OFFSET
1,2
COMMENTS
First two terms in row n: n,(n-1)^2; last term: 1.
Period of alternating row sums: (1,1,0).
For a discussion and guide to related arrays, see A208510.
Apparently this is A071920 without the marginal zeros, read by downwards antidiagonals, or T(n,k) = A071922(n,k). - R. J. Mathar, May 17 2014
LINKS
Andrew Howroyd, Table of n, a(n) for n = 1..1275 (rows 1..50)
FORMULA
u(n,x) = x*u(n-1,x) + v(n-1,x) + 1, v(n,x) = x*u(n-1,x) + (x+1)*v(n-1,x) + 1, where u(1,x)=1, v(1,x)=1.
T(n,k) = Sum_{j=0..floor(k/2)} binomial(k,2*j)*binomial(n-j,k). - Detlef Meya, Dec 05 2023
EXAMPLE
First five rows:
1
2...1
3...4....1
4...9....7....1
5...16...22...11...1
First three polynomials u(n,x): 1, 2 + x, 3 + 4x + x^2.
MATHEMATICA
u[1, x_] := 1; v[1, x_] := 1; z = 16;
u[n_, x_] := x*u[n - 1, x] + v[n - 1, x] + 1;
v[n_, x_] := x*u[n - 1, x] + (x + 1)*v[n - 1, x] + 1;
Table[Expand[u[n, x]], {n, 1, z/2}]
Table[Expand[v[n, x]], {n, 1, z/2}]
cu = Table[CoefficientList[u[n, x], x], {n, 1, z}];
TableForm[cu]
Flatten[%] (* A210219 *)
Table[Expand[v[n, x]], {n, 1, z}]
cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
TableForm[cv]
Flatten[%] (* A210220 *)
(* alternative program *)
T[n_, k_] := Sum[Binomial[k, 2*j]*Binomial[n-j, k], {j, 0, Floor[k/2]}]; Flatten[Table[T[n, k], {n, 1, 11}, {k, 1, n}]] (* Detlef Meya, Dec 05 2023 *)
PROG
(PARI) T(n, k) = sum(j=0, k\2, binomial(k, 2*j)*binomial(n-j, k)) \\ Andrew Howroyd, Jan 01 2024
CROSSREFS
KEYWORD
nonn,tabl
AUTHOR
Clark Kimberling, Mar 19 2012
STATUS
approved