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Smallest odd number k such that k*n*2^n - 1 is a prime number.
As N increases sum , (Sum_{k, n=1 to ..N} k) / sum (Sum_{n, n=1 to ..N} tends to n) approaches 0.833.
1*1*2^1 - 1 = unity, 3*1*2^1 - 1 = 5 , which is prime , so a(1) = 3.
1*2*2^2 - 1 = 7 , which is prime , so a(2) = 1.
1*3*2^3 - 1 = 23 , which is prime , so a(3) = 1.
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Conjecture: a(n) exists for every n.
Conjecture: a(n) exists for every n.The conjecture follows from Dirichlet's theorem on primes in arithmetic progressions. - Robert Israel, Jan 05 2016
Conjecture: a(n) exists for every n.The conjecture follows from Dirichlet's theorem on primes in arithmetic progressions. - _Robert Israel_, Jan 05 2016
Q:= proc(m) local k;
for k from 1 by 2 do if isprime(k*m-1) then return k fi od
end proc:
seq(Q(n*2^n), n=1..100); # Robert Israel, Jan 05 2016
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