Dirichlet's theorem on arithmetic progressions

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← Previous revision Revision as of 10:29, 1 May 2026 Line 54: Line 54: |12''n'' + 11 || 11, 23, 47, 59, 71, 83, 107, 131, 167, 179, ... || {{OEIS link|id=A068231}} |12''n'' + 11 || 11, 23, 47, 59, 71, 83, 107, 131, 167, 179, ... || {{OEIS link|id=A068231}} |} |} We can generate some forms of primes by using an iterative method. For example, we can generate primes of the form {{nowrap|4''n'' + 3}} by using the following method: We can generate some forms of primes by using an [[iterative method]]. For example, we can generate primes of the form {{nowrap|4''n'' + 3}} by using the following method:

Let {{nowrap|1=''a''₀ = 4(1) + 3 = 7}}. Then we let {{nowrap|1=''a''₁ = 4''a''₀ + 3 = 4(7) + 3 = 31}}, which is prime. We continue by computing {{nowrap|1=4(7)(31) + 3 = 871 = 13(67)}}. Because {{nowrap|1=4(7)(31) + 3}} is of the form {{nowrap|4''n'' + 3}}, either 13 or 67 is of the form {{nowrap|4''n'' + 3}}. We have that {{nowrap|1=67 = 4(16) + 3}} and is prime, so {{nowrap|1=''a''₃ = 67}}. We then continue this process to find successive primes of the form {{nowrap|4''n'' + 3}} (Silverman 2013). Let {{nowrap|1=''a''₀ = 4(1) + 3 = 7}}. Then we let {{nowrap|1=''a''₁ = 4''a''₀ + 3 = 4(7) + 3 = 31}}, which is prime. We continue by computing {{nowrap|1=4(7)(31) + 3 = 871 = 13(67)}}. Because {{nowrap|1=4(7)(31) + 3}} is of the form {{nowrap|4''n'' + 3}}, either 13 or 67 is of the form {{nowrap|4''n'' + 3}}. We have that {{nowrap|1=67 = 4(16) + 3}} and is prime, so {{nowrap|1=''a''₃ = 67}}. We then continue this process to find successive primes of the form {{nowrap|4''n'' + 3}} (Silverman 2013). Line 74: Line 74: contains a proportion 1/(''q'' − 1) of the primes. contains a proportion 1/(''q'' − 1) of the primes.

When compared to each other, progressions with a quadratic nonresidue remainder have typically slightly more elements than those with a quadratic residue remainder ([[Chebyshev's bias]]). When compared to each other, progressions with a quadratic nonresidue remainder have typically slightly more elements than those with a [[quadratic residue]] remainder ([[Chebyshev's bias]]).

== History == == History ==