a(N,k)  tabf head (staircase) for  A092078  W. Lang, Mar 19, 2004
 
   N   n,m   |k: 1   2   3   4   5   6   7   8   9  10  11  12  13  14  15  16  17  18  19  20  21  22  

   1   1,1   |   1
   
   2   2,1   |   1
   3   2,2   |   1 | 0
   
   4   3,1   |   1
   5   3,2   |   0 | 1
   6   3,3   |   1 | 0 | 0
   
   7   4,1   |   1
   8   4,2   |   1 | 1
   9   4,3   |   0 | 1 | 0
  10   4,4   |   1 | 0   0 | 0 | 0
  
  11   5,1   |   1
  12   5,2   |   0 | 2
  13   5,3   |   0 | 2 | 0
  14   5,4   |   0 | 1   0 | 0 | 0
  15   5,5   |   1 | 0   0 | 0   0 | 0 | 0
    
  16   6,1   |   1
  17   6,2   |   1 | 2
  18   6,3   |   1 | 1 | 1
  19   6,4   |   0 | 1   1 | 0 | 0
  20   6,5   |   0 | 1   0 | 0   0 | 0 | 0
  21   6,6   |   1 | 0   0   0 | 0   0   0 | 0   0 | 0   0
               
  22   7,1   |   1
  23   7,2   |   0 | 3
  24   7,3   |   0 | 3 | 1
  25   7,4   |   0 | 2   0 | 1 | 0
  26   7,5   |   0 | 1   1 | 0   0 | 0 | 0
  27   7,6   |   0 | 1   0   0 | 0   0   0 | 0   0 | 0 | 0
  28   7,7   |   1 | 0   0   0 | 0   0   0   0 | 0   0   0 | 0  0 | 0 | 0

  29   8,1   |   1
  30   8,2   |   1 | 3
  31   8,3   |   0 | 3 | 2
  32   8,4   |   1 | 1   1 | 2 | 0
  33   8,5   |   0 | 1   1 | 1   0 | 0 | 0
  34   8,6   |   0 | 1   1   0 | 0   0   0 | 0   0 | 0 | 0
  35   8,7   |   0 | 1   0   0 | 0   0   0   0 | 0   0   0 | 0   0 | 0 | 0
  36   8,8   |   1 | 0   0   0   0 | 0   0   0   0   0 | 0   0   0   0   0 | 0   0   0 | 0   0 | 0 | 0

  etc.

   N   n,m   |k: 1   2   3   4   5   6   7   8   9  10  11  12  13  14  15  16  17  18  19  20  21  22 
 

  Cf. Abramowitz-Stegun, pp. 831-2

  Example: a(18,1) = a(18,2) = a(18,3) = 1: All three partitions of 3 appear once as exponents of partitions of n=6 into m=3 parts. 
           
           Namely: (3) in [2^3], (1,2) in [1^2,4^1], and (1,1,1) in [1^1,2^1,3^1].

  Proposition: binomial(n-1,m-1) = sum(a(N(n,m),k)*m!/product((m^{(k)}_i)!),k=1..p(n,m)), where p(n,m) is the number of partitions of 
  n into m parts, and the product is over the parts of the k-th partition of m in the Abramowitz-Stegun order (see the A036040 W. Lang
  link).
  Proof: From the multifactorial representation of the Lah numbers A008297 (which constitute a Jabotinsky triangle (cf. A039692)) 
  and their explicit formula. 

  Examples for the Proposition: n=5, m=3 (N=13):  2*3!/(1!*2!) = 6 = binomial(4,2). 
                                n=6, m=2 (N=17):  1*(2!/2!) + 2*(2!/(1!*2!)) = 1 + 4 = 5 = binomial(5,1).