data ex(drop= i j);
array  num(9) _temporary_  (1,2,3,4,5,6,7,8,9);
array n(9) $;
do i=1 to dim(num);
   do j=1 to i;
       n(j)=cats(num(j),'*',num(i),'=',num(j)*num(i));
   end;
   output;
end;
run;

data _null_;
array  num(9) _temporary_  (1,2,3,4,5,6,7,8,9);
length str $200;
do i=dim(num) to 1 by -1;
        call missing(str);
    do j=1 to i;
        str=cat(strip(str)," ",cats(num(j),'*',num(i),'=',num(j)*num(i)));
    end;
        put str;
end;
run;

谢谢您的回复，我好好研究一下数组和输出结构!

谢谢指导，不过您这个输出的结果好像不是下三角的

这是非数组的方法。
DATA  D9;
DO  I=1 TO 9;
  DO  J=1 TO I;
    X=I*J;
    PUT @(3*J) X @;
  /* @(expression) moves the pointer to the column */
  /* that is given by the value of expression.     */
  END;
  PUT;
/* The null PUT statement releases an output line that is */
/* being held by a previous PUT statement with a trailing @. */
END;
RUN;