A020918 -id:A020918 - 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)

A054335 A convolution triangle of numbers based on A000984 (central binomial coefficients of even order).

Original entry on oeis.org

1, 2, 1, 6, 4, 1, 20, 16, 6, 1, 70, 64, 30, 8, 1, 252, 256, 140, 48, 10, 1, 924, 1024, 630, 256, 70, 12, 1, 3432, 4096, 2772, 1280, 420, 96, 14, 1, 12870, 16384, 12012, 6144, 2310, 640, 126, 16, 1, 48620, 65536, 51480, 28672, 12012, 3840, 924, 160, 18, 1

Offset: 0

Wolfdieter Lang, Mar 13 2000

In the language of the Shapiro et al. reference (given in A053121) such a lower triangular (ordinary) convolution array, considered as a matrix, belongs to the Bell-subgroup of the Riordan-group. The g.f. for the row polynomials p(n,x) (increasing powers of x) is 1/(sqrt(1-4*z)-x*z).
The column sequences are for m=0..20: A000984, A000302 (powers of 4), A002457, A002697, A002802, A038845, A020918, A038846, A020920, A040075, A020922, A045543, A020924, A054337, A020926, A054338, A020928, A054339, A020930, A054340, A020932.
Riordan array (1/sqrt(1-4*x),x/sqrt(1-4*x)). - Paul Barry, May 06 2009
The matrix inverse is apparently given by deleting the leftmost column from A206022. - R. J. Mathar, Mar 12 2013

			Triangle begins:
    1;
    2,    1;
    6,    4,   1;
   20,   16,   6,   1;
   70,   64,  30,   8,  1;
  252,  256, 140,  48, 10,  1;
  924, 1024, 630, 256, 70, 12, 1; ...
Fourth row polynomial (n=3): p(3,x) = 20 + 16*x + 6*x^2 + x^3.
From _Paul Barry_, May 06 2009: (Start)
Production matrix begins
    2,   1;
    2,   2,  1;
    0,   2,  2,  1;
   -2,   0,  2,  2,  1;
    0,  -2,  0,  2,  2,  1;
    4,   0, -2,  0,  2,  2, 1;
    0,   4,  0, -2,  0,  2, 2, 1;
  -10,   0,  4,  0, -2,  0, 2, 2, 1;
    0, -10,  0,  4,  0, -2, 0, 2, 2, 1; (End)

G. C. Greubel, Rows n = 0..100 of triangle, flattened
Paul Barry, Embedding structures associated with Riordan arrays and moment matrices, arXiv preprint arXiv:1312.0583 [math.CO], 2013.
Milan Janjić, Pascal Matrices and Restricted Words, J. Int. Seq., Vol. 21 (2018), Article 18.5.2.

Cf. A000984, A035324, A054336.

Row sums: A026671.

T:= function(n, k)
    if k mod 2=0 then return Binomial(2*n-k, n-Int(k/2))*Binomial(n-Int(k/2),Int(k/2))/Binomial(k,Int(k/2));
    else return 4^(n-k)*Binomial(n-Int((k-1)/2)-1, Int((k-1)/2));
    fi;
  end;
Flat(List([0..10], n-> List([0..n], k-> T(n, k) ))); # G. C. Greubel, Jul 20 2019

T:= func< n, k | (k mod 2) eq 0 select Binomial(2*n-k, n-Floor(k/2))* Binomial(n-Floor(k/2),Floor(k/2))/Binomial(k,Floor(k/2)) else 4^(n-k)*Binomial(n-Floor((k-1)/2)-1, Floor((k-1)/2)) >;
[[T(n,k): k in [0..n]]: n in [0..10]]; // G. C. Greubel, Jul 20 2019

A054335 := proc(n,k)
    if k <0 or k > n then
        0 ;
    elif type(k,odd) then
        kprime := floor(k/2) ;
        binomial(n-kprime-1,kprime)*4^(n-k) ;
    else
        kprime := k/2 ;
        binomial(2*n-k,n-kprime)*binomial(n-kprime,kprime)/binomial(k,kprime) ;
    end if;
end proc: # R. J. Mathar, Mar 12 2013
# Uses function PMatrix from A357368. Adds column 1,0,0,0,... to the left.
PMatrix(10, n -> binomial(2*(n-1), n-1)); # Peter Luschny, Oct 19 2022

Flatten[ CoefficientList[#1, x] & /@ CoefficientList[ Series[1/(Sqrt[1 - 4*z] - x*z), {z, 0, 9}], z]] (* or *)
a[n_, k_?OddQ] := 4^(n-k)*Binomial[(2*n-k-1)/2, (k-1)/2]; a[n_, k_?EvenQ] := (Binomial[n-k/2, k/2]*Binomial[2*n-k, n-k/2])/Binomial[k, k/2]; Table[a[n, k], {n, 0, 9}, {k, 0, n}] // Flatten (* Jean-François Alcover, Sep 08 2011, updated Jan 16 2014 *)

T(n, k) = if(k%2==0, binomial(2*n-k, n-k/2)*binomial(n-k/2,k/2)/binomial(k,k/2), 4^(n-k)*binomial(n-(k-1)/2-1, (k-1)/2));
for(n=0,10, for(k=0,n, print1(T(n,k), ", "))) \\ G. C. Greubel, Jul 20 2019

def T(n, k):
    if (mod(k,2)==0): return binomial(2*n-k, n-k/2)*binomial(n-k/2,k/2)/binomial(k,k/2)
    else: return 4^(n-k)*binomial(n-(k-1)/2-1, (k-1)/2)
[[T(n,k) for k in (0..n)] for n in (0..10)] # G. C. Greubel, Jul 20 2019

a(n, 2*k+1) = binomial(n-k-1, k)*4^(n-2*k-1), a(n, 2*k) = binomial(2*(n-k), n-k)*binomial(n-k, k)/binomial(2*k, k), k >= 0, n >= m >= 0; a(n, m) := 0 if n

Column recursion: a(n, m)=2*(2*n-m-1)*a(n-1, m)/(n-m), n>m >= 0, a(m, m) := 1.

G.f. for column m: cbie(x)*(x*cbie(x))^m, with cbie(x) := 1/sqrt(1-4*x).

G.f.: 1/(1-x*y-2*x/(1-x/(1-x/(1-x/(1-x/(1-... (continued fraction). - Paul Barry, May 06 2009

Sum_{k>=0} T(n,2*k)*(-1)^k*A000108(k) = A000108(n+1). - Philippe Deléham, Jan 30 2012

Sum_{k=0..floor(n/2)} T(n-k,n-2*k) = A098615(n). - Philippe Deléham, Feb 01 2012

T(n,k) = 4*T(n-1,k) + T(n-2,k-2) for k>=1. - Philippe Deléham, Feb 02 2012

Vertical recurrence: T(n,k) = 1*T(n-1,k-1) + 2*T(n-2,k-1) + 6*T(n-3,k-1) + 20*T(n-4,k-1) + ... for k >= 1 (the coefficients 1, 2, 6, 20, ... are the central binomial coefficients A000984). - Peter Bala, Oct 17 2015

A020920 Expansion of 1/(1-4*x)^(9/2).

Original entry on oeis.org

1, 18, 198, 1716, 12870, 87516, 554268, 3325608, 19122246, 106234700, 573667380, 3024791640, 15628090140, 79342611480, 396713057400, 1957117749840, 9540949030470, 46021048264620, 219878341708740, 1041528987041400, 4895186239094580, 22844202449108040

Offset: 0

N. J. A. Sloane

Also convolution of A000984 with A038846, also convolution of A000302 with A020918, also convolution of A002457 with A038845, also convolution of A002697 with A002802. - Rui Duarte, Oct 08 2011

Vincenzo Librandi, Table of n, a(n) for n = 0..200

Cf. A000302, A000332, A000984, A001622, A002457, A002697, A002802, A020918, A038845, A038846, A046521 (fifth column).

List([0..30], n-> Binomial(n+4, 4)*Binomial(2*(n+4), n+4)/70) # G. C. Greubel, Jul 20 2019

[(2*n+7)*(2*n+5)*(2*n+3)*(2*n+1)*Binomial(2*n, n)/105: n in [0..30]]; // Vincenzo Librandi, Jul 05 2013

seq(binomial(2*n+8, n+4)*binomial(n+4, n)/70, n=0..30); # Zerinvary Lajos, May 05 2007

CoefficientList[Series[1/(1-4x)^(9/2), {x,0,30}], x] (* Vincenzo Librandi, Jul 05 2013 *)

vector(30, n, n--; m=n+4; binomial(m, 4)*binomial(2*m, m)/70) \\ G. C. Greubel, Jul 20 2019

[binomial(n+4, 4)*binomial(2*(n+4), n+4)/70 for n in (0..30)] # G. C. Greubel, Jul 20 2019

a(n) = binomial(n+4, 4)*A000984(n+4)/A000984(4), where A000984 are the central binomial coefficients. - Wolfdieter Lang
a(n) = Sum_{ a+b+c+d+e+f+g+h+i=n} f(a)*f(b)*f(c)*f(d)*f(e)*f(f)*f(g) *f(h)*f(i) with f(n)=A000984(n). - Philippe Deléham, Jan 22 2004
a(n) = A000332(n+4)*A000984(n+4)/70. - Zerinvary Lajos, May 05 2007
From Rui Duarte, Oct 08 2011: (Start)
a(n) = ((2n+7)(2n+5)(2n+3)(2n+1)/(7*5*3*1)) * binomial(2n, n).
a(n) = binomial(2n+8, 8) * binomial(2n, n) / binomial(n+4, 4).
a(n) = binomial(n+4, 4) * binomial(2n+8, n+4) / binomial(8, 4). (End)
Boas-Buck recurrence: a(n) = (18/n)*Sum_{k=0..n-1} 4^(n-k-1)*a(k), n >= 1, a(0) = 1. Proof from a(n) = A046521(n+4, 4). See a comment there.
From Amiram Eldar, Mar 25 2022: (Start)
Sum_{n>=0} 1/a(n) = 1148/5 - 42*sqrt(3)*Pi.
Sum_{n>=0} (-1)^n/a(n) = 700*sqrt(5)*log(phi) - 11284/15, where phi is the golden ratio (A001622). (End)

A068763 Irregular triangle of the Fibonacci polynomials of A011973 multiplied diagonally by the Catalan numbers.

Original entry on oeis.org

1, 1, 1, 2, 2, 5, 6, 1, 14, 20, 6, 42, 70, 30, 2, 132, 252, 140, 20, 429, 924, 630, 140, 5, 1430, 3432, 2772, 840, 70, 4862, 12870, 12012, 4620, 630, 14, 16796, 48620, 51480, 24024, 4620, 252, 58786, 184756, 218790

Offset: 0

Wolfdieter Lang, Mar 04 2002

The row length sequence of this array is [1,2,2,3,3,4,4,5,5,...] = A008619(n+2), n>=0.
The row polynomials p(n,x) := Sum_{m=0..floor((n+1)/2)} a(n,m)*x^m produce, for x = (b-a^2)/a^2 (not 0), the two parameter family of sequences K(a,b; n) := (a^(n+1))*p(n,(b-a^2)/a^2) with g.f. K(a,b; x) := (1-sqrt(1-4*x*(a+x*(b-a^2))))/(2*x).
Some members are: K(1,1; n)=A000108(n) (Catalan), K(1,2; n)=A025227(n-1), K(2,1; n)=A025228(n-1), K(1,3; n)=A025229(n-1), K(3,1; n)=A025230(n-1). For a=b=2..10 the sequences K(a,a; n)/a are A068764-A068772.
The column sequences (without leading 0's) are: A000108 (Catalan), A000984 (central binomial), A002457, 2*A002802, 5*A020918, 14*A020920, 42*A020922, ...
a(n,m) is the number of ways to designate exactly m cherries over all binary trees with n internal nodes.  A cherry is an internal node whose descendants are both external nodes.  Cf. A091894 which gives the number of binary trees with m cherries. - Geoffrey Critzer, Jul 24 2020
This irregular triangle is essentially that of A011973 with its diagonals multiplied by the Catalan numbers of A000108. The diagonals of this triangle are then rows of the Pascal matrix A007318 multiplied by the Catalan numbers. - Tom Copeland, Dec 23 2023

			The irregular triangle begins:
   n\m    0     1     2     3    4   5
   0:     1
   1:     1     1
   2:     2     2
   3:     5     6     1
   4:    14    20     6
   5:    42    70    30     2
   6:   132   252   140    20
   7:   429   924   630   140    5
   8:  1430  3432  2772   840   70
   9:  4862 12870 12012  4620  630  14
  10: 16796 48620 51480 24024 4620 252
  ...
p(3,x) = 5 + 6*x + x^2.

P. Flajolet and R. Sedgewick, Analytic Combinatorics, 2009; see page 213.
Tad White, Quota Trees, arXiv:2401.01462 [math.CO], 2024. See p. 20.

Cf. A025227(n-1) (row sums).
Cf. A000007(n) (alternating row sums).
Cf. A000108, A001263, A007318, A011973, A033832, A034867, A133437 and A134264.

nn = 10; b[z_] := (1 - Sqrt[1 - 4 z])/(2 z);Map[Select[#, # > 0 &] &,
CoefficientList[Series[v b[v z] /. v -> (1 + u z ), {z, 0, nn}], {z, u}]] // Grid (* Geoffrey Critzer, Jul 24 2020 *)

a(n, m) = binomial(n+1-m, m)*C(n-m) if 0 <= m <= floor((n+1)/2), otherwise 0, with C(n) := A000108(n) (Catalan).
G.f. for column m=1, 2, ...: (x^(2*m-1))*C(m-1)/(1-4*x)^((2*m-1)/2); m=0: c(x), g.f. for A000108 (Catalan).
G.f. for row polynomials p(n, x): c(z) + x*z*c(x*(z^2)/(1-4*z))/sqrt(1-4*z) = (1-sqrt(1-4*z*(1+x*z)))/(2*z), where c(x) is the g.f. of A000108 (Catalan).
G.f. for triangle: (1 - sqrt(1 - 4*x (1 + y*x)))/(2*x). - Geoffrey Critzer, Jul 24 2020
The alternating row sums are {1,repeat 0} = {A000007(n)}{n>=0}. From the second comment p(n, x=-1) = K(1,0; n), with g.f. K(1,0; x) = 1. Or from the g.f. of the triangle with y = -1. Thanks to Aaron Li for asking for a proof. - _Wolfdieter Lang, Jan 21 2023
The series expansion of f(x) = (1 + 2sx - sqrt(1 + 4sx + 4d^2x^2))/(2x) at x = 0 is (s^2 - d^2) x + (2 d^2s - 2 s^3) x^2 + (d^4 - 6 d^2 s^2 + 5 s^4) x^3  + (-6 d^4 s + 20 d^2 s^3 - 14 s^5) x^4 + ..., containing the coefficients of this array. With s = (a+b)/2 and d = (a-b)/2, then f(x)/ab = g(x) = (1 + (a+b)x - sqrt((1+(a+b)x)^2 - 4abx^2))/(2abx) = x - (a + b) x^2 + (a^2 + 3 a b + b^2) x^3 - (a^3 + 6 a^2 b + 6 a b^2 + b^3) x^4 + ..., containing the Narayana polynomials of A001263, which can be simply transformed into A033282. The compositional inverse about the origin of g(x) is g^(-1)(x) = x/((1-ax)(1-bx)) = x/((1-(s+d)x)(1-(s-d)x)) = x + (a + b) x^2 + (a^2 + a b + b^2) x^3 + (a^3 + a^2 b + a b^2 + b^3) x^4 + ..., containing the complete homogeneous symmetric polynomials h_n(a,b) = (a^n - b^n)/(a-b), which are the polynomials of A034867 when expressed in s and d, e.g., ((s + d)^7 - (s - d)^7)/(2 d) = d^6 + 21 d^4 s^2 + 35 d^2 s^4 + 7 s^6. A133437 and A134264 for compositional inversion of o.g.f.s can be used to relate the sets of polynomials above. - Tom Copeland, Nov 28 2023

Title changed by Tom Copeland, Dec 23 2023

A020922 Expansion of 1/(1-4*x)^(11/2).

Original entry on oeis.org

1, 22, 286, 2860, 24310, 184756, 1293292, 8498776, 53117350, 318704100, 1848483780, 10418726760, 57302997180, 308554600200, 1630931458200, 8480843582640, 43464323361030, 219878341708740, 1099391708543700, 5439095821216200, 26651569523959380, 129450480544945560

Offset: 0

N. J. A. Sloane

Also convolution of A000984 with A040075, also convolution of A000302 with A020920, also convolution of A002457 with A038846, also convolution of A002697 with A020918, also convolution of A002802 with A038845. - Rui Duarte, Oct 08 2011

Vincenzo Librandi, Table of n, a(n) for n = 0..200

Cf. A000302, A000984, A001622, A002457, A002697, A002802, A020918, A020920, A038845, A038846, A040075, A046521 (sixth column).

List([0..30], n-> Binomial(n+5, 5)*Binomial(2*n+10, n+5)/252); # G. C. Greubel, Jul 20 2019

[(2*n+9)*(2*n+7)*(2*n+5)*(2*n+3)*(2*n+1)*Binomial(2*n, n)/945: n in [0..30]] // Vincenzo Librandi, Jul 05 2013

CoefficientList[Series[1/(1-4x)^(11/2), {x,0,30}], x] (* Vincenzo Librandi, Jul 05 2013 *)

vector(30, n, n--; m=n+5; binomial(m, 5)*binomial(2*m, m)/252) \\ G. C. Greubel, Jul 20 2019

[binomial(n+5, 5)*binomial(2*n+10, n+5)/252 for n in (0..30)] # G. C. Greubel, Jul 20 2019

a(n) = binomial(n+5, 5)*A000984(n+5)/A000984(5), where A000984 are central binomial coefficients. - Wolfdieter Lang
From Rui Duarte, Oct 08 2011: (Start)
a(n) = ((2n+9)(2n+7)(2n+5)(2n+3)(2n+1)/(9*7*5*3*1)) * binomial(2n, n).
a(n) = binomial(2n+10, 10) * binomial(2n, n) / binomial(n+5, 5).
a(n) = binomial(n+5, 5) * binomial(2n+10, n+5) / binomial(10, 5).
a(n) = Sum_{ i_1+i_2+i_3+i_4+i_5+i_6+i_7+i_8+i_9+i_10+i_11 = n } f(i_1)* f(i_2)*f(i_3)*f(i_4)*f(i_5)*f(i_6)*f(i_7)*f(i_8)*f(i_9)*f(i_10)*f(i_11) with f(k)=A000984(k). (End)
Boas-Buck recurrence: a(n) = (22/n)*Sum_{k=0..n-1} 4^(n-k-1)*a(k), n >= 1, a(0) = 1. Proof from a(n) = A046521(n+5, 5). See a comment there. - Wolfdieter Lang, Aug 10 2017
From Amiram Eldar, Mar 25 2022: (Start)
Sum_{n>=0} 1/a(n) = 162*sqrt(3)*Pi - 30816/35.
Sum_{n>=0} (-1)^n/a(n) = 4500*sqrt(5)*log(phi) - 33888/7, where phi is the golden ratio (A001622). (End)

A002803 a(n) = (2n+4)!/(4!n!(n+1)!).

Original entry on oeis.org

1, 15, 140, 1050, 6930, 42042, 240240, 1312740, 6928350, 35565530, 178474296, 878850700, 4259045700, 20359174500, 96172862400, 449608131720, 2082743551350, 9569730173850, 43651400793000, 197809768856700

Offset: 0

N. J. A. Sloane

C. Jordan, Calculus of Finite Differences. Budapest, 1939, p. 449.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

T. D. Noe, Table of n, a(n) for n = 0..200
T. R. S. Walsh and A. B. Lehman, Counting rooted maps by genus. III: Nonseparable maps, J. Combinatorial Theory Ser. B 18 (1975), 222-259.

Cf. A020918.

Table[(2*n+4)!/(4!*n!*(n+1)!), {n, 0, 20}] (* Vladimir Joseph Stephan Orlovsky, Dec 13 2008 *)

G.f.: 2F1(5/2,3;2;4x) =(1+x)/(1-4x)^(7/2). - R. J. Mathar, Aug 09 2015
a(n) = A020918(n)+A020918(n-1). - R. J. Mathar, Aug 09 2015
D-finite with recurrence n*(n+1)*a(n) -2*(n+2)*(2*n+3)*a(n-1)=0. - R. J. Mathar, Feb 08 2021

Showing 1-10 of 15 results. Next

Also convolution of A001791(n+1), n >= 0, with A038845; also convolution of A008549(n+1), n >= 0, with A002802; also convolution of A029760 with A002697; also convolution of A038806(n+1), n >= 0, with A002457; also convolution of A038836 with A000302 (powers of 4); also convolution of A041001 with A000984 (central binomial coefficients).

Also minimal 3-covers of a labeled n-set that cover 3 points of that set uniquely (if offset is 3). Cf. A057524 for unlabeled case. - Vladeta Jovovic, Sep 02 2000
Also convolution of A020918 with A000984 (central binomial coefficients).
Let M=[1,0,0,i;0,1,i,0;0,i,1,0;i,0,0,1], i=sqrt(-1). Then 1/det(I-xM) = 1/(1-4x)^4. - Paul Barry, Apr 27 2005
With a different offset, number of n-permutations (n=4) of 5 objects u, v, w, z, x with repetition allowed, containing exactly three u's. Example: a(1)=16 because we have uuuv, uuvu, uvuu, vuuu, uuuw, uuwu, uwuu, wuuu, uuuz, uuzu, uzuu, zuuu, uuux, uuxu, uxuu and xuuu. - Zerinvary Lajos, May 19 2008
From A152818. a(n) = A006044/6. - Paul Curtz, Jan 07 2009
Also convolution of A000302 with A038845, also convolution of A002457 with A002802, also convolution of A002697. - Rui Duarte, Oct 08 2011

Also convolution of A020920 with A000984 (central binomial coefficients).
With a different offset, number of n-permutations (n=5) of 5 objects u, v, w, z, x with repetition allowed, containing exactly four (4)u's. Example: a(1)=20 because we have uuuuv, uuuvu, uuvuu, uvuuu, vuuuu, uuuuw, uuuwu, uuwuu, uwuuu, wuuuu, uuuuz, uuuzu, uuzuu, uzuuu, zuuuu, uuuux, uuuxu, uuxuu, uxuuu and xuuuu. - Zerinvary Lajos, May 19 2008
Also convolution of A000302 with A038846, also convolution of A002457 with A020918, also convolution of A002697 with A038845, also convolution of A002802 with A002802. [Rui Duarte, Oct 08 2011]

cp's OEIS Frontend

A045724 Convolution of Catalan numbers A000108 with A020918.

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A041005 Convolution of Catalan numbers A000108(n+1), n >= 0, with A020918.

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

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A038846 4-fold convolution of A000302 (powers of 4); expansion of g.f. 1/(1-4*x)^4.

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A040075 5-fold convolution of A000302 (powers of 4); expansion of 1/(1-4*x)^5.

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A054335 A convolution triangle of numbers based on A000984 (central binomial coefficients of even order).

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A002803 a(n) = (2n+4)!/(4!n!(n+1)!).