[seqfan] Re: Confused about A144204

[franktaw at netscape.net]

Antidiagonal is a term relating to tables of numbers in the quarter 
plane.  I don't know that the term is currently used in any context 
other than the OEIS.

For example, taking the function n^k, we have an array that starts like:

1 1  1  1   1
0 1  2  3   4
0 1  4  9  16
0 1  8 27  64
0 1 16 64 256

extending infinitely down and to the right.

The diagonals of this sequence start at either edge, and continue 
infinitely down and to the right.  So the main diagonal is n^n: 
1,1,4,27,256,....  Other diagonals include n^(n+1), (n+1)^n, etc.

The antidiagonals are the diagonals in the other direction: down and to 
the left (or up and to the right, but the preferred enumeration in the 
OEIS is down and to the left).  These are always finite, and 
enumerating the antidiagonals enumerates the entire table:

1; 1,0; 1,1,0; 1,2,1,0; 1,3,4,1,0; ...

(In fact, this is A003992.)

Franklin T. Adams-Watters


-----Original Message-----
From: Alonso Del Arte <alonso.delarte at gmail.com>

I was thinking about the triangle determined by a(n, k) = nk - (n +
k), so I had Mathematica calculate it to ten rows with this code:
a[n_, k_] := a[n, k] = n*k - (n + k); ColumnForm[Table[a[n, k], {n,
10}, {k, n}], Center]

        -1
        -1, 0
        -1, 1, 3
        -1, 2, 5, 8
        -1, 3, 7, 11, 15
        -1, 4, 9, 14, 19, 24
        -1, 5, 11, 17, 23, 29, 35
        -1, 6, 13, 20, 27, 34, 41, 48
        -1, 7, 15, 23, 31, 39, 47, 55, 63
        -1, 8, 17, 26, 35, 44, 53, 62, 71, 80

I looked up the fifth row and A144204 came up as a result. But the
definition for that sequence is "Arises in complete intersection
threefolds, Array A[k,n] = (n+k-2)*(n-1) - 1 by antidiagonals." I'm
not exactly sure how the formula (n + k - 2)(n - 1) - 1 gives the
triangle quoted above. Just plugging it into the code above gives a
somewhat different triangle. I do admit that I'm confused by
"antidiagonal," which should be a fairly basic term.


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