I'm trying to write a program that returns the Pell numbers sequence based on a given number.

For example (pellNumb 6) should return a list (0 1 2 5 12 29 70)

This is my code so far.
I am able of calculating the numbers, but I am not able of skipping the double recursion.

(defun base (n)

  (if (= n 0)

      0

      (if (= n 1) 

          1))) 



(defun pellNumb (n)

  (if (or (= n 0) (= n 1))

      (base n)

      (let ((x (pellNumb (- n 2))))

        (setq y (+ (* 2 (pellNumb (- n 1))) x))

        (print y))))

The output for (pellNumb 4) is 2 2 5 12, and this is because i'm recursing to (pellNumb 2) twice.

Is there a way to skip that, and store these values in a list ?

Thanks!

Get the nth number

Yes, there is a way - use multiple values:

(defun pell-numbers (n)

  "Return the n-th Pell number, n-1 number is returned as the 2nd value.

See https://oeis.org/A000129, https://en.wikipedia.org/wiki/Pell_number"

  (check-type n (integer 0))

  (cond ((= n 0) (values 0 0))

        ((= n 1) (values 1 0))

        (t (multiple-value-bind (prev prev-1) (pell-numbers (1- n))

             (values (+ (* 2 prev) prev-1)

                     prev)))))

(pell-numbers 10)

==> 2378 ; 985

This is a standard trick for recursive sequences which depend on several previous values, such as the Fibonacci.

Performance

Note that your double recursion means that (pell-numbers n) has exponential(!) performance (computation requires O(2^n) time), while my single recursion is linear (i.e., O(n)).
Moreover, Fibonacci numbers have a convenient property which allows a logarithmic recursive implementation, i.e., taking O(log(n)) time.

Get all the numbers up to n in a list

If you need all numbers up to the nth, you need a simple loop:

(defun pell-numbers-loop (n)

  (loop repeat n

    for cur = 1 then (+ (* 2 cur) prev)

    and prev = 0 then cur

    collect cur))

(pell-numbers-loop 10)

==> (1 2 5 12 29 70 169 408 985 2378)

If you insist on recursion:

(defun pell-numbers-recursive (n)

  (labels ((pnr (n)

             (cond ((= n 0) (list 0))

                   ((= n 1) (list 1 0))

                   (t (let ((prev (pnr (1- n))))

                        (cons (+ (* 2 (first prev)) (second prev))

                              prev))))))

    (nreverse (pnr n))))

(pell-numbers-recursive 10)

==> (0 1 2 5 12 29 70 169 408 985 2378)

Note that the recursion is non-tail, so the loop version is probably more efficient.

One can, of course, produce a tail recursive version:

(defun pell-numbers-tail (n)

  (labels ((pnt (i prev)

             (if (= i 0)

                 prev ; done

                 (pnt (1- i)

                      (cond ((null prev) (list 0)) ; n=0

                            ((null (cdr prev)) (cons 1 prev)) ; n=1

                            (t

                             (cons (+ (* 2 (or (first prev) 1))

                                      (or (second prev) 0))

                                   prev)))))))

    (nreverse (pnt (1+ n) ()))))

(pell-numbers-tail 10)

==> (0 1 2 5 12 29 70 169 408 985 2378)