a(n,m)  tabf head (staircase) for A116863 (C_2(n), quadratic Casimir polynomials for A_n)


  n=1, A_1=su(2):      [1,2]

  n=2, A_2=su(3):      [2,2,2,6,6]= 2*[1,1,1,3,3]

  n=3, A_3=su(4):      [3,4,2,4,4,3,12,16,12]
 
  n=4, A_4=su(5):      [4,6,4,2,6,8,4,6,6,4,20,30,30,20]= 2*[2,3,2,1,3,4,2,3,3,2,10,15,15,10]

  n=5, A_5=su(6):      [5,8,6,4,2,8,12,8,4,9,12,6,8,8,5,30,48,54,48,30]

  n=6, A_6=su(7):      [6,10,8,6,4,2,10,16,12,8,4,12,18,12,6,12,16,8,10,10,6,42,70,84,84,70,42]
                    =2*[3, 5,4,3,2,1, 5, 8, 6,4,2, 6, 9, 6,3, 6, 8,4, 5, 5,3,21,35,42,42,35,21]  

  n=7, A_7=su(8):      [7,12,10,8,6,4,2,12,20,16,12,8,4,15,24,18,12,6,16,24,16,8,15,20,10,12,12,7,56,96,120,128,120,96,56]  

  n=8, A_8=su(9):      [8,14,12,10,8,6,4,2,14,24,20,16,12,8,4,18,30,24,18,12,6,20,32,24,16,8,20,30,20,10,18,24,12,14,14,8,72,
                        126,162,180,180,162,126,72]
                    =2*[4, 7, 6, 5,4,3,2,1, 7,12,10, 8, 6,4,2, 9,15,12, 9, 6,3,10,16,12, 8,4,10,15,10, 5, 9,12, 6, 7, 7,4, 36,
                         63, 81, 90, 90, 81, 63,36]
 
  n=9, A_9=su(10):     [9,16,14,12,10,8,6,4,2,16,28,24,20,16,12,8,4,21,36,30,24,18,12,6,24,40,32,24,16,8,25,40,30,20,10,24,
                        36,24,12,21,28,14,16,16,9,90,160,210,240,250,240,210,160,90]


   n=10, A_10=su(11):  [10,18,16,14,12,10,8,6,4,2,18,32,28,24,20,16,12,8,4,24,42,36,30,24,18,12,6,28,48,40,32,24,16,8,30,50,
                        40,30,20,10,30,48,36,24,12,28,42,28,14,24,32,16,18,18,10,110,198,264,308,330,330,308,264,198,110]]
                    =2*[ 5, 9, 8, 7, 6, 5,4,3,2,1, 9,16,14,12,10, 8, 6,4,2,12,21,18,15,12, 9, 6,3,14,24,20,16,12, 8,4,15,25,
                        20,15,10, 5,15,24,18,12, 6,14,21,14, 7,12,16, 8, 9, 9, 5, 55, 99,132,154,165,165,154,132, 99, 55]

   etc.
  
  ###########################################################################################################################
 
  For each n the first  n*(n+1)/2= A000217(n) numbers belong to quadratic monomials in the ordering given below  and the last 
  n numbers belong to the linear monomials in rising index order.  
  The ordering of the quadratic monomials is as follows:
  12,13,14,..., 1n; 22,23,...,2n;33,34,...,3n;...;nn. 
  Here ij stands for a[i] * a[j].
  The last n numbers are the coefficients of a[1], a[2],...,a[n].

  ############################################################################################################################

  The corresponding polynomials are, 
  for example for A_4=su(5), with 2*[2,3,2,1,3,4,2,3,3,2,10,15,15,10] (in accordance with the above mentioned order),
  
   2*(2*a[1]^2+3*a[1]*a[2]+2*a[1]*a[3]+a[1]*a[4]+
                    +3*a[2]^2+4*a[2]*a[3]+2*a[2]*a[4]+
                                    +3*a[3]^2+3*a[3]*a[4]+
                                                  +2*a[4]^2
   +10*a[1]+15*a[2]+15*a[3]+10*a[4])


   or for  A_5=su(6) with  [5,8,6,4,2,8,12,8,4,9,12,6,8,8,5,30,48,54,48,30]: 

    5*a[1]^2+8*a[1]*a[2]+6*a[1]*a[3]+4*a[1]*a[4]+2*a[1]*a[5]+
                +8*a[2]^2+12*a[2]*a[3]+8*a[2]*a[4]+4*a[2]*a[5]+
                              +9*a[3]^2+12*a[3]*a[4]+6*a[3]*a[5]+
                                             +8*a[4]^2+8*a[4]*a[5]+
                                                           +5*a[5]^2+
    +30*a[1]+48*a[2]+54*a[3]+48*a[4]+30*a[5]
  
  ############################################################################################################################

  The sum of the coefficients for n=1..10 are [3, 18, 60, 150, 315, 588, 1008, 1620, 2475, 3630]=
                                           =3*[1, 6, 20, 50, 105, 196, 336, 540, 825, 1210],  
  
  and the conjecture for the general case is  3*(n+1)^2*((n+1)^2-1)/12 = 3*A002415(n+1).
 
  ################################################ eof ########################################################################