OFFSET
1,2
COMMENTS
Principal diagonal: A172073.
Antidiagonal sums: A002419.
Row 1, (1,2,3,4,5,...)**(1,4,7,10,13,...): A002411.
Row 2, (1,2,3,4,5,...)**(4,7,10,13,16,...): A077414.
Row 3, (1,2,3,4,5,...)**(7,10,13,16,...): (k^3 + 7*k^2 + 6*k)/2.
Row 4, (1,2,3,4,5,...)**(10,13,16,...): (k^3 + 10*k^2 + 9*k)/2.
For a guide to related arrays, see A212500.
LINKS
Clark Kimberling, Antidiagonals n = 1..45, flattened
FORMULA
T(n,k) = 4*T(n,k-1)-6*T(n,k-2)+4*T(n,k-3)-T(n,k-4).
G.f. for row n: f(x)/g(x), where f(x) = x*(3*n - 2 - (3*n - 5)*x) and g(x) = (1 - x)^4.
EXAMPLE
Northwest corner (the array is read by falling antidiagonals):
1....6....18...40....75....126
4....15...36...70....120...189
7....24...54...100...165...252
10...33...72...130...210...315
13...42...90...160...255...378
MATHEMATICA
b[n_]:=n; c[n_]:=3n-2;
t[n_, k_]:=Sum[b[k-i]c[n+i], {i, 0, k-1}]
TableForm[Table[t[n, k], {n, 1, 10}, {k, 1, 10}]]
Flatten[Table[t[n-k+1, k], {n, 12}, {k, n, 1, -1}]]
r[n_]:=Table[t[n, k], {k, 1, 60}] (* A213761 *)
Table[t[n, n], {n, 1, 40}] (* A172073 *)
s[n_]:=Sum[t[i, n+1-i], {i, 1, n}]
Table[s[n], {n, 1, 50}] (* A002419 *)
CROSSREFS
KEYWORD
AUTHOR
Clark Kimberling, Jul 04 2012
STATUS
approved