%I #45 Feb 16 2021 08:12:27
%S 1,1,1,1,1,1,1,1,2,1,1,1,3,5,1,1,1,4,11,14,1,1,1,5,19,45,42,1,1,1,6,
%T 29,100,197,132,1,1,1,7,41,185,562,903,429,1,1,1,8,55,306,1257,3304,
%U 4279,1430,1,1,1,9,71,469,2426,8925,20071,20793,4862,1
%N Square array of Narayana polynomials N_n evaluated at the integers, A(n,k) = N_n(k), n>=0, k>=0, read by antidiagonals.
%C Mirror image of A008550. - _Philippe Deléham_, Sep 26 2014
%H Seiichi Manyama, <a href="/A243631/b243631.txt">Antidiagonals n = 0..139, flattened</a>
%F T(n, k) = 2F1([1-n, -n], [2], k), 2F1 the hypergeometric function.
%F T(n, k) = P(n,1,-2*n-1,1-2*k)/(n+1), P the Jacobi polynomials.
%F T(n, k) = sum(j=0..n-1, binomial(n,j)^2*(n-j)/(n*(j+1))*k^j), for n>0.
%F For a recurrence see the second Maple program.
%F The o.g.f. of column n is gf(n) = 2/(sqrt((n-1)^2*x^2-2*(n+1)*x+1)+(n-1)*x+1). - _Peter Luschny_, Nov 17 2014
%F T(n, k) ~ (sqrt(k)+1)^(2*n+1)/(2*sqrt(Pi)*k^(3/4)*n^(3/2)). - _Peter Luschny_, Nov 17 2014
%F The n-th row can for n>=1 be computed by a linear recurrence, a(x) = sum(k=1..n, (-1)^(k+1)*binomial(n,k)*a(x-k)) with initial values a(k) = p(n,k) for k=0..n and p(n,x) = sum(j=0..n-1, binomial(n-1,j)*binomial(n,j)*x^j/(j+1)) (implemented in the fourth Maple script). - _Peter Luschny_, Nov 19 2014
%F (n+1) * T(n,k) = (k+1) * (2*n-1) * T(n-1,k) - (k-1)^2 * (n-2) * T(n-2,k) for n>1. - _Seiichi Manyama_, Aug 08 2020
%F Sum_{k=0..n} T(k, n-k) = Sum_{k=0..n} 2F1([-k, 1-k], [2], n-k) = A132745(n). - _G. C. Greubel_, Feb 16 2021
%e [0] [1] [2] [3] [4] [5] [6] [7]
%e [0] 1, 1, 1, 1, 1, 1, 1, 1
%e [1] 1, 1, 1, 1, 1, 1, 1, 1
%e [2] 1, 2, 3, 4, 5, 6, 7, 8 .. A000027
%e [3] 1, 5, 11, 19, 29, 41, 55, 71 .. A028387
%e [4] 1, 14, 45, 100, 185, 306, 469, 680 .. A090197
%e [5] 1, 42, 197, 562, 1257, 2426, 4237, 6882 .. A090198
%e [6] 1, 132, 903, 3304, 8925, 20076, 39907, 72528 .. A090199
%e [7] 1, 429, 4279, 20071, 65445, 171481, 387739, 788019 .. A090200
%e A000108, A001003, A007564, A059231, A078009, A078018, A081178
%e First few rows of the antidiagonal triangle are:
%e 1;
%e 1, 1;
%e 1, 1, 1;
%e 1, 1, 2, 1;
%e 1, 1, 3, 5, 1;
%e 1, 1, 4, 11, 14, 1;
%e 1, 1, 5, 19, 45, 42, 1; - _G. C. Greubel_, Feb 16 2021
%p # Computed with Narayana polynomials:
%p N := (n,k) -> binomial(n,k)^2*(n-k)/(n*(k+1));
%p A := (n,x) -> `if`(n=0, 1, add(N(n,k)*x^k, k=0..n-1));
%p seq(print(seq(A(n,k), k=0..7)), n=0..7);
%p # Computed by recurrence:
%p Prec := proc(n,N,k) option remember; local A,B,C,h;
%p if n = 0 then 1 elif n = 1 then 1+N+(1-N)*(1-2*k)
%p else h := 2*N-n; A := n*h*(1+N-n); C := n*(h+2)*(N-n);
%p B := (1+h-n)*(n*(1-2*k)*(1+h)+2*k*N*(1+N));
%p (B*Prec(n-1,N,k) - C*Prec(n-2,N,k))/A fi end:
%p T := (n, k) -> Prec(n,n,k)/(n+1);
%p seq(print(seq(T(n,k), k=0..7)), n=0..7);
%p # Array by o.g.f. of columns:
%p gf := n -> 2/(sqrt((n-1)^2*x^2-2*(n+1)*x+1)+(n-1)*x+1):
%p for n from 0 to 11 do PolynomialTools:-CoefficientList(convert( series(gf(n), x, 12), polynom), x) od; # _Peter Luschny_, Nov 17 2014
%p # Row n by linear recurrence:
%p rec := n -> a(x) = add((-1)^(k+1)*binomial(n,k)*a(x-k), k=1..n):
%p ini := n -> seq(a(k) = A(n,k), k=0..n): # for A see above
%p row := n -> gfun:-rectoproc({rec(n),ini(n)},a(x),list):
%p for n from 1 to 7 do row(n)(8) od; # _Peter Luschny_, Nov 19 2014
%t MatrixForm[Table[JacobiP[n,1,-2*n-1,1-2*x]/(n+1), {n,0,7},{x,0,7}]]
%t Table[Hypergeometric2F1[1-k, -k, 2, n-k], {n,0,12}, {k,0,n}]//Flatten (* _G. C. Greubel_, Feb 16 2021 *)
%o (Sage)
%o def NarayanaPolynomial():
%o R = PolynomialRing(ZZ, 'x')
%o D = [1]
%o h = 0
%o b = True
%o while True:
%o if b :
%o for k in range(h, 0, -1):
%o D[k] += x*D[k-1]
%o h += 1
%o yield R(expand(D[0]))
%o D.append(0)
%o else :
%o for k in range(0, h, 1):
%o D[k] += D[k+1]
%o b = not b
%o NP = NarayanaPolynomial()
%o for _ in range(8):
%o p = next(NP)
%o [p(k) for k in range(8)]
%o (Sage)
%o def A243631(n,k): return 1 if n==0 else sum( binomial(n,j)^2*k^j*(n-j)/(n*(j+1)) for j in [0..n-1])
%o flatten([[A243631(k,n-k) for k in [0..n]] for n in [0..12]]) # _G. C. Greubel_, Feb 16 2021
%o (Magma)
%o A243631:= func< n,k | n eq 0 select 1 else (&+[ Binomial(n,j)^2*k^j*(n-j)/(n*(j+1)): j in [0..n-1]]) >;
%o [A243631(k,n-k): k in [0..n], n in [0..12]]; // _G. C. Greubel_, Feb 16 2021
%Y Cf. A001263, A008550 (mirror), A204057 (another version), A242369 (main diagonal), A099169 (diagonal), A307883, A336727.
%Y Rows[2-7]: A000027, A028387, A090197, A090198, A090199, A090200.
%Y Columns[1-7]: A000108, A001003, A007564, A059231, A078009, A078018, A081178.
%Y Cf. A132745.
%K nonn,tabl
%O 0,9
%A _Peter Luschny_, Jun 08 2014