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911 results from messages in primenumbers

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  • ---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...
    david.broadhurst@... Nov 5, 2014
  • 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
    david.broadhurst@... Oct 30, 2014
  • ---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
    david.broadhurst@... Oct 30, 2014
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  • ---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...
    david.broadhurst@... Oct 8, 2014
  • ---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...
    david.broadhurst@... Oct 7, 2014
  • 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...
    david.broadhurst@... Sep 30, 2014
  • ---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
    david.broadhurst@... Sep 29, 2014
  • ---In primenumbers@^$1, wrote : > N - ( k^2 - l^2 ) = u .... [1] > (k^2 - m^2) - N = v .... [2] > Then logically > N = k^2 - (l - ( u / (u+v) ) )^2 .... [3] Simply bad logic: [1] defines u; [2] defines v; [3] cannot possibly be derived from [1] and [2], since it has, in general, a denominator (l^2-m^2)^2. For proper logic, see Fermat's factorization method - Wikipedia, the free...
    david.broadhurst@... Sep 28, 2014
  • Paul Underwood used up his monthly gremlin allowance with an 8-selfridge test that was very easy to fool: {ferfro(x,n)=gcd(x^3-x,n)==1&&kronecker(x^2-4,n)==-1&& Mod(2*x,n)^(n-1)==1&&Mod(Mod(1,n)*(L+x),L^2-x*L+1)^(n+1)==1+2*x^2;} {tst(n,x,y)=gcd(x^2-y^2,n)==1&&ferfro(x,n)&&ferfro(y,n);} {n=126256669511639877109; x=5031702531319296552; y=9025414833578502318; if(tst(n,x,y)&&!isprime(n...
    david.broadhurst@... Aug 4, 2014
  • ---In primenumbers@^$1, wrote : http://scribd.com/doc/168629586/Exact-Recurrence-Formula-for-Prime-Numbers > However, this formula cannot be used for direct > calculations due to rapidly growing huge precision > required. I like your use of a truncated Euler product and a truncated sum for zeta(4*prime(n)) to produce this impractical formula for prime(n+1) given all of the first n...
    david.broadhurst@... Jul 23, 2014