arises from Q=f(6457935020528)=7.80*10^130 roughly
with maxprime=5881217=prime[405194].
• ... Definition: For positive integer n let p be the largest prime divisor of n*(n+1) and S(n) = log(n)/log(p) be the figure of merit for the smoothness of the
Message 1 of 2 , Apr 20, 2012
View Source
"WarrenS" <warren.wds@...> wrote:

> max prime involved in either 2573827

Definition: For positive integer n let p be the largest
prime divisor of n*(n+1) and S(n) = log(n)/log(p) be
the figure of merit for the smoothness of the
consecutive integers n and n+1.

S(n) =~ log(7.5828465*10^124)/log(2573827) =~ 19.480

Here is a 135-digit example with greater merit:

{f(y)=
(y^2-11^4)*(y^2-35^2)*(y^2-47^2)*
(y^2-94^2)*(y^2-146^2)*(y^2-148^2)/
67440294559676054016000 - 1;}

n = f(12971885307194);

{if(type(n)=="t_INT",F=factor(n*(n+1))[,1];
p=F[#F];print("n has "#Str(n)" digits");
print("max prime is "p);
default(realprecision,5);
print("S(n) =~ "log(n)/log(p)));}

n has 135 digits
max prime is 6244451
S(n) =~ 19.797

David
Your message has been successfully submitted and would be delivered to recipients shortly.
• Changes have not been saved
Press OK to abandon changes or Cancel to continue editing
• Your browser is not supported
Kindly note that Groups does not support 7.0 or earlier versions of Internet Explorer. We recommend upgrading to the latest Internet Explorer, Google Chrome, or Firefox. If you are using IE 9 or later, make sure you turn off Compatibility View.