A046748 -id:A046748 - cp's OEIS frontend

A046521 Array T(i,j) = binomial(-1/2-i,j)*(-4)^j, i,j >= 0 read by antidiagonals going down.

1, 2, 1, 6, 6, 1, 20, 30, 10, 1, 70, 140, 70, 14, 1, 252, 630, 420, 126, 18, 1, 924, 2772, 2310, 924, 198, 22, 1, 3432, 12012, 12012, 6006, 1716, 286, 26, 1, 12870, 51480, 60060, 36036, 12870, 2860, 390, 30, 1, 48620, 218790, 291720, 204204, 87516, 24310

Offset: 0

Wolfdieter Lang

Or, a triangle related to A000984 (central binomial) and A000302 (powers of 4).
This is an example of a Riordan matrix. See the Shapiro et al. reference quoted under A053121 and Notes 1 and 2 of the Wolfdieter Lang reference, p. 306.
As a number triangle, this is the Riordan array (1/sqrt(1-4x),x/(1-4x)). - Paul Barry, May 30 2005
The A- and Z- sequences for this Riordan matrix are (see the Wolfdieter Lang link under A006232 for the D. G. Rogers, D. Merlini et al. and R. Sprugnoli references on Riordan A- and Z-sequences with a summary): A-sequence [1,4,0,0,0,...] and Z-sequence 4+2*A000108(n)*(-1)^(n+1)=[2, 2, -4, 10, -28, 84, -264, 858, -2860, 9724, -33592, 117572, -416024, 1485800, -5348880, 19389690, -70715340, 259289580, -955277400, 3534526380], n >= 0. The o.g.f. for the Z-sequence is 4-2*c(-x) with the Catalan number o.g.f. c(x). - Wolfdieter Lang, Jun 01 2007
As a triangle, T(2n,n) is A001448. Row sums are A046748. Diagonal sums are A176280. - Paul Barry, Apr 14 2010
From Wolfdieter Lang, Aug 10 2017: (Start)
The row polynomials R(n, x) of Riordan triangles R = (G(x), F(x)), with F(x)= x*Fhat(x), belong to the class of Boas-Buck polynomials (see the reference). Hence they satisfy the Boas-Buck identity (we use the notation of Rainville, Theorem 50, p. 141):
  (E_x - n*1)*R(n, x) = -Sum_{k=0..n-1} (alpha(k)*1 + beta(k)*E_x)*R(n-1.k, x), for n >= 0, where E_x = x*d/dx (Euler operator). The Boas-Buck sequences are given by alpha(k) := [x^k] ((d/dx)log(G(x))) and beta(k) := [x^k] (d/dx)log(Fhat(x)).
This entails a recurrence for the sequence of column m of the Riordan triangle T, n > m >= 0: T(n, m) = (1/(n-m))*Sum_{k=m..n-1} (alpha(n-1-k) + m*beta(n-1-k))*T(k, m), with input T(m,m).
For the present case the Boas-Buck identity for the row polynomials is (E_x - n*1)*R(n, x) = -Sum_{k=0..n-1} 2^(2*k+1)*(1 + 2*E_x)*R(n-1-k, x), for n >= 0. For the ensuing recurrence for the columns m of the triangle T see the formula and example section. (End)
From Peter Bala, Mar 04 2018: (Start)
The following two remarks are particular cases of more general results for Riordan arrays of the form (f(x), x/(1 - k*x)).
1) Let R(n,x) denote the n-th row polynomial of this triangle. The polynomial R(n,4*x) has the e.g.f. Sum_{k = 0..n} T(n,k)*(4*x)/k!. The e.g.f. for the n-th diagonal of the triangle (starting at n = 0 for the main diagonal) equals exp(x) * the e.g.f. for the polynomial R(n,4*x). For example, when n = 3 we have exp(x)*(20 + 30*(4*x) + 10*(4*x)^2/2! + (4*x)^3/3!) = 20 + 140*x + 420*x^2/2! + 924*x^3/3! + 1716*x^4/4! + ....
2) Let P(n,x) = Sum_{k = 0..n} T(n,k)*x^(n-k) denote the n-th row polynomial in descending powers of x. P(n,x) is the n-th degree Taylor polynomial of (1 + 4*x)^(n-1/2) about 0. For example, for n = 4 we have (1 + 4*x)^(7/2) = 70*x^4 + 140*x^3 + 70*x^2 + 14*x + 1 + O(x^5).
Let C(x) = (1 - sqrt(1 - 4*x))/(2*x) denote the o.g.f. of the Catalan numbers A000108. The derivatives of C(x) are determined by the identity (-1)^n * x^n/n! * (d/dx)^n(C(x)) = 1/(2*x)*( 1 - P(n,-x)/(1 - 4*x)^(n-1/2) ), n = 0,1,2,.... See Lang 2002. Cf. A283150 and A283151. (End)

			Array begins:
  1,  2,   6,  20,   70, ...
  1,  6,  30, 140,  630, ...
  1, 10,  70, 420, 2310, ...
  1, 14, 126, 924, 6006, ...
Recurrence from A-sequence: 140 = a(4,1) = 20 + 4*30.
Recurrence from Z-sequence: 252 = a(5,0) = 2*70 + 2*140 - 4*70 + 10*14 - 28*1.
From _Paul Barry_, Apr 14 2010: (Start)
As a number triangle, T(n, m) begins:
n\k       0      1       2       3      4      5     6    7   8  9 10 ...
0:        1
1:        2      1
2:        6      6       1
3:       20     30      10       1
4:       70    140      70      14      1
5:      252    630     420     126     18      1
6:      924   2772    2310     924    198     22     1
7:     3432  12012   12012    6006   1716    286    26    1
8:    12870  51480   60060   36036  12870   2860   390   30   1
9:    48620 218790  291720  204204  87516  24310  4420  510  34  1
10:  184756 923780 1385670 1108536 554268 184756 41990 6460 646 38  1
... [Reformatted and extended by _Wolfdieter Lang_, Aug 10 2017]
Production matrix begins
      2, 1,
      2, 4, 1,
     -4, 0, 4, 1,
     10, 0, 0, 4, 1,
    -28, 0, 0, 0, 4, 1,
     84, 0, 0, 0, 0, 4, 1,
   -264, 0, 0, 0, 0, 0, 4, 1,
    858, 0, 0, 0, 0, 0, 0, 4, 1,
  -2860, 0, 0, 0, 0, 0, 0, 0, 4, 1 (End)
Boas-Buck recurrence for column m = 2, and n = 4: T(4, 2) = (2*(2*2+1)/2) * Sum_{k=2..3} 4^(3-k)*T(k, 2) = 5*(4*1 + 1*10) = 70. - _Wolfdieter Lang_, Aug 10 2017
From _Peter Bala_, Feb 15 2018: (Start)
With C(x) = (1 - sqrt( 1 - 4*x))/(2*x),
-x^3/3! * (d/dx)^3(C(x)) = 1/(2*x)*( 1 - (1 - 10*x + 30*x^2 - 20*x^3)/(1 - 4*x)^(5/2) ).
x^4/4! * (d/dx)^4(C(x)) = 1/(2*x)*( 1 - (1 - 14*x + 70*x^2 - 140*x^3 + 70*x^4 )/(1 - 4*x)^(7/2) ). (End)

Ralph P. Boas, jr. and R. Creighton Buck, Polynomial Expansions of analytic functions, Springer, 1958, pp. 17 - 21, (last sign in eq. (6.11) should be -).
Earl D. Rainville, Special Functions, The Macmillan Company, New York, 1960, ch. 8, sect. 76, 140 - 146.

Muniru A Asiru, Antidiagonals n = 0..50, flattened
Peter Bala, A 4-parameter family of embedded Riordan arrays
Peter Bala, A note on the diagonals of a proper Riordan Array
Paul Barry, Embedding structures associated with Riordan arrays and moment matrices, arXiv preprint arXiv:1312.0583 [math.CO], 2013.
Jonathan W. Bober, Factorial ratios, hypergeometric series, and a family of step functions, arXiv:0709.1977 [math.NT], 2007; J. London Math. Soc. (2) 79 2009, 422-444.
Wolfdieter Lang, First 10 rows.
Wolfdieter Lang, On polynomials related to derivatives of the generating function of Catalan numbers, Fib. Quart. 40,4 (2002) 299-313; T(n,m) is called B(n,m) there.
Benjamin Lovitz and Nathaniel Johnston, A hierarchy of eigencomputations for polynomial optimization on the sphere, arXiv:2310.17827 [math.OC], 2023. See pp. 35, 39. See also Math. Prog. (2025) Series A.
Helmut Prodinger, Some information about the binomial transform, The Fibonacci Quarterly, 32, 1994, 412-415.

Columns include: A000984 (m=0), A002457 (m=1), A002802 (m=2), A020918 (m=3), A020920 (m=4), A020922 (m=5), A020924 (m=6), A020926 (m=7), A020928 (m=8), A020930 (m=9), A020932 (m=10).
Row sums: A046748.
Cf. A001147, A006882, A007318, A068555, A111959, A283150, A283151.

Flat(List([0..9],n->List([0..n],m->Binomial(2*n,n)*Binomial(n,m)/Binomial(2*m,m)))); # Muniru A Asiru, Jul 19 2018

[Binomial(n+1,k+1)*Catalan(n)/Catalan(k): k in [0..n], n in [0..12]]; // G. C. Greubel, Jul 28 2024

t[i_, j_] := If[i < 0 || j < 0, 0, (2*i + 2*j)!*i!/(2*i)!/(i + j)!/j!]; Flatten[Reverse /@ Table[t[n, k - n] , {k, 0, 9}, {n, k, 0, -1}]][[1 ;; 51]] (* Jean-François Alcover, Jun 01 2011, after PARI prog. *)

T(i,j)=if(i<0 || j<0,0,(2*i+2*j)!*i!/(2*i)!/(i+j)!/j!)

def A046521(n,k): return binomial(n+1, k+1)*catalan_number(n)/catalan_number(k)
flatten([[A046521(n,k) for k in range(n+1)] for n in range(13)]) # G. C. Greubel, Jul 28 2024

T(n, m) = binomial(2*n, n)*binomial(n, m)/binomial(2*m, m), n >= m >= 0.
G.f. for column m: ((x/(1-4*x))^m)/sqrt(1-4*x).
Recurrence from the A-sequence given above: a(n,m) = a(n-1,m-1) + 4*a(n-1,m), for n >= m >= 1.
Recurrence from the Z-sequence given above: a(n,0) = Sum_{j=0..n-1} Z(j)*a(n-1,j), n >= 1; a(0,0)=1.
As a number triangle, T(n,k) = C(2*n,n)*C(n,k)/C(2*k,k) = C(n-1/2,n-k)*4^(n-k). - Paul Barry, Apr 14 2010
From Peter Bala, Apr 11 2012: (Start):
One of three infinite families of integral factorial ratio sequences of height 1 (see Bober, Theorem 1.2). The other two are A007318 and A068555.
The triangular array equals exp(S), where the infinitesimal generator S has [2,6,10,14,18,...] on the main subdiagonal and zeros elsewhere.
Recurrence equation for the square array: T(n+1,k) = (k+1)/(4*n+2)*T(n,k+1). (End)
T(n,k) = 4^(n-k)*A006882(2*n - 1)/(A006882(2*n - 2*k)*A006882(2*k - 1)) = 4^(n-k)*(2*n - 1)!!/((2*n - 2*k)!*(2*k - 1)!!). - Peter Bala, Nov 07 2016
Boas-Buck recurrence for column m, m > n >= 0: T(n, m) = (2*(2*m+1)/(n-m))*Sum_{k=m..n-1} 4^(n-1-k)*T(k, m), with input T(n, n) = 1. See a comment above. - Wolfdieter Lang, Aug 10 2017
From Peter Bala, Aug 13 2021: (Start)
Analogous to the binomial transform we have the following sequence transformation formula: g(n) = Sum_{k = 0..n} T(n,k)*b^(n-k)*f(k) iff f(n) = Sum_{k = 0..n} (-1)^(n-k)*T(n,k)*b^(n-k)*g(k). See Prodinger, bottom of p. 413, with b replaced with 4*b, c = 1 and d = 1/2.
Equivalently, if F(x) = Sum_{n >= 0} f(n)*x^n and G(x) = Sum_{n >= 0} g(n)*x^n are a pair of  formal power series then
G(x) = 1/sqrt(1 - 4*b*x) * F(x/(1 - 4*b*x)) iff F(x) = 1/sqrt(1 + 4*b*x) * G(x/(1 + 4*b*x)).
The m-th power of this array has entries m^(n-k)*T(n,k). (End)

A098615 G.f. A(x) satisfies: A(xG(x)) = G(x), where G(x) is the g.f. for A098614(n) = Fibonacci(n+1)Catalan(n).

Original entry on oeis.org

1, 1, 3, 5, 13, 25, 61, 125, 295, 625, 1447, 3125, 7151, 15625, 35491, 78125, 176597, 390625, 880125, 1953125, 4390901, 9765625, 21920913, 48828125, 109486993, 244140625, 547018941, 1220703125, 2733608905, 6103515625, 13662695645, 30517578125, 68294088535, 152587890625, 341399727335, 762939453125, 1706739347095, 3814697265625, 8532741458075, 19073486328125, 42660172763995, 95367431640625

Offset: 0

Paul D. Hanna, Oct 14 2004

nonn

G.f. satisfies: A(x) = x/(series reversion of x*G098614(x)), where G098614 is the g.f. for A098614 = {1*1, 1*1, 2*2, 3*5, 5*14, 8*42, 13*132, ...}.
Hankel transform is 2^n. Image of F(n+1) under the Riordan array (c(x^2),xc(x^2)), c(x) the g.f. of A000108. The sequence 0,1,1,3,5,... has general term Sum_{k=0..floor(n/2)} (C(n-1,k) - C(n-1,k-1))*F(n-2k). It is the image of the Fibonacci numbers under the transform of generating functions g(x)-> g(xc(x^2)), c(x) the g.f. of A000108. This sequence has Hankel transform -(-4)^((n-1)/2)(1-(-1)^n)/2. - Paul Barry, Oct 01 2007
The sequence of fractions 1, 1/2, 3/4, 5/8, 13/16, 25/32, ... or a(n)/2^n is the image of F(n+1) under the Chebyshev related (rational) Riordan array c((x/2)^2),(x/2)c((x/2)^2)) where c(x) is the g.f. of A000108. The Hankel transform of this fraction sequence is 1/(2^(n^2)). - Paul Barry, Jun 17 2008

Paul D. Hanna, Table of n, a(n) for n = 0..300
Paul Barry and A. Hennessy, Meixner-Type Results for Riordan Arrays and Associated Integer Sequences, J. Int. Seq. 13 (2010) # 10.9.4, example 30.
Cyril Banderier, Markus Kuba, and Michael Wallner, Analytic Combinatorics of Composition schemes and phase transitions with mixed Poisson distributions, arXiv:2103.03751 [math.PR], 2021.

Cf. A000045, A000108, A046748, A053121, A054335, A098614.

R:=PowerSeriesRing(Rationals(), 30);
Coefficients(R!( (x+Sqrt(1-4*x^2))/(1-5*x^2) )); // G. C. Greubel, Jul 31 2024

Array[Sum[Binomial[(# - 1)/2, (# - k)/2]*2^(# - k - 1)*((-1)^(# - k) + 1), {k, 0, #}] &, 42, 0] (* or *)
CoefficientList[Series[(Sqrt[1 - 4 x^2] + x)/(1 - 5 x^2), {x, 0, 41}], x] (* Michael De Vlieger, May 20 2021 *)

a(n):=sum(binomial((n-1)/2,(n-k)/2)*2^(n-k-1)*((-1)^(n-k)+1),k,0,n); /* Vladimir Kruchinin, Apr 16 2011 */

{ a(n) = polcoeff((sqrt(1-4*x^2+x^2*O(x^n))+x)/(1-5*x^2),n) }
for(n=0,50,print1(a(n),", "))

def A098615_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P( (x+sqrt(1-4*x^2))/(1-5*x^2) ).list()
A098615_list(30) # G. C. Greubel, Jul 31 2024

G.f.: (x + sqrt(1-4*x^2)) / (1-5*x^2).
G.f. satisfies: A(x) = sqrt(1 + 2*x*A(x) + 5*x^2*A(x)^2). - Paul D. Hanna, Nov 18 2014
a(2*n) = A046748(n).
a(2*n+1) = 5^n.
a(n) = Sum_{k=0..floor((n+1)/2)} (C(n,k) - C(n,k-1))*Fibonacci(n-2k+1). - Paul Barry, Oct 01 2007
G.f.: 1/(1-x-2x^2/(1-x^2/(1-x^2/(1-x^2/(1-x^2/(1-.... (continued fraction). - Paul Barry, Feb 09 2009
a(n) = Sum_{k=0..n} binomial((n-1)/2,(n-k)/2)*2^(n-k-1)*(1+(-1)^(n-k)). - Vladimir Kruchinin, Apr 16 2011
From Gary W. Adamson, Sep 22 2011: (Start)
a(n) is the upper left term in M^n, M = an infinite square production matrix as follows:
   1, 1, 1, 0, 0, 0, ...
   1, 0, 0, 1, 0, 0, ...
   1, 0, 0, 0, 1, 0, ...
   0, 1, 0, 0, 0, 1, ...
   0, 0, 1, 0, 0, 0, ...
   0, 0, 0, 1, 0, 0, ...
   0, 0, 0, 0, 1, 0, ...
   0, 0, 0, 0, 0, 1, ...
   ... (End)
a(n) = Sum_{k=0..floor(n/2)} A054335(n-k,n-2k). - Philippe Deléham, Feb 01 2012
a(n) = Sum_{k=0..n} A053121(n,k)*A000045(k+1). - Philippe Deléham, Feb 03 2012
n*a(n) +(n-1)*a(n-1) -3*(3*n-4)*a(n-2) -3*(3*n-7)*a(n-3) +20*(n-3)*a(n-4) +20*(n-4)*a(n-5) = 0. - R. J. Mathar, Jul 21 2017

A082761 Trinomial transform of the Fibonacci numbers (A000045).

Original entry on oeis.org

1, 4, 20, 104, 544, 2848, 14912, 78080, 408832, 2140672, 11208704, 58689536, 307302400, 1609056256, 8425127936, 44114542592, 230986743808, 1209462292480, 6332826779648, 33159111507968, 173623361929216, 909103725543424, 4760128905543680, 24924358531088384, 130505635564355584

Offset: 0

Emanuele Munarini, May 21 2003

Hankel transform of Sum_{k=0..n} (-1)^k*C(2k, k) (see A054108). - Paul Barry, Jan 13 2009
Hankel transform of A046748. - Paul Barry, Apr 14 2010
For positive n, a(n) equals the permanent of the (2n) X (2n) tridiagonal matrix with sqrt(2)'s along the three central diagonals. - John M. Campbell, Jul 12 2011
The limiting ratio is: Lim_{n -> oo} a(n)/a(n-1) = 1 + phi^3. - Bob Selcoe, Mar 18 2014
Invert transform of A052984. Invert transform is A083066. Binomial transform of A033887. Binomial transform is A163073. - Michael Somos, May 26 2014

			a(5) = 2848 = 5*(544) + 4 + 20 + 104.
G.f. = 1 + 4*x + 20*x^2 + 104*x^3 + 544*x^4 + 2848*x^5 + 14912*x^6 + ...

Vincenzo Librandi, Table of n, a(n) for n = 0..150
László Németh, The trinomial transform triangle, J. Int. Seqs., Vol. 21 (2018), Article 18.7.3. Also arXiv:1807.07109 [math.NT], 2018.
Index entries for linear recurrences with constant coefficients, signature (6,-4).

Cf. A000045, A001519, A027907, A033887, A046748, A052984, A054108, A069429, A083066, A147703, A163073.

[2^n * Fibonacci(2*n+1): n in [0..40]]; // Vincenzo Librandi, Jul 15 2011

a[ n_] := 2^n Fibonacci[ 2 n + 1]; (* Michael Somos, May 26 2014 *)
a[ n_] := If[ n < 0, SeriesCoefficient[ (2 - x) / (4 - 6 x + x^2), {x, 0, -1 - n}], SeriesCoefficient[ (1 - 2 x) / (1 - 6 x + 4 x^2), {x, 0, n}]]; (* Michael Somos, Oct 22 2017 *)
LinearRecurrence[{6,-4},{1,4},30] (* Harvey P. Dale, Jul 11 2014 *)

a(n)=fibonacci(2*n+1)<Charles R Greathouse IV, Jul 15 2011

{a(n) = if( n<0, n = -1 - n; 2^(-1-2*n), 1) * polcoeff( (1 - 2*x) / (1 - 6*x + 4*x^2) + x * O(x^n), n)}; /* Michael Somos, Oct 22 2017 */

[2^n*fibonacci(2*n+1) for n in range(41)] # G. C. Greubel, Jul 28 2024

a(n) = Sum_{k=0..2*n} A027907(n, k)*A000045(k+1).
From Paul Barry, Jul 16 2003: (Start)
Third binomial transform of (1, 1, 5, 5, 25, 25, ....).
a(n) = ((1+sqrt(5))(3+sqrt(5))^n-(1-sqrt(5))*(3-sqrt(5))^n)/(2*sqrt(5)). (End)
From R. J. Mathar, Nov 04 2008: (Start)
G.f.: (1-2*x)/(1-6*x+4*x^2).
a(n) = 6*a(n-1) - 4*a(n-2). (End)
a(n) = Sum_{k=0..n} A147703(n,k)*3^k. - Philippe Deléham, Nov 14 2008
For n>=2: a(n) = 5*a(n-1) + Sum_{i=1..n-2} a(i). - Bob Selcoe, Mar 18 2014
a(n) = a(-1-n) * 2^(2*n+1) for all n in Z. - Michael Somos, Mar 18 2014
a(n) = 2^n*Fibonacci(2*n+1), or 2^n*A001519(n+1). - Bob Selcoe, May 25 2014
From Michael Somos, May 26 2014: (Start)
a(n) - a(n-1) = A069429(n).
a(n+1) * a(n-1) - a(n)^2 = 4^n.
G.f.: 1 / (1 - 4*x / (1 - x / (1 - x))). (End)
E.g.f.: exp(3*x)*(5*cosh(sqrt(5)*x) + sqrt(5)*sinh(sqrt(5)*x))/5. - Stefano Spezia, May 24 2024

A046814 Row sums of triangle A046527.

Original entry on oeis.org

1, 2, 8, 37, 179, 881, 4369, 21746, 108444, 541362, 2704158, 13512392, 67534828, 337584992, 1687627800, 8437136085, 42182258715, 210899507685, 1054456597965, 5272139698215, 26360193558735, 131799177579015

Offset: 0

Wolfdieter Lang

G. C. Greubel, Table of n, a(n) for n = 0..1000

Cf. A000108, A046527, A046714, A046748.

[n le 1 select 1 else 5*Self(n-1) - 3*Catalan(n-1)/(2*n-3): n in [1..40]]; // G. C. Greubel, Jul 28 2024

CoefficientList[Series[(1-4*x)*(1-Sqrt[1-4*x])/(2*x*(1-5*x)), {x,0,40}], x] (* G. C. Greubel, Jul 28 2024 *)

@CachedFunction
def A046814(n): return 1 if n==0 else 5*A046814(n-1) - 3*catalan_number(n)/(2*n-1)
[A046814(n) for n in range(41)] # G. C. Greubel, Jul 28 2024

G.f.: c(x) * (1-4*x) / (1-5*x), where c(x) = g.f. for Catalan A000108.
a(n) = C(n) + A046714(n-1) with A046714(-1) = 0 and C(n) = A000108(n) are the Catalan numbers.
a(n) = C(n) + (5^n - A046748(n))/2.
a(n) = 5*a(n-1) - 3*C(n)/(2*n-1), a(0)=1.
D-finite with recurrence a(n) = (9*n-1)*a(n-1)/(n+1) - 10*(2*n-3)*a(n-2)/(n+1), n >= 2, a(0)=1, a(1)=2.

Offset corrected by Sean A. Irvine, Apr 25 2021

A046885 Row sums of triangle A046658.

Original entry on oeis.org

1, 4, 18, 85, 411, 2013, 9933, 49236, 244750, 1218888, 6077644, 30329434, 151439158, 756452890, 3779590010, 18888255205, 94405918355, 471899946985, 2359022096225, 11793343217935, 58960151969255, 294776293579255

Offset: 1

Wolfdieter Lang

G. C. Greubel, Table of n, a(n) for n = 1..1000

Cf. A000108, A046658, A046714, A046748.

[n le 1 select 1 else 5*Self(n-1) - Catalan(n-1): n in [1..40]]; // G. C. Greubel, Jul 28 2024

Rest@CoefficientList[Series[Sqrt[1-4*x]*(1-Sqrt[1-4*x])/(2*(1-5*x)), {x,0,40}], x] (* G. C. Greubel, Jul 28 2024 *)

@CachedFunction
def A046885(n): return 1 if n==1 else 5*A046885(n-1) - catalan_number(n-1)
[A046885(n) for n in range(1,41)] # G. C. Greubel, Jul 28 2024

a(n) = 2*5^(n-1) - A046714(n-1) = (A046748(n) - 5^(n-1))/2.
G.f.: x*(2 - c(x))/(1-5*x), where c(x) is the g.f. of A000108 (Catalan numbers).
Inhomogeneous recursion: a(n) = 5*a(n-1) - C(n-1), n >= 2, a(1)=1; C(n) = A000108(n) (Catalan).
Homogeneous recursion: a(n) = (3*(3*n-2)/n)*a(n-1) - (10*(2*n-3)/n)*a(n-2), n >= 3, a(1)=1, a(2)=4.

A339000 Triangle read by rows: T(n, k) = C(n, k)Sum_{j=0..n} C(n, k-j)C(n+j, j)/C(2*j, j).

Original entry on oeis.org

1, 1, 2, 1, 7, 5, 1, 15, 32, 13, 1, 26, 111, 123, 34, 1, 40, 285, 603, 429, 89, 1, 57, 610, 2094, 2748, 1408, 233, 1, 77, 1155, 5845, 12170, 11196, 4437, 610, 1, 100, 2002, 14014, 42355, 60686, 42255, 13587, 1597, 1, 126, 3246, 30030, 124137, 254756, 271961, 150951, 40736, 4181

Offset: 0

Vladimir Kruchinin, Nov 18 2020

			Triangle begins as:
  1;
  1,  2;
  1,  7,   5;
  1, 15,  32,   13;
  1, 26, 111,  123,   34;
  1, 40, 285,  603,  429,   89;
  1, 57, 610, 2094, 2748, 1408, 233;

G. C. Greubel, Rows n = 0..50 of the triangle, flattened

Cf. A000045 (Fibonacci), A001519, A008459, A046748 (row sums).

b:=Binomial;
A339000:= func< n,k | b(n,k)*(&+[b(n,k-j)*b(n+j,j)/b(2*j,j): j in [0..n]]) >;
[A339000(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Jul 31 2024

T[n_, k_]:= With[{B=Binomial}, B[n,k]*Sum[B[n,k-j]*B[n+j,j]/B[2*j,j], {j,0,n}]];
Table[T[n,k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Jul 31 2024 *)

T(n,m):=(binomial(n,m))*sum(((binomial(n,m-k))*(binomial(n+k,k)) )/(binomial(2*k,k)),k,0,n);

b=binomial
def A339000(n,k): return b(n,k)*sum(b(n,k-j)*b(n+j,j)//b(2*j,j) for j in range(n+1))
flatten([[A339000(n,k) for k in range(n+1)] for n in range(13)]) # G. C. Greubel, Jul 31 2024

G.f.: A008459(x,y)/(1-x*y*A008459(x,y)^2).

T(n,n) = Fibonacci(2*n+1).

cp's OEIS Frontend

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A339000 Triangle read by rows: T(n, k) = C(n, k)*Sum_{j=0..n} C(n, k-j)*C(n+j, j)/C(2*j, j).

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A339000 Triangle read by rows: T(n, k) = C(n, k)Sum_{j=0..n} C(n, k-j)C(n+j, j)/C(2*j, j).