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601 - 610 of 2617 for Prime Generating PolynomialSearch Results
Honaker's problem asks for all consecutive prime number triples (p,q,r) with p<q<r such that p|(qr+1). Caldwell and Cheng (2005) showed that the only Honaker triplets for ...
Legendre's formula counts the number of positive integers less than or equal to a number x which are not divisible by any of the first a primes, (1) where |_x_| is the floor ...
p^x is an infinitary divisor of p^y (with y>0) if p^x|_(y-1)p^y, where d|_kn denotes a k-ary Divisor (Guy 1994, p. 54). Infinitary divisors therefore generalize the concept ...
Given an arithmetic progression of terms an+b, for n=1, 2, ..., the series contains an infinite number of primes if a and b are relatively prime, i.e., (a,b)=1. This result ...
Fermat's 4n+1 theorem, sometimes called Fermat's two-square theorem or simply "Fermat's theorem," states that a prime number p can be represented in an essentially unique ...
A generalization of Fermat's little theorem. Euler published a proof of the following more general theorem in 1736. Let phi(n) denote the totient function. Then a^(phi(n))=1 ...
Let a_n>=0 and suppose sum_(n=1)^inftya_ne^(-an)∼1/a as a->0^+. Then sum_(n<=x)a_n∼x as x->infty. This theorem is a step in the proof of the prime number theorem, but has ...
Let U(P,Q) and V(P,Q) be Lucas sequences generated by P and Q, and define D=P^2-4Q. (1) Then {U_((n-(D/n))/2)=0 (mod n) when (Q/n)=1; V_((n-(D/n))/2)=D (mod n) when (Q/n)=-1, ...
Let p be prime and r = r_mp^m+...+r_1p+r_0 (0<=r_i<p) (1) k = k_mp^m+...+k_1p+k_0 (0<=k_i<p), (2) then (r; k)=product_(i=0)^m(r_i; k_i) (mod p). (3) This is proved in Fine ...
Let q be a positive integer, then Gamma_0(q) is defined as the set of all matrices [a b; c d] in the modular group Gamma Gamma with c=0 (mod q). Gamma_0(q) is a subgroup of ...
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