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189 quadratically smooth below 10^5000 bits

Jaroslaw Wroblewski — Wed, 10 Feb 2010 07:25:53 GMT

Following the same construction as for 180, I got: AP189: pp97^2 + 24*61#*i, for i=-126...62 where pp97=97#/61# q[-93]=1 (mod 315) q[0]=1 (mod 763048) Bits

Re: 193 quadratically smooth below 10^5000 bits

djbroadhurst — Tue, 09 Feb 2010 22:19:08 GMT

... Jaroslaw Wroblewski wrote:> ... Yes, it's difficult to reconcile all the desiderata. As I emphasized, your successful AP180 came

Re: 193 quadratically smooth below 10^5000 bits

Jaroslaw Wroblewski — Tue, 09 Feb 2010 20:57:43 GMT

Oooops, I am very very sorry, I take it back. I forgot that N must be a square and in this case it is not. This construction is incorrect. So the last bid is

193 quadratically smooth below 10^5000 bits

Jaroslaw Wroblewski — Tue, 09 Feb 2010 20:23:07 GMT

AP193: 113#/61# + 18*61#*i for i =-126...66 q[0]=1 (mod 480720240) 9.088221267417938382229896738704326597987823 E4986 bits Taking p[0] as a square is not as

Re: 180 quadratically smooth below 10^5000 bits

djbroadhurst — Tue, 09 Feb 2010 19:57:27 GMT

... That's neat! I had forgotten that we can have q[0] odd when p[0] is a square. Here is my check:

180 quadratically smooth below 10^5000 bits

Jaroslaw Wroblewski — Tue, 09 Feb 2010 07:58:26 GMT

Here is 180 below 10^5000 bits: AP180 is pp97^2 + 4*61#*i, for i=-133...46 where pp97 = 97# / 61# The exceptional congruences are: q[-95]=1 (mod 315) q[0]=1

Re: 158 quadratically smooth below 10^5000 bits ?

Jaroslaw Wroblewski — Tue, 09 Feb 2010 06:02:35 GMT

I do confirm it is OK according to my script (written independently in Mathematica) and the number of bits matches within the accuracy you gave. Note that

Re: 26 consecutive quadratically smooth numbers

Jaroslaw Wroblewski — Tue, 09 Feb 2010 05:19:37 GMT

2010/2/8 djbroadhurst ... My original script was set to check odd prime divisors in power at most 2. After modyfying it, I do confirm

158 quadratically smooth below 10^5000 bits ?

djbroadhurst — Tue, 09 Feb 2010 01:11:18 GMT

... It seems to me that we can get 158 consecutive quadratically smooth numbers below 10^5000 bits, by using an *almost* coprime AP158: 58*73#*i + 1 , for i =

Re: 26 consecutive quadratically smooth numbers

djbroadhurst — Mon, 08 Feb 2010 21:37:57 GMT

... My check gave .... [109, [47628, 2270]] 1.0560232981680036084416389573629934112 E78 bits Will you buy my factor of 3^5 in n-109 ? I know that it's

Re: 26 consecutive quadratically smooth numbers

djbroadhurst — Mon, 08 Feb 2010 09:38:11 GMT

... So Dmitry did help, after all :-) David

Re: 25 consecutive quadratically smooth numbers

djbroadhurst — Mon, 08 Feb 2010 09:16:01 GMT

... Ah, sorry, my fault! That was the infamous bug in using eulerphi to reduce a huge exponent in a PowerMod. The term p = 5161 in the AP is indeed divisible

26 consecutive quadratically smooth numbers

Jaroslaw Wroblewski — Mon, 08 Feb 2010 08:08:40 GMT

Here is a 26: AP26: 121+840*i for i=-9...16 p[-9]=7439; q[-9]=7440 (mod 14878) p[-8]=6599; q[-8]=6600 (mod 13198) p[-7]=5759; q[-7]=106 (mod 210) Note the

Re: 25 consecutive quadratically smooth numbers

Jaroslaw Wroblewski — Mon, 08 Feb 2010 05:43:30 GMT

In the first place, thank you, David, for setting up so nice and stimulating challenge. ... There is something fishy here. You claim that n+31 is divisible by

Re: 25 consecutive quadratically smooth numbers

djbroadhurst — Sun, 07 Feb 2010 23:33:41 GMT

... Mike Oakes is "hors de combat" at present, for personal reasons. When he returns, "Deo volente", I think that he may be as happy as am I that Jarek was so