 (define (e-euler k) 
   (+ 2.0 (cont-frac (lambda (i) 1) 
                     (lambda (i) 
                       (if (= (remainder i 3) 2) 
                           (/ (+ i 1) 1.5) 
                           1)) 
                     k)))

 ;Alternative solution. 
  
 ; There is a repeating pattern with a cycle of 3 in the sequence of values of d. 
 ; The first value is 1, the second a power of 2 and the third is again a 1 
 ; To find out the relative place of a value in a cycle take the index of the  
 ; value in the sequence modulus 3.       
 ; Further note the second value within a cycle goes up from 2 to 2+2 to 2+2+2 ... 
  
 (define (d i) ;  
   (define (d-iter  value times-two j) 
     (let ((j-mod-3 (modulo j 3))) 
       (if (> j  i) 
           value 
           (d-iter 
            (cond ((= j-mod-3 1) 1) 
                  ((= j-mod-3 2) times-two) 
                  ((= j-mod-3 0) 1)) 
            (if (= j-mod-3  2) 
                (+ times-two 2) 
                times-two) 
            (+ j 1))))) 
   (d-iter 0 2 1))         
  
 (define (euler-e)   ;multiplication by 1.0 forces fraction to real 
   (* 1.0 (+ (finite-cont-frac n d 100) 2))) 
  
 (define (enum f n)  ;displays the function values upto and including n 
   (define (enum-iter list-of-values i) 
     (if (= i  n) 
         (display (reverse list-of-values)) 
         (enum-iter (cons (f i) list-of-values) (+ i 1)))) 
   (enum-iter '() 1))

sicp-ex-1.38