Euler Project 27

Euler published the remarkable quadratic formula:

n² + n + 41

It turns out that the formula will produce 40 primes for the consecutive values n = 0 to 39. However, when n = 40, 40² + 40 + 41 = 40(40 + 1) + 41 is divisible by 41, and certainly when n = 41, 41² + 41 + 41 is clearly divisible by 41.

Using computers, the incredible formula  n²  79n + 1601 was discovered, which produces 80 primes for the consecutive values n = 0 to 79. The product of the coefficients, 79 and 1601, is 126479.

Considering quadratics of the form:

n² + an + b, where |a|  1000 and |b|  1000

where |n| is the modulus/absolute value of n
e.g. |11| = 11 and |4| = 4

Find the product of the coefficients, a and b, for the quadratic expression that produces the maximum number of primes for consecutive values of n, starting with n = 0.

My Solution Actually, we can improve the effect, because the b should be prime, but I just loop if from 1 to 999.

(defun  primep (n)

(cond ((minusp n) 'nil)

     ((= 2 n) t)

     ((evenp n) 'nil)

              ((= 1 n) 'nil)

              (t (let ((limit (floor (sqrt n))))

                      (do ((i 3 (incf i 2)))

                          ((> i limit) t)

                          (if (= 0 (mod n i)) (return-from primep 'nil)))))))

(defun ConsecutivePrime(i j)

  (let ((num 0))

    (loop for k from 0 to 1000

 do(if (null (primep (+ (* k k) (+ j (* i k)))))

       (return num)

(incf num)))

   num))

(defun Euler27()

  (let ((currenta 0)

         (currentb 0)

         (maxnum 0))

    (loop for a from -998 to 999

        do(loop for b from 1 to 999

             do(let ((currentaccount (ConsecutivePrime a b)))

                      (if (progn (> currentaccount maxnum))

                          (progn (setq currenta a)

                                    (setq currentb b)

                                    (setq maxnum currentaccount))))))

    (* currentb currenta)))