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## 911 results from messages in primenumbers

• ### Re: Two Interesting Pairs of Odd Positive Primes

---In primenumbers@^\$1, wrote : > Consider 4 numbers of primes (L1,L2,L3,L4) > where L2=(L1+1)/2 and L3=(L2+1)/2 > and L3=(L2+1)/2 and L4=(L3+1)/2 and L1, L2, L3, L4 > are the largest member of a twin prime pair There is a long history of studying such "bi-twin" chains: http://en.wikipedia.org/wiki/Bi-twin_chain http://www.primenumbers.net/Henri/fr-us/BiTwinRec.htm http://mathworld...
• ### Re: Two Interesting Pairs of Odd Positive Primes

w_sindelar@^\$1... wrote: > I tried the opposite of the above, where P is the smallest member > of a twin prime pair and M is also the smallest member of a twin > member of a twin prime pair. It does not work. Obviously not. If P is prime, P+2 is prime and P > 5, then M = (P+1)/2 is divisible by 3 and greater than 3. Hence M cannot be prime. David
• ### Re: Another extention Pitago number

---In primenumbers@^\$1, wrote : > (x+y)^2=v(v+u)+z(z+t) > (z+t)^2=u(u+v)+y(y+x) > (u+v)^2=t(t+z)+x(x+y) The first two equations imply the third. This Diophantine system of two equations in 6 variables does not seem to be closely related to the subject of this list, namely prime numbers. But perhaps I am mistaken? David
• ### Re: old theorem; proof of new idea!

---In primenumbers@^\$1, wrote : > if 'm' is from the set of natural numbers, then every odd > prime divisor 'd' of c^(b^m) +/- 1 implies that > q == +/-1 (mod b^(m+1)) Here there is no definition of q. If it is intended that q = d, then the claim is clearly wrong. For example: 2^(3^5)-1 is divisible by 7, but neither 6 nor 8 is divisible by a power of 3 greater than 3^2; 2^(3^5)+1...
• ### Re: On reducibility of 1+x^2+x^3+x^5+...+x^p

---In primenumbers@^\$1, wrote : > I'm conjecturing that the reducibility (over integers) of > the above only holds when the number of primes congruent to > 1 and 3 modulo 4 meet For odd primes, the "when" is trivial to prove: simply set x=I. The "only" seems hard, to me. Here's a test for odd primes p < 10^3: {n1=n3=0;z=1+x^2; forprime(p=3,10^3,z+=x^p;m=p%4; n1+=m==1;n3+=m==3;pol=z...
• ### Re: First polynomial time algorithm to output Ramanujan's "highly composite numbers."

Warren replied off list and then asked me to post his reply. Here it is: >> I have not implemented my HCN algorithm, I merely proved it works (and even that, sketchily; there are some parts I've oversimplified). So, I do not know how fast it will be in practice. However... whatever Flammenkamp's techniques are, they could be hybridized with my rigorous techniques, resulting in a...
• ### Re: First polynomial time algorithm to output Ramanujan's "highly composite numbers."

---In primenumbers@^\$1, wrote : > ....there are algorithms to output > the first K of Ramanujan's "highly composite numbers" (HCNs) > ... in time bounded by polynomial(K). How long does it take, with K=779674, to check Achim Flammenkamp's 10-year-old list? http://wwwhomes.uni-bielefeld.de/achim/highly.txt David