a(n,m)  tabf head (staircase) for A115201
  
  Number of even parts of partitions of n in Abramowitz-Stegun (A-St) order.
 
  A permutation is odd, resp. even, if the partition corresponding to its cycle structure has an odd, resp. even number of even parts.
 
  Therefore, the permutation (123)(4)(5), corresponding to the partition (1,1,3), is even and belongs to A_5, the (invariant) subgroup 

  of the symmetric group S_5, called the alternating group A_5. 


  n\m   1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42

 
  1     0  

  2     1  0 

  3     0  1  0

  4     1  0  2  1  0

  5     0  1  1  0  2  1  0

  6     1  0  2  0  1  1  3  0  2  1  0

  7     0  1  1  1  0  2  0  2  1  1  3  0  2  1  0  

  8     1  0  2  0  2  1  1  1  3  1  0  2  0  2  4  1  1  3  0  2  1  0 

  9     0  1  1  1  1  0  2  0  2  2  2  0  1  1  1  3  1  3  0  2  0  2  4  1  1  3  0  2  1  0

 10     1  0  2  0  2  0  1  1  1  1  3  1  3  1  0  2  0  2  2  2  0  4  2  1  1  1  3  1  3  5  0  2  0  2  4  1  1  3  0  2  1  0

   .
   .
   .
       
  n\m   1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42



 The sequence of row lengths is p(n):=A000041(n) (partition numbers).

 One could add the row for n=0 with a 1, if the part 0 is considered for n=0, and only for this n.

 For the ordering of this tabf array a(n,m) see Abramowitz-Stegun ref. pp. 831-2. 

 E.g. a(4,4) refers to the fourth partition of n=4 in this ordering, namely (1^2,2)=[1,1,2], whence a(4,4)=1 because the number of 

 even parts is 1.  a(5,4)=1 because the partition (1,2^2)=(1,2,2) has an even number, namely 2, of even parts.



 The sequence of row sums give A066898: [0, 1, 1, 4, 5, 11, 15, 28, 38, 62,...].
  


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