... By my reckoning, these are record-holders: [0.586422053204725619783563934076247530509997851734, 8997] ...
22474
Kermit Rose
kermit1941
Jan 23, 2011 4:29 pm
Hello I've blindly implemented elliptic curve factoring algorithm, and applied it to factoring numbers of form 2**k - 1. It seems to have gotten stuck at k =...
22475
Kermit Rose
kermit1941
Jan 23, 2011 10:04 pm
For what primes p, is 2 a cubic residue?...
22476
djbroadhurst
Jan 23, 2011 10:21 pm
... See http://oeis.org/A040028 and the list in http://oeis.org/A040028/b040028.txt David...
22477
Brian
ballbt
Jan 24, 2011 12:25 am
Very interesting results. For a random integer in the range of 2^63 to 2^64 I get identical results with BPSW & (Frobenius using method B*) 115,788 nanoseconds...
22478
djbroadhurst
Jan 24, 2011 12:43 am
... Pari-GP's "ispseudoprime" implements BPSW as follows: 1) trial division: if a factor, then composite, else 2) Fermat test: if fails, then composite, else ...
22479
Di Maria Giovanni
calimero22
Jan 24, 2011 3:42 am
Hi All does it exist a formula to count immediately the bits 0 and 1 in a number, without to compute them manually? Example: 1001011101 ('1' are 6, '0' are 4)....
22480
Kevin Acres
codefinda
Jan 24, 2011 3:58 am
Just confirming David's figures for k=47135 and 66204: V[47135] = 0.58642205320472558747016559381071589001859987395296 V[66204] =...
22481
djbroadhurst
Jan 24, 2011 5:11 am
... After k = 832586, I obtained these: [0.586422053204725587309002972404045019568721890267, 1248879] [0.586422053204725587309002972404045018326967898140,...
22482
mikeoakes2
Jan 24, 2011 1:06 pm
... The best introduction is this book:- "Rational Points on Elliptic Curves", by Joseph Silverman and John Tate, 281 pp, Springer 1992 Available new for...
22483
djbroadhurst
Jan 24, 2011 3:38 pm
... It's fairly easy to cope with k < 1G. Barring rogues, ! [0.586422053204725587309002398411001918298379628392, 957981916] is the 85th record-holder. The next...
22484
djbroadhurst
Jan 24, 2011 5:12 pm
... With 64-bit linux, the Pari-GP limit for the magnitude of a floating point number is 2^(2^61), i.e. about 694,127,911,065,419,642 decimal digits: ...
22485
Phil
philiplouisb...
Jan 24, 2011 7:12 pm
Search for "Hamming weight" at Wikipedia or elsewhere....
22486
Phil
philiplouisb...
Jan 24, 2011 7:17 pm
Pollard rho is sufficient to factor 2^93-1 = 7 * 2147483647 * 658812288653553079. Explanations and sample code for several factorization algorithms, including...
22487
Kermit Rose
kermit1941
Jan 24, 2011 7:54 pm
Hello Giovanni. Here is the algorithm that I would have used to count bits. Suppose you wish to count the bits in positive integer z. Presume that z is a copy...
22488
whygee@...
yasep16
Jan 24, 2011 8:06 pm
On Mon, 24 Jan 2011 14:54:30 -0500, Kermit Rose <kermit@...> ... don't you have something more ... smart ? what you all are speaking about is the old...
22489
Di Maria Giovanni
calimero22
Jan 24, 2011 8:27 pm
Hello Kermitthank you very much.Your code works fine, but.... it's no immediate. I's an algorithm tha ends when 'z' will be 0.Thank you in every...
22490
djbroadhurst
Jan 24, 2011 9:50 pm
... You have written down the obvious identity 2^93-1 = (2^3-1) * (2^31-1) * ((2^93-1)/(2^3-1)/(2^31-1)) Why do we need Mr Pollard to tell us that? David...
22491
Mark
marku606
Jan 25, 2011 2:45 am
Here's a the first part of a news article about it. Mark ScienceDaily (Jan. 24, 2011) — For centuries, some of the greatest names in math have tried to make...
22492
djbroadhurst
Jan 25, 2011 3:13 am
... http://www.aimath.org/news/partition/brunier-ono gives an "amusing" derivation of p(1)=1. David...
22493
Dimiter Skordev
dskordev
Jan 25, 2011 7:55 am
Hi, Some expressions which can be regarded as formulas can be written for the count you mention, but applying them needs approximately the same amount of...
22494
Dimiter Skordev
dskordev
Jan 25, 2011 8:23 am
I am sorry for the typo - of course, it must be "greatest"; instead of "greates"! Dimiter...
22495
Mark
marku606
Jan 25, 2011 3:05 pm
... Now, why didn't I think of that? Wait, the answer has dawned on me ......
22496
andrew_j_walker
Jan 27, 2011 6:53 am
http://www-graphics.stanford.edu/~seander/bithacks.html has this and much more! :-) Andrew...
22497
Sebastian Martin Ruiz
s_m_ruiz
Jan 29, 2011 10:49 pm
Hello all: Let's consider the triplet of prime numbers: Tn,k = {p (n+k), pn, p (n-k)} k <n positive integers. We will say that Tn,k is symmetric if p(n+k) +p...
In fact that would follow if the Goldbach conjecture is true, or at least a stronger version of it: "Any even positive integer greater than 6 is the sum of at...
22500
Jaroslaw Wroblewski
jarek372000
Jan 31, 2011 6:18 am
Here is another prime 17-tuplet (and a new 17 Simultaneous Primes record): 597173610793158380251873601 + d, d = 0, 6, 8, 12, 18, 20, 26, 32, 36, 38, 42, 48,...
22501
Kermit Rose
kermit1941
Jan 31, 2011 8:45 pm
I've implemented the algebraic factoring routine for getting the algebraic factors of the difference of two powers, where one power is significantly larger...
22502
djbroadhurst
Jan 31, 2011 9:27 pm
... Exercise: Find the algebraic factors of 16^137-1. Comment: The maximum number of digits in any such factor is 42, so if your routine produces a larger ...
