A103365 -id:A103365 - cp's OEIS frontend

A115369 Decimal expansion of first zero of BesselJ(1,z).

3, 8, 3, 1, 7, 0, 5, 9, 7, 0, 2, 0, 7, 5, 1, 2, 3, 1, 5, 6, 1, 4, 4, 3, 5, 8, 8, 6, 3, 0, 8, 1, 6, 0, 7, 6, 6, 5, 6, 4, 5, 4, 5, 2, 7, 4, 2, 8, 7, 8, 0, 1, 9, 2, 8, 7, 6, 2, 2, 9, 8, 9, 8, 9, 9, 1, 8, 8, 3, 9, 3, 0, 9, 5, 1, 9, 0, 1, 1, 4, 7, 0, 2, 1, 4, 1, 1, 2, 8, 7, 4, 7, 5, 7, 4, 2, 3, 1, 2, 6, 7, 2, 4, 4, 7

Offset: 1

Eric W. Weisstein, Jan 21 2006

Also the first root of the sinc(2,x) function, that is, the radial component of the 2D Fourier transform of a 2-dimensional unit disc. - Stanislav Sykora, Nov 14 2013

Also the first root of the derivative of BesselJ_0. - Jean-François Alcover, Jul 01 2015

			3.8317059702075123156...

Stanislav Sykora, K-Space Images of n-Dimensional Spheres and Generalized Sinc Functions
Eric Weisstein's World of Mathematics, Bessel Function Zeros
Index entries for transcendental numbers

Cf. A115368, A115370, A115371, A115372, A115373.

Cf. A103365, A103366, A238390, A180874.

BesselJZero[1, 1] // N[#, 105]& // RealDigits // First (* Jean-François Alcover, Feb 06 2013 *)

solve(x=3,4,besselj(1,x)) \\ Charles R Greathouse IV, Feb 19 2014

besseljzero(1) \\ Charles R Greathouse IV, Aug 06 2022

A103364 Matrix inverse of the Narayana triangle A001263.

Original entry on oeis.org

1, -1, 1, 2, -3, 1, -7, 12, -6, 1, 39, -70, 40, -10, 1, -321, 585, -350, 100, -15, 1, 3681, -6741, 4095, -1225, 210, -21, 1, -56197, 103068, -62916, 19110, -3430, 392, -28, 1, 1102571, -2023092, 1236816, -377496, 68796, -8232, 672, -36, 1, -27036487, 49615695, -30346380, 9276120, -1698732, 206388

Offset: 1

Paul D. Hanna, Feb 02 2005

The first column is A103365. The second column is A103366. Row sums are all zeros (for n > 1). Absolute row sums form A103367.

Let E(y) = Sum_{n >= 0} y^n/(n!*(n+1)!) = (1/sqrt(y))*BesselI(1,2*sqrt(y)). Then this triangle is the generalized Riordan array (1/E(y), y) with respect to the sequence n!*(n+1)! as defined in Wang and Wang. - Peter Bala, Aug 07 2013

			Rows begin:
        1;
       -1,        1;
        2,       -3,       1;
       -7,       12,      -6,       1;
       39,      -70,      40,     -10,     1;
     -321,      585,    -350,     100,   -15,     1;
     3681,    -6741,    4095,   -1225,   210,   -21,   1;
   -56197,   103068,  -62916,   19110, -3430,   392, -28,   1;
  1102571, -2023092, 1236816, -377496, 68796, -8232, 672, -36, 1;
  ...
From _Peter Bala_, Aug 09 2013: (Start)
The real zeros of the row polynomials R(n,x) appear to converge to zeros of E(alpha*x) as n increases, where alpha = -3.67049 26605 ... ( = -(A115369/2)^2).
Polynomial | Real zeros to 5 decimal places
= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
R(5,x)     | 1, 3.57754, 3.81904
R(10,x)    | 1, 3.35230, 7.07532,  9.14395
R(15,x)    | 1, 3.35231, 7.04943, 12.09668, 15.96334
R(20,x)    | 1, 3.35231, 7.04943, 12.09107, 18.47845, 24.35255
= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Function   |
= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
E(alpha*x) | 1, 3.35231, 7.04943, 12.09107, 18.47720, 26.20778, ...
Note: The n-th zero of E(alpha*x) may be calculated in Maple 17 using the instruction evalf( BesselJZeros(1,n)/ BesselJZeros(1,1))^2 ). (End)

G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened
W. Wang and T. Wang, Generalized Riordan array, Discrete Mathematics, Vol. 308, No. 24, 6466-6500.

Cf. A000108, A001263, A055133, A086646, A103365, A103366, A103367, A104033.

T[n_, 1]:= Last[Table[(-1)^(n - 1)*(CoefficientList[Series[x/BesselJ[1, 2*x], {x, 0, 1000}], x])[[k]]*(n)!*(n - 1)!, {k, 1, 2*n - 1, 2}]]
T[n_, n_] := 1; T[2, 1] := -1; T[3, 1] := 2; T[n_, k_] := T[n, k] = T[n - 1, k - 1]*n*(n - 1)/(k*(k - 1)); Table[T[n, k], {n, 1, 50}, {k, 1, n}] // Flatten (* G. C. Greubel, Jan 04 2016 *)
T[n_, n_] := 1; T[n_, 1]/;n>1 := T[n, 1] = -Sum[T[n, j], {j, 2, n}]; T[n_, k_]/;1Oliver Seipel, Jan 01 2025 *)

PARI
```
T(n,k)=if(n
				
```

From Peter Bala, Aug 07 2013: (Start)
Let E(y) = Sum_{n >= 0} y^n/(n!*(n+1)!) = (1/sqrt(y))* BesselI(1,2*sqrt(y)). Generating function: E(x*y)/E(y) = 1 + (-1 + x)*y/(1!*2!) + (2 - 3*x + x^2)*y^2/(2!*3!) + (-7 + 12*x - 6*x^2 + x^3)*y^3/(3!*4!) + .... The n-th power of this array has a generating function E(x*y)/E(y)^n. In particular, the matrix inverse A001263 has a generating function E(y)*E(x*y).
Recurrence equation for the row polynomials: R(n,x) = x^n - Sum_{k = 0..n-1} 1/(n-k+1)*binomial(n,k)*binomial(n+1,k+1) *R(k,x) with initial value R(0,x) = 1.
Let alpha denote the root of E(x) = 0 that is smallest in absolute magnitude. Numerically, alpha = -3.67049 26605 ... = -(A115369/2)^2. It appears that for arbitrary complex x we have lim_{n->oo} R(n,x)/R(n,0) = E(alpha*x). Cf. A055133, A086646 and A104033.
A stronger result than pointwise convergence may hold: the convergence may be uniform on compact subsets of the complex plane. This would explain the observation that the real zeros of the polynomials R(n,x) seem to converge to the zeros of E(alpha*x) as n increases. Some numerical examples are given below. (End)
From Werner Schulte, Jan 04 2017, corrected May 05 2025: (Start)
T(n,k) = T(n-1,k-1)*n*(n-1)/(k*(k-1)) for 1 < k <= n;
T(n,k) = T(n+1-k,1)*A001263(n,k) for 1 <= k <= n;
Sum_{k=1..n} T(n,k)*A000108(k) = 1 for n > 0. (End)

A238390 E.g.f.: x / BesselJ(1, 2*x) (even powers only).

Original entry on oeis.org

1, 1, 4, 35, 546, 13482, 485892, 24108513, 1576676530, 131451399794, 13609184032808, 1712978776719938, 257612765775847132, 45620136452519144700, 9396239458048330569840, 2227147531572856811691105, 601916577165056911293330930, 183994483721828524163677628370

Offset: 0

Vaclav Kotesovec, Mar 01 2014

nonn

Aerated, the e.g.f. is e^(a.t) =  1/AC(i*t) = 1/[I_1(2i*t)/(it)] = 1/Sum_{n>=0} (-1)^n t^(2n) / [n!(n+1)!] = a_0 + a_2 t^2/2! + a_4 t^4/4! + ... = 1 + t^2/2! + 4 t^4/4! + 35 t^6/6! + ..., where AC(t) is the e.g.f. for the aerated Catalan numbers c_n of A126120 and I_n(t) are the modified Bessel functions of the first kind (i = sqrt(-1)). The signed, aerated sequence b_n = (i)^n a_n has the e.g.f. e^(b.t) = 1/AC(t) and, therefore, (i*a. + c.)^n = Sum_{k=0..n} binomial(n,k) i^k a_k c_(n-k) vanishes except for n=0 for which it's unity. - Tom Copeland, Jan 23 2016
With q(n) = A126120(n+1) and q(0) = 0, d(2n) =  (-1)^n A238390(n) and zero for odd arguments, and r(2n+1) = (-1)^n A180874(n+1) and zero for even arguments, then r(n) = (q. + d.)^n = Sum_{k=0..n} binomial(n,k) q(k) d(n-k), relating these sequences (and A000108) through binomial convolutions. Then also, (r. + c. + d.)^n = r(n). See A180874 for proofs and for relations to A097610. For quick reference, q = (0, 1, 0, 2, 0, 5, 0, 14, ..), d = (1, 0, -1, 0, 4, 0, -35, 0, ..), and r = (0, 1, 0, -1, 0, 5, 0, -56, ..). - Tom Copeland, Jan 28 2016
Aerated and signed, this sequence contains the moments m(n) of the Appell polynomial sequence UMT(n,h1,h2) that is the umbral compositional inverse of the Appell sequence of Motzkin polynomials MT(n,h1,h2) of A097610 with exp[x UMT(.,h1,h2)] = e^(x*h1) / AC(x*y) where y = sqrt(h2) and AC is defined above. UMT(n,h1,h2) = (m.y + h1)^n with (m.)^(2n) = m(2n) = (-1)^n A238390(n) and zero otherwise. Consequently, the associated lower triangular matrices A007318(n,k)*m(n-k) and A007318(n,k)*A126120(n-k) form an inverse pair (cf. also A133314), and MT(n,UMT(.,h1,h2),h2) = h1^n = UMT(n,MT(.,h1,h2),h2). - Tom Copeland, Jan 30 2016

G. C. Greubel, Table of n, a(n) for n = 0..98

Cf. A103365, A180874, A115369, A000275, A002190.

Cf. A000108, A007318, A097610, A126120, A133314.

S:= series(x/BesselJ(1,2*x),x,102):
seq((2*j)!*coeff(S,x,2*j),j=0..50); # Robert Israel, Jan 31 2016

Table[(CoefficientList[Series[x/BesselJ[1, 2*x], {x, 0, 40}], x] * Range[0, 40]!)[[n]], {n, 1, 41, 2}]

a(n) ~ c * (n!)^2 / (sqrt(n) * r^n), where r = BesselJZero[1, 1]^2/16 = 0.91762316513274332857623611, and c = 1/(Sqrt[Pi]*BesselJ[2, BesselJZero[1, 1]]) = 1.4008104828035425937394082168... - Vaclav Kotesovec, Mar 01 2014, updated Apr 01 2018

A103366 Second column of triangle A103364, which equals the matrix inverse of the Narayana triangle (A001263).

Original entry on oeis.org

1, -3, 12, -70, 585, -6741, 103068, -2023092, 49615695, -1487006785, 53477384268, -2272859942574, 112699083156751, -6447858833644515, 421601806346674320, -31242589674606301224, 2604618355967848204587, -242686626407684239621113, 25124942798118355122058980

Offset: 2

Paul D. Hanna, Feb 02 2005

sign

Vaclav Kotesovec, Table of n, a(n) for n = 2..200

Cf. A001263, A103364, A103365, A103367, A115369.

a(n)=if(n<2,0,(matrix(n,n,m,j,binomial(m-1,j-1)*binomial(m,j-1)/j)^-1)[n,2])

a(n) ~ (-1)^n * 2^(2*n-5) * n! * (n-1)! / (BesselJ(2, BesselJZero(1, 1)) * BesselJZero(1, 1)^(2*n-4)), where BesselJZero(1, 1) = A115369. - Vaclav Kotesovec, Jul 11 2025

A155926 G.f. satisfies: A(x) = B(xA(x)) where A(x) = Sum_{n>=0} a(n)x^n/[n!(n+1)!/2^n] and B(x) = Sum_{n>=0} x^n/[n!(n+1)!/2^n].

Original entry on oeis.org

1, 1, 4, 37, 621, 16526, 640207, 34039027, 2379382609, 211619306134, 23337543447296, 3125553148981176, 499716551101393705, 94016487294245251308, 20561796731966531616954, 5172827581575899147920471

Offset: 0

Paul D. Hanna, Jan 30 2009

nonn

			G.f.: A(x) = 1 + x + 4*x^2/3 + 37*x^3/18 + 621*x^4/180 + 16526*x^5/2700 +...+ a(n)*x^n/[n!*(n+1)!/2^n] +...
B(x) = 1 + x + 1/3*x^2 + 1/18*x^3 + 1/180*x^4 +...+ x^n/[n!*(n+1)!/2^n] +... where
A(x) = B(x*A(x)) and B(x) = A(x/B(x)) ;
1/B(x) = 1 - x + 2*x^2/3 - 7*x^3/18 + 39*x^4/180 - 321*x^5/2700 +...+ (-1)^n*A103365(n)*x^n/[n!*(n+1)!/2^n] +...
Also, A(x) = C(x*A(x)^2) where:
C(x) = 1 + x - 2*x^2/3 + 19*x^3/18 - 379*x^4/180 + 12726*x^5/2700 +...+ A155927(n)*x^(n+1)/[n!*(n+1)!/2^n] +...
A(x)^2 = 1 + 2*x + 11*x^2/3 + 122*x^3/18 + 2302*x^4/180 + 66482*x^5/2700 +...

Cf. A105558, A105556, A001263, A103365, A006472, A090181, A155927, A155928 (A^2).

{a(n)=local(F=sum(k=0,n,x^k/(k!*(k+1)!/2^k))+x*O(x^n));polcoeff(serreverse(x/F)/x,n)*n!*(n+1)!/2^n}

{a(n)=local(N=matrix(n+1, n+1, m, j, if(m>=j, binomial(m-1, j-1)*binomial(m, j-1)/j))); sum(j=0, n, (N^n)[n+1, j+1])/(n+1)}

a(n) = A105558(n)/(n+1) = A105556(2n,n)/(n+1) = [N^(n+1)](n+1,1)/(n+1) for n>=0, where N^(n+1) is the (n+1)-th matrix power of the Narayana triangle N=A001263.
G.f.: A(x) = Series_Reversion[x/B(x)]/x where B(x) = A(x/B(x)) = Sum_{n>=0} x^n/[n!*(n+1)!/2^n].
G.f. satisfies: A(x) = C(x*A(x)^2) and C(x) = A(x/C(x)^2) where C(x) is the g.f. of A155927.

A281260 Triangular array of generalized Narayana numbers T(n,k) = 2binomial(n+1,k) binomial(n-2,k-1)/(n+1) for n >= 1 and 0 <= k <= n-1, read by rows.

Original entry on oeis.org

1, 0, 2, 0, 2, 3, 0, 2, 8, 4, 0, 2, 15, 20, 5, 0, 2, 24, 60, 40, 6, 0, 2, 35, 140, 175, 70, 7, 0, 2, 48, 280, 560, 420, 112, 8, 0, 2, 63, 504, 1470, 1764, 882, 168, 9, 0, 2, 80, 840, 3360, 5880, 4704, 1680, 240, 10, 0, 2, 99, 1320, 6930, 16632, 19404, 11088, 2970, 330, 11, 0, 2, 120, 1980, 13200, 41580

Offset: 1

Werner Schulte, Jan 18 2017

The current array is the case m = 1 of the generalized Narayana numbers N_m(n,k) := (m+1)/(n+1)*binomial(n+1,k)*binomial(n-m-1,k-1) for m >= 0, n >= m and 0 <= k <= n-m with N_m(n,0) = A000007(n-m). Case m = 0 gives the table of Narayana numbers A001263 without leading column N_0(n,0) = A000007(n). For m = 2 see A281293, and for m = 3 see A281297. For combinatorial interpretations see the link to: David Callan, Generalized Narayana Numbers.
The polynomials p(m,n,x) = Sum_{k=0..n-m} N_m(n,k)*x^k satisfy the recurrence equation: x*p"(m,n,x) = n*(n-m-1)*p(m,n-1,x) for n > m >= 0 (p" is the second derivative of p), i.e.: k*(k-1)*N_m(n,k) = n*(n-m-1)*N_m(n-1,k-1) for k > 0 and n > m >= 0. Furthermore: Sum_{k=0..n-m} binomial(n+1-k,m+1)*N_m(n,k) = binomial(n,m)*A088218(n-m) for n >= m >= 0.
There is a relationship of these N_m(n,k) to those N_r(n,k) of A145596 (see the second comment): N_m(n+1,k) = N_r(n,k)*binomial(k+r,r)/binomial(n,r) for k >= 1 and 1 <= m = r <= n, and alternatively: N_r(n,k) = N_m(n+1,k)*binomial(n,m)/ binomial(k+m,m).
Conjecture: Sum_{k=1..n-m} binomial(n+1-k,m) * N_m(n,k) * A103365(n+1-m-k) = (m+1)^2 * A000007(n-m-1) for n > m >= 0.

			The triangle begins:
n\k:  0  1    2     3      4      5      6      7      8     9   10  11  . . .
01 :  1
02 :  0  2
03 :  0  2    3
04 :  0  2    8     4
05 :  0  2   15    20      5
06 :  0  2   24    60     40      6
07 :  0  2   35   140    175     70      7
08 :  0  2   48   280    560    420    112      8
09 :  0  2   63   504   1470   1764    882    168      9
10 :  0  2   80   840   3360   5880   4704   1680    240    10
11 :  0  2   99  1320   6930  16632  19404  11088   2970   330   11
12 :  0  2  120  1980  13200  41580  66528  55440  23760  4950  440  12
etc.

Michael De Vlieger, Table of n, a(n) for n = 1..11325 (rows n = 1..150, flattened)
David Callan, Generalized Narayana Numbers
Vladimir Kruchinin, Dmitry Kruchinin, and Yuriy Shablya, On some properties of generalized Narayana numbers, Tomsk State University of Control Systems and Radioelectronics, (Tomsk, Russia 2019).
Feiyang Lin, F-polynomials for the R-Kronecker quiver, University of Minnesota, Research Experiences for Undergrads (2020).
Bo Wang and Candice X.T. Zhang, Interlacing property of a family of generating polynomials over Dyck paths, arXiv:2309.05903 [math.CO], 2023.
Yi Wang and Arthur L.B. Yang, Total positivity of Narayana matrices, arXiv:1702.07822 [math.CO], 2017.
James J. Y. Zhao, On the positive zeros of generalized Narayana polynomials related to the Boros-Moll polynomials, arXiv:2108.03590 [math.CO], 2021.

Cf. A000007, A001263, A033184, A088218, A103365, A108838, A145596, A281293, A281297.

Table[2 Binomial[n + 1, k] Binomial[n - 2, k - 1]/(n + 1), {n, 1, 12}, {k, 0, n - 1}] // Flatten (* Michael De Vlieger, Jan 19 2017 *)

Row sums are A033184(n+1,2).
The same triangle as A108838 with reversed rows but without leading column.
G.f.: ((x*y-x-1)*sqrt(x^2*y^2+(-2*x^2-2*x)*y+x^2-2*x+1)+x^2*y^2+(-2*x^2-2*x)*y+x^2+1)/(2*x). - Vladimir Kruchinin, Oct 11 2020
G.f. satisfies x*A(x,y)^2-(x^2*y^2+((-2)*x^2-2*x)*y+x^2+1)*A(x,y)+x=0. - Vladimir Kruchinin, Oct 11 2020

A103367 Absolute row sums of triangle A103364, which equals the matrix inverse of the Narayana triangle (A001263).

Original entry on oeis.org

1, 2, 6, 26, 160, 1372, 15974, 245142, 4817712, 118198568, 3542890648, 127417949496, 5415490994368, 268526379444104, 15363229400769566, 1004545432884250126, 74441340170270921952, 6205992783298302119536

Offset: 2

Paul D. Hanna, Feb 02 2005

nonn

The ratio of the absolute row sum to the absolute value of column 1 for row n of triangle A103364, as n grows, approaches the limit given by limit_{n->inf} Sum_{k=1,n} |A103364(n,k)| / |A103365(n)| = 4.37281937295201487280058335227924496590808747668123198019676685492873...

Cf. A001263, A103364, A103365, A103366.

{a(n)=if(n<1,0,sum(k=1,n,abs(matrix(n,n,m,j,binomial(m-1,j-1)*binomial(m,j-1)/j)^-1)[n,k]))}

A155927 G.f. satisfies: A(x) = B(x/A(x)) where A(x) = Sum_{n>=0} a(n)x^n/[n!(n+1)!/2^n] and B(x) = A(xB(x)) = Sum_{n>=0} x^n/[n!(n+1)!/2^n].

Original entry on oeis.org

1, 1, -2, 19, -379, 12726, -641465, 45181627, -4232016719, 508271819428, -76108505872794, 13896010073569130, -3038043685025188631, 783439451518414509612, -235289860249420249309140

Offset: 0

Paul D. Hanna, Jan 31 2009

sign

			G.f.: A(x) = 1 + x - 2*x^2/3 + 19*x^3/18 - 379*x^4/180 + 12726*x^5/2700 +...+ a(n)*x^n/[n!*(n+1)!/2^n] +...
G.f. satisfies: A(x) = B(x/A(x)) and B(x) = A(x*B(x)) where:
B(x) = 1 + x + 1/3*x^2 + 1/18*x^3 + 1/180*x^4 +...+ x^n/[n!*(n+1)!/2^n] +...
G.f. satisfies: A(x) = F(x/A(x)^2) and F(x) = A(x*F(x)^2) where:
F(x) = 1 + x + 4*x^2/3 + 37*x^3/18 + 621*x^4/180 + 16526*x^5/2700 +...+ A155926(n)*x^(n+1)/[n!*(n+1)!/2^n] +...
G.f. satisfies: A(x) = 1/G(x/A(x)) and G(x) = 1/A(x/G(x)) where:
G(x) = 1 - x + 2*x^2/3 - 7*x^3/18 + 39*x^4/180 - 321*x^5/2700 +...+ A103365(n)*x^(n+1)/[n!*(n+1)!/2^n] +...

Cf. A155926, A103365.

{a(n)=local(B=sum(k=0,n,x^k/(k!*(k+1)!/2^k))+x*O(x^n));polcoeff(x/serreverse(x*B),n)*n!*(n+1)!/2^n}

G.f. satisfies: A(x) = F(x/A(x)^2) and F(x) = A(x*F(x)^2) where F(x) = Sum_{n>=0} A155926(n)*x^n/[n!*(n+1)!/2^n].

G.f. satisfies: A(x) = 1/G(x/A(x)) and G(x) = 1/A(x/G(x)) where G(x) = Sum_{n>=0} A103365(n)*x^n/[n!*(n+1)!/2^n].

A280580 Triangle read by rows: T(n,k) = binomial(2n,2k)binomial(2n-2*k,n-k)/(n+1-k) for 0<=k<=n.

Original entry on oeis.org

1, 1, 1, 2, 6, 1, 5, 30, 15, 1, 14, 140, 140, 28, 1, 42, 630, 1050, 420, 45, 1, 132, 2772, 6930, 4620, 990, 66, 1, 429, 12012, 42042, 42042, 15015, 2002, 91, 1, 1430, 51480, 240240, 336336, 180180, 40040, 3640, 120, 1, 4862, 218790, 1312740, 2450448, 1837836, 612612, 92820, 6120, 153, 1

Offset: 0

Werner Schulte, Jan 05 2017

			Triangle begins:
n\k:     0      1       2       3       4      5     6    7  8  . . .
  0:     1
  1:     1      1
  2:     2      6       1
  3:     5     30      15       1
  4:    14    140     140      28       1
  5:    42    630    1050     420      45      1
  6:   132   2772    6930    4620     990     66     1
  7:   429  12012   42042   42042   15015   2002    91    1
  8:  1430  51480  240240  336336  180180  40040  3640  120  1
  etc.
T(3,2) = binomial(6,4) * binomial(2,1) / (3+1-2) = 15 * 2 / 2 = 15. - _Indranil Ghosh_, Feb 15 2017

Indranil Ghosh, Rows 0..100 of triangle, flattened

Row sums are A026945.
Triangle related to A000108, A001006, A001263, and A039599.
Cf. A000007, A001003, A005568, A103365, A238390.

T(n,k) = A001263(n+1,k+1)*A000108(n)/A000108(k) for 0 <= k <= n.
T(n,k) = binomial(2*n,2*k)*A000108(n-k) for 0 <= k <= n.
T(n,k) = A039599(n,k)*binomial(n+1+k,2*k+1)/(n+1-k) for 0 <= k <= n.
Recurrences: T(n,0) = A000108(n) and (1) T(n,k) = T(n,k-1)*(n+1-k)*(n+2-k)/ (2*k*(2*k-1)) for 0 < k <= n, (2) T(n,k) = T(n-1,k-1)*n*(2*n-1)/(k*(2*k-1)).
The row polynomials p(n,x) = Sum_{k=0..n} T(n,k)*x^(2*k) satisfy the recurrence equation p"(n,x) = 2*n*(2*n-1)*p(n-1,x) with initial value p(0,x) = 1 ( n > 0, p" is the second derivative of p ), and Sum_{n>=0} p(n,x)*t^(2*n)/((2*n)!) = cosh(x*t)*(Sum_{n>=0} A000108(n)*t^(2*n)/((2*n)!)).
Conjectures: (1) Sum_{k=0..n} (-1)^k*T(n,k)*A238390(k) = A000007(n);
  (2) Antidiagonal sums equal A001003(n);
  (3) Matrix inverse equals T(n,k)*A103365(n+1-k).
Sum_{k=0..n} (n+1-k)*T(n,k) = A002426(2*n) = A082758(n).
Sum_{k=0..n} T(n,k)*A000108(k) = A000108(n)*A000108(n+1) = A005568(n).
Matrix product: Sum_{i=0..n} T(n,i)*T(i,k) = T(n,k)*A000108(n+1-k) for 0<=k<=n.
T(n,k) = A097610(2*n,2*k) for 0 <= k <= n.
Sum_{k=0..n} (k+1)*T(n,k)*A000108(k) = binomial(2*n+1,n)*A000108(n).

A281000 Triangle read by rows: T(n,k) = binomial(2n+1, 2k+1)binomial(2n-2*k, n-k)/(n+1-k) for 0 <= k <= n.

Original entry on oeis.org

1, 3, 1, 10, 10, 1, 35, 70, 21, 1, 126, 420, 252, 36, 1, 462, 2310, 2310, 660, 55, 1, 1716, 12012, 18018, 8580, 1430, 78, 1, 6435, 60060, 126126, 90090, 25025, 2730, 105, 1, 24310, 291720, 816816, 816816, 340340, 61880, 4760, 136, 1, 92378, 1385670, 4988412, 6651216, 3879876, 1058148, 135660, 7752, 171, 1

Offset: 0

Werner Schulte, Jan 12 2017

			Triangle begins:
n\k:      0       1       2       3       4      5     6    7  8  . . .
  0:      1
  1:      3       1
  2:     10      10       1
  3:     35      70      21       1
  4:    126     420     252      36       1
  5:    462    2310    2310     660      55      1
  6:   1716   12012   18018    8580    1430     78     1
  7:   6435   60060  126126   90090   25025   2730   105    1
  8:  24310  291720  816816  816816  340340  61880  4760  136  1
  etc.
T(3,2) = binomial(7,5) * binomial(2,1) / (3+1-2) = 21 * 2 / 2 = 21. - _Indranil Ghosh_, Feb 15 2017

Indranil Ghosh, Rows 0..100 of triangle, flattened

Row sums are A099250.
Triangle related to A000108, A097610, A280580.
Cf. A000007, A001003, A001006, A002426, A103365, A125558, A273055.

Table[Binomial[2n+1,2k+1] Binomial[2n-2k,n-k]/(n+1-k),{n,0,10},{k,0,n}]// Flatten (* Harvey P. Dale, Nov 25 2018 *)

T(n,k) = A097610(2*n+1, 2*k+1) = binomial(2*n+1, 2*k+1)*A000108(n-k) = A280580(n,k)*(2*n+1)/(2*k+1) for 0 <= k <= n.
Recurrences: T(n,0) = (2*n+1)*A000108(n) and
  (1) T(n,k) = T(n,k-1)*(n+1-k)*(n+2-k)/(2*k*(2*k+1)) for 0 < k <= n,
  (2) T(n,k) = T(n-1, k-1)*n*(2*n+1)/(k*(2*k+1)).
The row polynomials p(n,x) = Sum_{k=0..n} T(n,k)*x^(2*k+1) satisfy the recurrence equation p"(n,x) = (2*n+1)*2*n*p(n-1,x) with initial value p(0,x) = x (n > 0, p" is the second derivative of p), and Sum_{n>=0} p(n,x)*t^(2*n+1)/ ((2*n+1)!) = sinh(x*t)*(Sum_{n>=0} A000108(n)*t^(2*n)/((2*n)!)).
Conjectures:
  (1) Antidiagonal sums equal A001003(n+1);
  (2) Sum_{k=0..n} (-1)^k*T(n,k)*(2*k+1)*A000108(k)*A103365(k) = A000007(n);
  (3) Matrix inverse equals T(n,k)*A103365(n+1-k).
Sum_{k=0..n} (n+1-k)*T(n,k) = A002426(2*n+1) = A273055(n).
Sum_{k=0..n} T(n,k)*(2*k+1)*A000108(k) = (2*n+1)*A000108(n)*A000108(n+1) = A125558(n+1).
Matrix product: Sum_{i=0..n} T(n,i)*T(i,k) = T(n,k)*A000108(n+1-k) for 0<=k<=n.

cp's OEIS Frontend

A115369 Decimal expansion of first zero of BesselJ(1,z).

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A103364 Matrix inverse of the Narayana triangle A001263.

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A238390 E.g.f.: x / BesselJ(1, 2*x) (even powers only).

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A103366 Second column of triangle A103364, which equals the matrix inverse of the Narayana triangle (A001263).

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A155926 G.f. satisfies: A(x) = B(x*A(x)) where A(x) = Sum_{n>=0} a(n)*x^n/[n!*(n+1)!/2^n] and B(x) = Sum_{n>=0} x^n/[n!*(n+1)!/2^n].

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A281260 Triangular array of generalized Narayana numbers T(n,k) = 2*binomial(n+1,k)* binomial(n-2,k-1)/(n+1) for n >= 1 and 0 <= k <= n-1, read by rows.

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A103367 Absolute row sums of triangle A103364, which equals the matrix inverse of the Narayana triangle (A001263).

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A155927 G.f. satisfies: A(x) = B(x/A(x)) where A(x) = Sum_{n>=0} a(n)*x^n/[n!*(n+1)!/2^n] and B(x) = A(x*B(x)) = Sum_{n>=0} x^n/[n!*(n+1)!/2^n].

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A155926 G.f. satisfies: A(x) = B(xA(x)) where A(x) = Sum_{n>=0} a(n)x^n/[n!(n+1)!/2^n] and B(x) = Sum_{n>=0} x^n/[n!(n+1)!/2^n].

A281260 Triangular array of generalized Narayana numbers T(n,k) = 2binomial(n+1,k) binomial(n-2,k-1)/(n+1) for n >= 1 and 0 <= k <= n-1, read by rows.

A155927 G.f. satisfies: A(x) = B(x/A(x)) where A(x) = Sum_{n>=0} a(n)x^n/[n!(n+1)!/2^n] and B(x) = A(xB(x)) = Sum_{n>=0} x^n/[n!(n+1)!/2^n].

A280580 Triangle read by rows: T(n,k) = binomial(2n,2k)binomial(2n-2*k,n-k)/(n+1-k) for 0<=k<=n.

A281000 Triangle read by rows: T(n,k) = binomial(2n+1, 2k+1)binomial(2n-2*k, n-k)/(n+1-k) for 0 <= k <= n.