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%I #16 Aug 12 2017 12:58:12
%S 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,
%T 45,170,405,666,783,666,405,170,45,6,7,63,287,840,1736,2646,3039,2646,
%U 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
%N 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.
%H G. C. Greubel, <a href="/A264766/b264766.txt">Table of n, a(n) for the first 50 rows, flattened</a>
%F 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.
%F 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.
%F T(n,k) = T(n,2*n-k) for 0 <= k <= 2*n.
%F 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.
%F Recurrence: p(0,x) = 1, and p(n+1,x) = (1+x+x^2)*p(n,x)+(1+x)^(2*n+2), n >= 0.
%F 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).
%F 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.
%F T(n,n-1) = binomial(2*n+2,n) - A027907(n+1,n) for n > 0.
%F T(n+1,n)/(n+2) = A000108(n+2) - A001006(n+1) for n >= 0 (see also A058987).
%F Row sums: p(n,1) = A005061(n+1) for n >= 0.
%F Alternating row sums: p(n,-1) = 1 for n >= 0.
%F p(n,-2) = Sum_{k=0..2*n} T(n,k)*(-2)^k = A003462(n+1) for n >= 0.
%F T(n,k) = Sum_{j=0..k} (-1)^j*A260056(n,j)*binomial(2*n-j,k-j) for 0 <= k <= 2*n.
%F A260056(n,k) = Sum_{j=0..k} (-1)^j*T(n,j)*binomial(2*n-j,k-j) for 0 <= k <= 2*n.
%F p(n,-1-x) = Sum{k=0..2*n} A260056(n,k)*x^(2*n-k) for n >= 0.
%F p(n,-x/(1+x))*(1+x)^(2*n) = Sum_{k=0..2*n} A260056(n,k)*x^k for n >= 0.
%F Sum_{n>=0) p(n,x)*t^n = 1/((1-t*(1+x)^2)*(1-t*(1+x+x^2))).
%F p(n,x)*x = (1+x)^(2*n+2) - (1+x+x^2)^(n+1), n >= 0.
%F T(n,k) = binomial(2*n+2,k+1) - A027907(n+1,k+1) for 0 <= k <= 2*n.
%e The irregular triangle T(n,k) begins:
%e n\k: 0 1 2 3 4 5 6 7 8 9 10 11 12
%e 0: 1
%e 1: 2 3 2
%e 2: 3 9 13 9 3
%e 3: 4 18 40 51 40 18 4
%e 4: 5 30 90 165 201 165 90 30 5
%e 5: 6 45 170 405 666 783 666 405 170 45 6
%e 6: 7 63 287 840 1736 2646 3039 2646 1736 840 287 63 7
%e etc.
%e The polynomial corresponding to row 2 is p(2,x) = 3 + 9*x + 13*x^2 + 9*x^3 + 3*x^4.
%t 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 *)
%o (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));
%o tabf(nn) = for (n=0, nn, for (k=0, 2*n, print1(T(n, k), ", ");); print();); \\ _Michel Marcus_, Nov 24 2015
%Y Cf. A000108, A000984, A001006, A002426, A003462, A005061, A027907, A058987, A163774, A260056.
%K nonn,easy,tabf
%O 0,2
%A _Werner Schulte_, Nov 23 2015