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Square array read by antidiagonals: Hilbert transform of the Narayana numbers A001263.
5

%I #20 Jan 22 2018 21:24:24

%S 1,1,1,1,3,1,1,6,5,1,1,10,16,7,1,1,15,40,31,9,1,1,21,85,105,51,11,1,1,

%T 28,161,295,219,76,13,1,1,36,280,721,771,396,106,15,1,1,45,456,1582,

%U 2331,1681,650,141,17,1

%N Square array read by antidiagonals: Hilbert transform of the Narayana numbers A001263.

%C Refer to A145905 for the definition of the Hilbert transform of a lower triangular array. For the Hilbert transform of A008459, the array of type B Narayana numbers, see A108625.

%C This seems to be a duplicate of A273350. - _Alois P. Heinz_, Jun 04 2016. This could probably be proved by showing that the g.f.s are the same. - _N. J. A. Sloane_, Jul 02 2016

%F Row n generating function: 1/(n+1) * 1/(1-x) * Jacobi_P(n,1,1,(1+x)/(1-x)) = N_n(x)/(1-x)^n where N_n(x) denotes the shifted Narayana polynomial N_n(x) = sum{k = 1..n} A001263(k)*x^(k-1) of degree n-1.

%F Conjectural column n generating function: N_n(x^2)/(1-x)^(2n+1).

%F The entries in row n are given by the values of a polynomial function p_n(x) at x = 0,1,2,... . The first few are p_1(x) = 2x + 1, p_2(x) = (5x^2 + 5x + 2)/2, p_3(x) = (2x + 1)*(7x^2 + 7x + 6)/6 and p_4(x) = (7x^4 + 14x^3 + 21x^2 + 14x + 4)/4. These polynomials appear to have their zeros on the line Re x = -1/2; that is, the polynomials p_n(-x) appear to satisfy a Riemann hypothesis. The corresponding result for A108625 is true (see A142995 for details).

%F Contribution from _Paul Barry_, Jan 06 2009: (Start)

%F The g.f. for the corresponding number triangle is:

%F 1/(1-x-xy-x^2y/(1-x-x^2y/(1-x-xy-x^2y/(1-x-x^2y/(1-x-xy-x^2y.... (a continued fraction). (End)

%F This g.f. satisfies x^2*y*g^2 - (1-x-x*y)*g + 1 = 0. - _R. J. Mathar_, Jun 16 2016

%F G.f.: ((y-1)*sqrt(((x^2+2*x+1)*y-x^2+2*x-1)/(y-1))+(-x-1)*y-x+1)/(2*x^2*y). - _Vladimir Kruchinin_, Jan 15 2018

%F T(n,m) = 1/(n+1)*Sum_{i=0..m+1} C(n+1,i-1)*C(n+1,i)*C(n+m-i+1,m+1-i). - _Vladimir Kruchinin_, Jan 15 2018

%e The array begins

%e n\k|..0.....1.....2.....3.....4.....5

%e =====================================

%e 0..|..1.....1.....1.....1.....1.....1

%e 1..|..1.....3.....5.....7.....9....11

%e 2..|..1.....6....16....31....51....76

%e 3..|..1....10....40...105...219...396

%e 4..|..1....15....85...295...771..1681

%e 5..|..1....21...161...721..2331..6083

%e ...

%e Row 2: (1 + 3x + x^2)/(1 - x)^3 = 1 + 6x + 16x^2 + 31x^3 + ... .

%e Row 3: (1 + 6x + 6x^2 + x^3)/(1 - x)^4 = 1 + 10x + 40x^2 + 105x^3 + ... .

%t Table[1/(# + 1)*Sum[Binomial[# + 1, i - 1] Binomial[# + 1, i] Binomial[# + k - i + 1, k + 1 - i], {i, 0, k + 1}] &[m - k], {m, 0, 9}, {k, 0, m}] // Flatten (* _Michael De Vlieger_, Jan 15 2018 *)

%o (Maxima)

%o taylor(((y-1)*sqrt(((x^2+2*x+1)*y-x^2+2*x-1)/(y-1))+(-x-1)*y-x+1)/(2*x^2*y),x,0,10,y,0,10);

%o T(n,m,k):=1/(n+1)*sum(binomial(n+1,i-1)*binomial(n+1,i)*binomial(n+m-i+1,m+1-i),i,0,m+1); /* _Vladimir Kruchinin_, Jan 15 2018 */

%Y Cf. A001263, A005891 (row 2), A063490 (row 3), A108625 (Hilbert transform of h-vectors of type B associahedra).

%Y Cf. also A273350.

%K easy,nonn,tabl

%O 0,5

%A _Peter Bala_, Oct 31 2008