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Special notice only
• ### Re: limit for a serie

0.2614972128476427837554268386086958590515666482612 ---In primenumbers@^\$1, wrote : Sum[1/Prime[i],{i,1,n]-Log[Log[n]]
• ### Re: Concealed Property of the Larger Prime of a TPM

> 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...
• ### Re: Primitive part definition?

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...
• ### Re: [PrimeNumbers] Helfgott's improvement of Erathostenes' sieve

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
• ### Re: Connection between a Prime, its Number of Digits Plus One, its Sum of digits

> 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...
• ### Iteration of x^2 --> x + a mod p

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...
• ### Re: n-->n^2+a

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...