OFFSET
0,2
COMMENTS
The triangle is related to the triangle of trinomial coefficients.
LINKS
G. C. Greubel, Table of n, a(n) for the first 100 rows, flattened
FORMULA
T(n,0) = n+1, and T(n,k) = 0 for k < 0 or k > 2*n, and T(n+1,k) = T(n,k) + T(n,k-1) + T(n,k-2) for k > 0.
T(n,k) = Sum_{j=0..n} A027907(j,k) for 0 <= k <= 2*n.
T(n,k) = Sum_{j=0..k} (-1)^(k-j)*A027907(n+1,j+1) for 0 <= k <= 2*n.
T(n,k) = T(n,2*n-1-k) + (-1)^k for 0 <= k < 2*n.
p(n,x) = Sum_{k=0..2*n} T(n,k)*x^k = Sum_{k=0..n} (1+x+x^2)^k for n >= 0.
p(n,x) = ((1+x+x^2)^(n+1)-1)/(x+x^2), p(n,0) = p(n,-1) = n+1 for n >= 0.
p(n+1,x) = (1+x+x^2)*p(n,x)+1 for n >= 0.
Sum_{n>=0} p(n,x)*t^n = 1/((1-t)*(1-t*(1+x+x^2))).
T(n,2*n) = 1, and T(n,n) = A113682(n) for n >= 0.
If d(n) is n-th antidiagonal sum of the triangle then: d(n) = A008937(n+1), and d(n+2)-d(n) = A001590(n+5) for n >= 0.
Conjecture: If a(n) is n-th antidiagonal alternating sum of the triangle then: a(n) = A004524(n+3).
Sum_{k=0..2*n} (-1)^k*T(n,k)^2 = (3^(n+1)-1)/2 for n >= 0.
Sum_{k=0..2*n} (-1)^k*(y*k+1)*T(n,k) = Sum{k=0..n} y*k+1 = (n+1)*(y*n+2)/2 for real y and n >= 0.
Conjecture of linear recurrence for column k: Sum_{m=0..k+2} (-1)^m*T(n+m,k)* binomial(k+2,m) = 0 for k >= 0 and n >= 0.
EXAMPLE
The irregular triangle T(n,k) begins:
n\k: 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 ...
0 1;
1 2 1 1;
2 3 3 4 2 1;
3 4 6 10 9 7 3 1;
4 5 10 20 25 26 19 11 4 1;
5 6 15 35 55 71 70 56 34 16 5 1;
6 7 21 56 105 161 196 197 160 106 55 22 6 1;
7 8 28 84 182 322 462 554 553 463 321 183 83 29 7 1;
etc.
The polynomial corresponding to row 2 is p(2,x) = 3+3*x+4*x^2+2*x^3+x^4.
MATHEMATICA
A027907[n_, k_] := Sum[Binomial[n, j]*Binomial[j, k - j], {j, 0, n}]; Table[ Sum[A027907[j, k], {j, 0, n}], {n, 0, 10}, {k, 0, 2*n} ] // Flatten (* G. C. Greubel, Mar 07 2017 *)
CROSSREFS
KEYWORD
nonn,easy,tabf
AUTHOR
Werner Schulte, Nov 08 2015
STATUS
approved