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  • 0.2614972128476427837554268386086958590515666482612 ---In primenumbers@^$1, wrote : Sum[1/Prime[i],{i,1,n]-Log[Log[n]]
    david.broadhurst@... Aug 18
  • > the number J of consecutive primes from the > Mth prime to the (2*M)th prime inclusive, > will always equal M+1 This is tautologically true for every positive integer M. Proof: By definition, there are precisely 2*M primes that are less than or equal to the (2*M)th prime. Precisely (M-1) of these are less than the Mth prime. Hence, by definition, J = 2*M - (M-1) = M+1. > I think...
    david.broadhurst@... Dec 3, 2016
  • Since you seem badly confused, I have written a consistent and detailed reply. Chris Caldwell's The Prime Glossary: primitive part The Prime Glossary: primitive part Welcome to the Prime Glossary: a collection of definitions, information and facts all related to prime numbers. This pages contains the entry titled 'primitive ... View on primes.utm.edu Preview by Yahoo agrees with my...
    david.broadhurst@... Dec 2, 2016
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  • I imagine that the S.A. article was a plug for Helfgott's seminar on November 2, after which there may be more informed comment. Georg-August-Universität Göttingen - Göttingen-Hannover Number Theory Workshops Georg-August-Universität Göttingen - Göttingen-Han... Webseiten der Georg-August-Universität Göttingen View on www.uni-goettingen.de Preview by Yahoo
    david.broadhurst@... Oct 17, 2016
  • > I think the following statement is true. > no prime integer P > 11 exists, such that its > SL equals its SR and that both are prime. Proof: Assume that P > 9, that SL = SR and that SL is prime. Then 2*SL = (Nd+1)*Sd. Since SL is prime and Nd+1 > 2, we conclude that Sd = 2 and Nd+1 = SL. Hence P = 10^(Nd-1)+1 and Nd+1 is an odd prime. Since Nd-1 is odd, P is divisible by 11. But P...
    david.broadhurst@... Oct 17, 2016
  • Iteration of x^2 --> x + a mod p Let a be an integer, p be a prime, and consider iterations of the map x --> x^2 + a, acting on integers, or on their residues modulo p. If, for some initial integer x, iteration generates an infinite chain of integers coprime to p, then say that p "complies" with a. If not, then say that p "obstructs" a. It follows that p obstructs a if and only if...
    david.broadhurst@... Sep 30, 2016
  • Update: I believe that p=9557113 is an obstacle for James' case a=88. In this case, the longest chains of iterates coprime to p have length 13906. There are 4 such chains, starting with n=+/-1475332 mod p and n=+/-2388843 mod p. No prime p<10^7 is an obstacle for James' case a=54. ---In primenumbers@^$1, wrote : By my reckoning, the next value with no obstacle at primes less than...
    david.broadhurst@... Sep 26, 2016
  • By my reckoning, the next value with no obstacle at primes less than 10^6 is a=190.
    david.broadhurst@... Sep 26, 2016
  • PS: This conundrum was resolved by a physicist, Wendell Furry, in 1942: http://www.nature.com/nature/journal/v150/n3795/abs/150120a0.html Unfortunately the paper is restricted to subscribers to Nature. His method involves polylogarithmic integrals. By using dilogarithms, the asymptotic discrepancy was reduced to 5 parts per million. Higher-order trilogarithmic corrections are...
    david.broadhurst@... Aug 12, 2016
  • Kermit suggested: > The approximate number of primes between m^2 and (m+1)^2 is > (2m+1) * product over all positive primes, p, less than or equal > to m of (p-1)/p No. That is Chebyschev's conundrum. As m tends to infinity, the left hand side is asymptotic to m/log(m), by the prime number theorem, while the right hand side is asymptotic to C*m/log(m) with C=2*exp(-Euler) = 1...
    david.broadhurst@... Aug 12, 2016