OFFSET
1,5
COMMENTS
Array with recursion T(i,j) = T(i-1,j) + T(i,j-1), and boundaries T(0,n) = T(n,0) = a(n). Here a(n) is the array T read by antidiagonals. Require that a(0)=a(1)=1.
LINKS
Alex Meiburg, Table of n, a(n) for n = 1..19999
EXAMPLE
The array looks like
1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 2, ...
1, 2, 3, 4, 6, 7, 8, 11, 14, 15, 17, ...
1, 3, 6, 10, 16, 23, 31, 42, 56, 71, 88, ...
1, 4, 10, 20, 36, 59, 90, 132, 188, 259, 347, ...
2, 6, 16, 36, 72, 131, 221, 353, 541, 800, ...
1, 7, 23, 59, 131, 262, 483, 836, 1377, ...
1, 8, 31, 90, 221, 483, 966, 1802, ...
3, 11, 42, 132, 353, 836, 1802, ...
3, 14, 56, 188, 541, 1377, ...
1, 15, 71, 259, 800, ...
2, 17, 88, 347, ...
... [Table corrected and reformatted by Jon E. Schoenfield, Jan 14 2018]
The defining property is that when this array is read by antidiagonals we get 1,1,1,1,2,1,... which is both the sequence itself and the top row and first column of the array.
MATHEMATICA
t[a_, b_] := (t[a, b] = t[a, b - 1] + t[a - 1, b]);
t[0, x_] := a[x]; t[x_, 0] := a[x];
a[0] = 1; a[1] = 1;
a[x_] := With[{k = Floor[(Sqrt[8 x + 1] - 1)/2]},
t[x - k (k + 1)/2, (k + 1) (k + 2)/2 - x - 1]]
a /@ Range[60]
TableForm[ Table[t[i, j], {i, 0, 5}, {j, 0, 12}]]
CROSSREFS
KEYWORD
AUTHOR
Alex Meiburg, Dec 29 2017
STATUS
approved