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A185778
Second weight array of Pascal's triangle (formatted as a rectangle), by antidiagonals.
2
1, -1, -1, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 1, 2, 1, 0, 0, 0, 0, 1, 3, 3, 1, 0, 0, 0, 0, 1, 4, 6, 4, 1, 0, 0, 0, 0, 1, 5, 10, 10, 5, 1, 0, 0, 0, 0, 1, 6, 15, 20, 15, 6, 1, 0, 0, 0, 0, 1, 7, 21, 35, 35, 21, 7, 1, 0, 0, 0, 0, 1, 8, 28, 56, 70, 56, 28, 8, 1, 0, 0, 0, 0, 1, 9, 36, 84, 126, 126, 84, 36, 9, 1, 0, 0
OFFSET
1,5
COMMENTS
Using "->" to mean "is the weight array of" as defined at A144112:
A185779->A144225->A007318->A014430->A077023->A185779, where each of these is formatted as a rectangle (e.g., A007318 is Pascal's triangle). Read in reverse order, each is the accumulation array of the preceding array. It appears that successive weight arrays of A185779 contain Pascal's triangle except for initial terms.
FORMULA
(See the Mathematica code.)
EXAMPLE
Northwest corner:
1....-1....0....0....0....0....0,...0
-1....2....0....0....0....0....0....0
0.....0....0....1....1....1....1....1
0.....0....1....2....3....4....5....6
0.....0....1....3....6....10...15...21
0.....0....1....4....10...20...35...56
MATHEMATICA
(* This code produces three arrays: A144225, A007318, A185778. *)
f[n_, 0]:=0; f[0, k_]:=0; (* Used to make the weight array *)
f[1, 1]:=1; f[n_, 1]:=0; f[1, k_]:=0
f[n_, 2]:=1; f[2, k_]:=1;
f[n_, k_]:=-1+(n+k-4)!/((n-2)!*(k-2)!)/; k>1&&n>1;
TableForm[Table[f[n, k], {n, 1, 10}, {k, 1, 15}]] (* A144225 *)
s[n_, k_]:=Sum[f[i, j], {i, 1, n}, {j, 1, k}]; (* accumulation array of {f(n, k)} *)
TableForm[Table[s[n, k], {n, 1, 10}, {k, 1, 15}]] (* A007318, Pascal's triangle formatted as a rectangle *)
w[m_, n_]:=f[m, n]+f[m-1, n-1]-f[m, n-1]-f[m-1, n]/; Or[m>0, n>0];
TableForm[Table[w[n, k], {n, 1, 10}, {k, 1, 15}]] (* A185778 *)
Table[w[n-k+1, k], {n, 14}, {k, n, 1, -1}]//Flatten
KEYWORD
sign,tabl
AUTHOR
Clark Kimberling, Feb 03 2011
STATUS
approved