const multipliers = divisors(3*5*7*11).grep { _ > 1 } func sff(N) { N.is_prime && return 1 # n is prime N.is_square && return N.isqrt # n is square multipliers.each {|k| var P0 = isqrt(k*N) # P[0]=floor(sqrt(N) var Q0 = 1 # Q[0]=1 var Q = (k*N - P0*P0) # Q[1]=N-P[0]^2 & Q[i] var P1 = P0 # P[i-1] = P[0] var Q1 = Q0 # Q[i-1] = Q[0] var P = 0 # P[i] var Qn = 0 # P[i+1] var b = 0 # b[i] while (!Q.is_square) { # until Q[i] is a perfect square b = idiv(isqrt(k*N) + P1, Q) # floor(floor(sqrt(N+P[i-1])/Q[i]) P = (b*Q - P1) # P[i]=b*Q[i]-P[i-1] Qn = (Q1 + b*(P1 - P)) # Q[i+1]=Q[i-1]+b(P[i-1]-P[i]) (Q1, Q, P1) = (Q, Qn, P) } b = idiv(isqrt(k*N) + P, Q) # b=floor((floor(sqrt(N)+P[i])/Q[0]) P1 = (b*Q0 - P) # P[i-1]=b*Q[0]-P[i] Q = (k*N - P1*P1)/Q0 # Q[1]=(N-P[0]^2)/Q[0] & Q[i] Q1 = Q0 # Q[i-1] = Q[0] loop { b = idiv(isqrt(k*N) + P1, Q) # b=floor(floor(sqrt(N)+P[i-1])/Q[i]) P = (b*Q - P1) # P[i]=b*Q[i]-P[i-1] Qn = (Q1 + b*(P1 - P)) # Q[i+1]=Q[i-1]+b(P[i-1]-P[i]) break if (P == P1) # until P[i+1]=P[i] (Q1, Q, P1) = (Q, Qn, P) } with (gcd(N,P)) {|g| return g if g.is_ntf(N) } } return 0 } [ 11111, 2501, 12851, 13289, 75301, 120787, 967009, 997417, 4558849, 7091569, 13290059, 42854447, 223553581, 2027651281, 11111111111, 100895598169, 1002742628021, 60012462237239, 287129523414791, 11111111111111111, 384307168202281507, 1000000000000000127, 9007199254740931, 922337203685477563, 314159265358979323, 1152921505680588799, 658812288346769681, 419244183493398773, 1537228672809128917 ].each {|n| var v = sff(n) if (v == 0) { say "The number #{n} is not factored." } elsif (v == 1) { say "The number #{n} is a prime." } else { say "#{n} = #{[n/v, v].sort.join(' * ')}" } }