Search Results for ""

The O'Nan group is the sporadic group O'N of order |O'N| = 460815505920 (1) = 2^9·3^4·5·7^3·11·19·31. (2) It is implemented in the Wolfram Language as ONanGroupON[].

The evolute of Cayley's sextic with parametrization x = 4acos^3(1/3theta)cost (1) y = 4acos^3(1/3theta)sint (2) is given by x_e = 1/4[2+3cos(2/3t)-cos(2t)] (3) y_e = ...

E(a,b)/p denotes the elliptic group modulo p whose elements are 1 and infty together with the pairs of integers (x,y) with 0<=x,y<p satisfying y^2=x^3+ax+b (mod p) (1) with a ...

A number s of trials in which the probability of success p_i varies from trial to trial. Let x be the number of successes, then var(x)=spq-ssigma_p^2, (1) where sigma_p^2 is ...

If a function f has a pole at z_0, then the negative power part sum_(j=-k)^(-1)a_j(z-z_0)^j (1) of the Laurent series of f about z_0 sum_(j=-k)^inftya_j(z-z_0)^j (2) is ...

sum_(n=0)^(infty)(-1)^n[((2n-1)!!)/((2n)!!)]^3 = 1-(1/2)^3+((1·3)/(2·4))^3+... (1) = _3F_2(1/2,1/2,1/2; 1,1;-1) (2) = [_2F_1(1/4,1/4; 1;-1)]^2 (3) = ...

The sum of reciprocal multifactorials can be given in closed form by the beautiful formula m(n) = sum_(n=0)^(infty)1/(n!...!_()_(k)) (1) = ...

The areas of the regions illustrated above can be found from the equations A+4B+4C=1 (1) A+3B+2C=1/4pi. (2) Since we want to solve for three variables, we need a third ...

The number of binary bits necessary to represent a number, given explicitly by BL(n) = 1+|_lgn_| (1) = [lg(n+1)], (2) where [x] is the ceiling function, |_x_| is the floor ...

The prime HP(n) reached starting from a number n, concatenating its prime factors, and repeating until a prime is reached. For example, for n=9, 9=3·3->33=3·11->311, so 311 ...