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Square array of Narayana polynomials N_n evaluated at the integers, A(n,k) = N_n(k), n>=0, k>=0, read by antidiagonals.
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%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