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URL: https://oeis.org/A264766

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A264766
Irregular symmetric triangle of coefficients T(n,k) of the polynomials p(n,x) = Sum_{k=0..n} binomial(n+1,k)*(1+x)^(2*k)*(-x)^(n-k) for 0 <= k <= 2*n.
1
1, 2, 3, 2, 3, 9, 13, 9, 3, 4, 18, 40, 51, 40, 18, 4, 5, 30, 90, 165, 201, 165, 90, 30, 5, 6, 45, 170, 405, 666, 783, 666, 405, 170, 45, 6, 7, 63, 287, 840, 1736, 2646, 3039, 2646, 1736, 840, 287, 63, 7, 8, 84, 448, 1554, 3864, 7224, 10424, 11763, 10424, 7224, 3864, 1554, 448, 84, 8, 9, 108, 660, 2646, 7686, 17010, 29520, 40851, 45481, 40851, 29520, 17010, 7686, 2646, 660, 108, 9, 10
OFFSET
0,2
FORMULA
T(n,k) = Sum_{j=0..n-d} (-1)^j*binomial(n+1,j+1)*binomial(2*n-2*j,k-j) if d = 0 or better d = abs(k-n), and 0 <= k <= 2*n.
Recurrence: 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-2) + T(n,k-1) + T(n,k) + binomial(2*n+2,k) for k > 0 and n >= 0.
T(n,k) = T(n,2*n-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)^(2*k)*(1+x+x^2)^(n-k) = Sum_{k=0..n} binomial(n+1,k)*(1+x+x^2)^k*x^(n-k) for n >= 0.
Recurrence: p(0,x) = 1, and p(n+1,x) = (1+x+x^2)*p(n,x)+(1+x)^(2*n+2), n >= 0.
T(n,n) = Sum_{j=0..n} (-1)^(n-j)*binomial(n+1,j)*binomial(2*j,j) = A000984(n+1)-A002426(n+1) for n >= 0 (see also A163774).
Sum_{n>=0} T(n,n)*x^(n+1) = 1/sqrt(1-4*x) - 1/sqrt(1-2*x-3*x^2) for abs(x) < 1/4.
T(n,n-1) = binomial(2*n+2,n) - A027907(n+1,n) for n > 0.
T(n+1,n)/(n+2) = A000108(n+2) - A001006(n+1) for n >= 0 (see also A058987).
Row sums: p(n,1) = A005061(n+1) for n >= 0.
Alternating row sums: p(n,-1) = 1 for n >= 0.
p(n,-2) = Sum_{k=0..2*n} T(n,k)*(-2)^k = A003462(n+1) for n >= 0.
T(n,k) = Sum_{j=0..k} (-1)^j*A260056(n,j)*binomial(2*n-j,k-j) for 0 <= k <= 2*n.
A260056(n,k) = Sum_{j=0..k} (-1)^j*T(n,j)*binomial(2*n-j,k-j) for 0 <= k <= 2*n.
p(n,-1-x) = Sum{k=0..2*n} A260056(n,k)*x^(2*n-k) for n >= 0.
p(n,-x/(1+x))*(1+x)^(2*n) = Sum_{k=0..2*n} A260056(n,k)*x^k for n >= 0.
Sum_{n>=0} p(n,x)*t^n = 1/((1-t*(1+x)^2)*(1-t*(1+x+x^2))).
p(n,x)*x = (1+x)^(2*n+2) - (1+x+x^2)^(n+1), n >= 0.
T(n,k) = binomial(2*n+2,k+1) - A027907(n+1,k+1) for 0 <= k <= 2*n.
EXAMPLE
The irregular triangle T(n,k) begins:
n\k: 0 1 2 3 4 5 6 7 8 9 10 11 12
0: 1
1: 2 3 2
2: 3 9 13 9 3
3: 4 18 40 51 40 18 4
4: 5 30 90 165 201 165 90 30 5
5: 6 45 170 405 666 783 666 405 170 45 6
6: 7 63 287 840 1736 2646 3039 2646 1736 840 287 63 7
etc.
The polynomial corresponding to row 2 is p(2,x) = 3 + 9*x + 13*x^2 + 9*x^3 + 3*x^4.
MATHEMATICA
T[n_, k_] := Sum[(-1)^j*Binomial[n + 1, j + 1]*Binomial[2*n - 2*j, k - j], {j, 0, n - Abs[k - n]}]; Table[T[n, k], {n, 0, 10}, {k, 0, 2*n}] // Flatten (* G. C. Greubel, Aug 12 2017 *)
PROG
(PARI) T(n, k) = sum(j=0, n-abs(k-n), (-1)^j*binomial(n+1, j+1)*binomial(2*n-2*j, k-j));
tabf(nn) = for (n=0, nn, for (k=0, 2*n, print1(T(n, k), ", "); ); print(); ); \\ Michel Marcus, Nov 24 2015
KEYWORD
nonn,easy,tabf
AUTHOR
Werner Schulte, Nov 23 2015
STATUS
approved