A046521 -id:A046521 - cp's OEIS frontend

A129818 Riordan array (1/(1+x), x/(1+x)^2), inverse array is A039599.

1, -1, 1, 1, -3, 1, -1, 6, -5, 1, 1, -10, 15, -7, 1, -1, 15, -35, 28, -9, 1, 1, -21, 70, -84, 45, -11, 1, -1, 28, -126, 210, -165, 66, -13, 1, 1, -36, 210, -462, 495, -286, 91, -15, 1, -1, 45, -330, 924, -1287, 1001, -455, 120, -17, 1, 1, -55, 495, -1716, 3003, -3003, 1820, -680, 153, -19, 1

Offset: 0

Philippe Deléham, Jun 09 2007

This sequence is up to sign the same as A129818. - T. D. Noe, Sep 30 2011
Row sums: A057078. - Philippe Deléham, Jun 11 2007
Subtriangle of the triangle given by (0, -1, 0, -1, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (1, 0, 1, -1, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Mar 19 2012
This triangle provides the coefficients of powers of x^2 for the even-indexed Chebyshev S polynomials (see A049310): S(2*n,x) = Sum_{k=0..n} T(n,k)*x^(2*k), n >= 0. - Wolfdieter Lang, Dec 17 2012
If L(x^n) := C(n) = A000108(n) (Catalan numbers), then the polynomials P_n(x) := Sum_{k=0..n} T(n,k)*x^k are orthogonal with respect to the inner product given by (f(x),g(x)) := L(f(x)*g(x)). - Michael Somos, Jan 03 2019

			Triangle T(n,k) begins:
  n\k  0   1    2     3     4     5    6    7    8   9 10 ...
   0:  1
   1: -1   1
   2:  1  -3    1
   3: -1   6   -5     1
   4:  1 -10   15    -7     1
   5: -1  15  -35    28    -9     1
   6:  1 -21   70   -84    45   -11    1
   7: -1  28 -126   210  -165    66  -13    1
   8:  1 -36  210  -462   495  -286   91  -15    1
   9: -1  45 -330   924 -1287  1001 -455  120  -17   1
  10:  1 -55  495 -1716  3003 -3003 1820 -680  153 -19  1
  ... Reformatted by _Wolfdieter Lang_, Dec 17 2012
Recurrence from the A-sequence A115141:
15 = T(4,2) = 1*6 + (-2)*(-5) + (-1)*1.
(0, -1, 0, -1, 0, 0, ...) DELTA (1, 0, 1, -1, 0, 0, ...) begins:
  1
  0,  1
  0, -1,   1
  0,  1,  -3,   1
  0, -1,   6,  -5,  1
  0,  1, -10,  15, -7,  1
  0, -1,  15, -35, 28, -9, 1. - _Philippe Deléham_, Mar 19 2012
Row polynomial for n=3 in terms of x^2: S(6,x) = -1 + 6*x^2 -5*x^4 + 1*x^6, with Chebyshev's S polynomial. See a comment above. - _Wolfdieter Lang_, Dec 17 2012
Boas-Buck type recurrence: -35 = T(5,2) = (5/3)*(-1*1 +1*(-5) - 1*15) = -3*7 = -35. - _Wolfdieter Lang_, Jun 03 2020

Vincenzo Librandi, Rows n = 1..101, flattened
Paul Barry, A Catalan Transform and Related Transformations on Integer Sequences, Journal of Integer Sequences, Vol. 8 (2005), Article 05.4.5.
Paul Barry and Aoife Hennessy, The Euler-Seidel Matrix, Hankel Matrices and Moment Sequences, J. Int. Seq. 13 (2010) # 10.8.2, example 15.
Aoife Hennessy, A Study of Riordan Arrays with Applications to Continued Fractions, Orthogonal Polynomials and Lattice Paths, Ph. D. Thesis, Waterford Institute of Technology, Oct. 2011.

Cf. A039599, A085478, A123970, A321620.

# The function RiordanSquare is defined in A321620.
RiordanSquare((1 - sqrt(1 - 4*x))/(2*x), 10):
LinearAlgebra[MatrixInverse](%); # Peter Luschny, Jan 04 2019

max = 10; Flatten[ CoefficientList[#, y] & /@ CoefficientList[ Series[ (1 + x)/(1 + (2 - y)*x + x^2), {x, 0, max}], x]] (* Jean-François Alcover, Sep 29 2011, after Wolfdieter Lang *)

@CachedFunction
def A129818(n,k):
    if n< 0: return 0
    if n==0: return 1 if k == 0 else 0
    h = A129818(n-1,k) if n==1 else 2*A129818(n-1,k)
    return A129818(n-1,k-1) - A129818(n-2,k) - h
for n in (0..9): [A129818(n,k) for k in (0..n)] # Peter Luschny, Nov 20 2012

T(n,k) = (-1)^(n-k)*A085478(n,k) = (-1)^(n-k)*binomial(n+k,2*k).
Sum_{k=0..n} T(n,k)*A000531(k) = n^2, with A000531(0)=0. - Philippe Deléham, Jun 11 2007
Sum_{k=0..n} T(n,k)*x^k = A033999(n), A057078(n), A057077(n), A057079(n), A005408(n), A002878(n), A001834(n), A030221(n), A002315(n), A033890(n), A057080(n), A057081(n), A054320(n), A097783(n), A077416(n), A126866(n), A028230(n+1) for x = 0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16, respectively. - Philippe Deléham, Nov 19 2009
O.g.f.: (1+x)/(1+(2-y)*x+x^2). - Wolfdieter Lang, Dec 15 2010
O.g.f. column k with leading zeros (Riordan array, see NAME): (1/(1+x))*(x/(1+x)^2)^k, k >= 0. - Wolfdieter Lang, Dec 15 2010
From Wolfdieter Lang, Dec 20 2010: (Start)
Recurrences from the Z- and A-sequences for Riordan arrays. See the W. Lang link under A006232 for details and references.
T(n,0) = -1*T(n-1,0), n >= 1, from the o.g.f. -1 for the Z-sequence (trivial result).
T(n,k) = Sum_{j=0..n-k} A(j)*T(n-1,k-1+j), n >= k >= 1, with A(j):= A115141(j) = [1,-2,-1,-2,-5,-14,...], j >= 0 (o.g.f. 1/c(x)^2 with the A000108 (Catalan) o.g.f. c(x)). (End)
T(n,k) = (-1)^n*A123970(n,k). - Philippe Deléham, Feb 18 2012
T(n,k) = -2*T(n-1,k) + T(n-1,k-1) - T(n-2,k), T(0,0) = T(1,1) = 1, T(1,0) = -1, T(n,k) = 0 if k < 0 or if k > n. - Philippe Deléham, Mar 19 2012
A039599(m,n) = Sum_{k=0..n} T(n,k) * C(k+m) where C(n) are the Catalan numbers. - Michael Somos, Jan 03 2019
Equals the matrix inverse of the Riordan square (cf. A321620) of the Catalan numbers. - Peter Luschny, Jan 04 2019
Boas-Buck type recurrence for column k >= 0 (see Aug 10 2017 comment in A046521 with references): T(n,k) = ((1 + 2*k)/(n - k))*Sum_{j = k..n-1} (-1)^(n-j)*T(j,k), with input T(n,n) = 1, and T(n,k) = 0 for n < k. - Wolfdieter Lang, Jun 03 2020

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)

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)

A001780 Expansion of 1/((1+x)(1-x)^9).

Original entry on oeis.org

1, 8, 37, 128, 367, 920, 2083, 4352, 8518, 15792, 27966, 47616, 78354, 125136, 194634, 295680, 439791, 641784, 920491, 1299584, 1808521, 2483624, 3369301, 4519424, 5998876, 7885280, 10270924, 13264896

Offset: 0

N. J. A. Sloane

Vincenzo Librandi, Table of n, a(n) for n = 0..2000
Jia Huang, Partially Palindromic Compositions, J. Int. Seq. (2023) Vol. 26, Art. 23.4.1. See pp. 4, 17.
Index entries for linear recurrences with constant coefficients, signature (8,-27,48,-42,0,42,-48,27,-8,1).

Cf. A001769, A158454 (signed column k=4), A001779 (first differences), A169796 (binomial trans.).

Vec(1/(1+x)/(1-x)^9+O(x^99)) \\ Charles R Greathouse IV, Apr 18 2012

a(n) = 431*n/168 + (-1)^n/512 + 391*n^3/288 + 26011*n^2/10080 + 797*n^4/1920 + 11*n^5/144 + n^6/120 + n^7/2016 + n^8/80640 + 511/512. - R. J. Mathar, Mar 15 2011
Boas-Buck recurrence: a(n) = (1/n)*Sum_{p=0..n-1} (9 + (-1)^(n-p))*a(p), n >= 1, a(0) = 1. See the Boas-Buck comment in A046521 (here for the unsigned column k = 4 with offset 0). - Wolfdieter Lang, Aug 10 2017
a(n)+a(n+1) = A000581(n+9) . - R. J. Mathar, Jan 06 2021

A001786 Expansion of 1/((1+x)*(1-x)^11).

Original entry on oeis.org

1, 10, 56, 230, 771, 2232, 5776, 13672, 30086, 62292, 122464, 230252, 416394, 727672, 1233584, 2035176, 3276559, 5159726, 7963384, 12066626, 17978389, 26373776, 38138464, 54422576, 76705564, 106873832, 147313024, 201017112, 271716644, 364028752, 483631776, 637467632, 833975341

Offset: 0

N. J. A. Sloane

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Jia Huang, Partially Palindromic Compositions, J. Int. Seq. (2023) Vol. 26, Art. 23.4.1. See pp. 4, 17.
Index entries for linear recurrences with constant coefficients, signature (10,-44,110,-165,132,0,-132,165,-110,44,-10,1).

Cf. A001780, A158454 (signed column k=5).

Eleventh column of A112465.

R:=PowerSeriesRing(Integers(), 50);
Coefficients(R!( 1/((1+x)*(1-x)^11) )); // G. C. Greubel, Apr 20 2025

CoefficientList[Series[1/((1+x)(1-x)^11),{x,0,50}],x] (* Vincenzo Librandi, Feb 24 2012 *)
LinearRecurrence[{10,-44,110,-165,132,0,-132,165,-110,44,-10,1},{1,10,56,230,771,2232, 5776,13672,30086,62292,122464,230252},30] (* Harvey P. Dale, Oct 22 2015 *)

def A001786_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P( 1/((1+x)*(1-x)^11) ).list()
print(A001786_list(50)) # G. C. Greubel, Apr 20 2025

Boas-Buck recurrence: a(n) = (1/n)*Sum_{p=0..n-1} (11 + (-1)^(n-p))*a(p), n >= 1, a(0) = 1. See the Boas-Buck comment in A046521 (here for the unsigned column k = 5 with offset 0). - Wolfdieter Lang, Aug 10 2017

A104578 A Padovan convolution triangle.

Original entry on oeis.org

1, 0, 1, 1, 0, 1, 1, 2, 0, 1, 1, 2, 3, 0, 1, 2, 3, 3, 4, 0, 1, 2, 6, 6, 4, 5, 0, 1, 3, 7, 12, 10, 5, 6, 0, 1, 4, 12, 16, 20, 15, 6, 7, 0, 1, 5, 17, 30, 30, 30, 21, 7, 8, 0, 1, 7, 24, 45, 60, 50, 42, 28, 8, 9, 0, 1, 9, 36, 70, 95, 105, 77, 56, 36, 9, 10, 0, 1, 12, 50, 111, 160, 175, 168, 112, 72

Offset: 0

Paul Barry, Mar 16 2005

A Padovan convolution triangle. See A000931 for the Padovan sequence.
Row sums are tribonacci numbers A000073(n+2). Antidiagonal sums are A008346. The first columns are A000931(n+3), A228577.
From Wolfdieter Lang, Oct 30 2018: (Start)
The alternating row sums give A001057(n+1), for n >= 0.
The inverse of this Riordan triangle is given in A319203.
The row polynomials R(n, x) := Sum_{k=0..n} T(n, k)*x^k, with R(-1, x) = 0, appear in the Cayley-Hamilton formula for nonnegative powers of a 3 X 3 matrix with Det M = sigma(3;3) = x1*x2*x3 = +1, sigma(3; 2) := x1*x2 + x1*x*3 + x2*x^3  = -1 and Tr M = sigma(3; 1) = x1 + x2 = x, where x1, x2, and  x3, are the eigenvalues of M, and sigma the elementary symmetric functions, as M^n = R(n-2, x)*M^2  + (R(n-3, x) + R(n-4, x))*M + R(n-3, x)*1_3, for n >= 3, where M^0 = 1_3 is the 3 X 3 unit matrix.
For the Cayley-Hamilton formula for 3 X 3 matrices with Det M = +1, sigma(3,2) = +1 and Tr(M) = x see A321196.
(End)

			From _Wolfdieter Lang_, Oct 30 2018: (Start)
The triangle T begins:
    n\k   0  1  2  3  4  5  6  7  8  9 10 ...
    --------------------------------------
    0:    1
    1:    0  1
    2:    1  0  1
    3:    1  2  0  1
    4:    1  2  3  0  1
    5:    2  3  3  4  0  1
    6:    2  6  6  4  5  0  1
    7:    3  7 12 10  5  6  0  1
    8:    4 12 16 20 15  6  7  0  1
    9:    5 17 30 30 30 21  7  8  0  1
   10:    7 24 45 60 50 42 28  8  9  0  1
   ...
Cayley-Hamilton formula for the tribonacci Q-matrix TQ(x) =[[x,1,1], [1,0,0], [0,1,0]] with Det(TQ) = +1, sigma(3, 2) = -1, and Tr(TQ) = x. For n = 3: TQ(x)^3 = R(1, x)*TQ(x)^2  + (R(0 x) + R(-1, x))*TQ(x) + R(0, x)*1_3 = x*TQ(x)^2 + TQ(x) + 1_3. For x = 1 see also A058265 (powers of the tribonacci constant).
Recurrence: T(6, 2) = T(5, 1) + T(4, 2) + T(3, 2) = 3 + 3 + 0 = 6.
Z- and A- recurrence with A319202 = {1, 0, 1, 1, -1, -3, 0, ...}:
  T(5, 0) = 0*1 + 1*2 + 1*3 + (-1)*0 + (-3)*1 = 2; T(5,2) = 1*2 + 0*3 + 1*0 + 1*1 = 3.
Boas-Buck type recurrence with b = {0, 2, 3, ...}: T(5, 2) = ((1+2)/(5-2)) * (3*1 + 2*0 + 0*3) = 1*3 = 3.
(End)

Tomislav Došlic and Luka Podrug, Tilings of a Honeycomb Strip and Higher Order Fibonacci Numbers, arXiv:2203.11761 [math.CO], 2022.
Milan Janjić, Words and Linear Recurrences, J. Int. Seq. 21 (2018), #18.1.4.

Cf. A000073, A000931, A001057, A001608, A008346, A058265, A228577, A319202, A319203 (inverse), A321196.

T[n_, k_] /; 0 <= k <= n := T[n, k] = T[n-1, k-1] + T[n-2, k] + T[n-3, k]; T[0, 0] = 1; T[, ] = 0; Table[T[n, k], {n, 0, 12}, {k, 0, n}] (* Jean-François Alcover, Jun 11 2019 *)

# uses[riordan_array from A256893]
riordan_array( 1/(1 - x^2 - x^3), x/(1 - x^2 - x^3), 8) # Peter Luschny, Nov 09 2018

Riordan array (1/(1 - x^2 - x^3), x/(1 - x^2 - x^3)).
T(n,k) = T(n-1,k-1) + T(n-2,k) + T(n-3,k), T(0,0)=1, T(n,k)=0 if k > n or if k < n. - Philippe Deléham, Jan 08 2014
From Wolfdieter Lang, Oct 30 2018: (Start)
The Riordan property  T = (G(x), x*G(x)) with G(x)= 1/(1-x^2-x^3) implies the following.
G.f. of row polynomials R(n, x) is G(x,z) = 1/(1- x*z - z^2 - z^3).
G.f. of column sequence k:  x^k/(1 - x^2 - x^3)^(k+1), k >= 0.
Boas-Buck recurrence (see the Aug 10 2017 remark in A046521, also for the reference):
T(n, k) = ((k+1)/(n-k))*Sum_{j=k..n-1} b(n-1-j)*T(j, k), for n >= 1, k = 0,1, ..., n-1, and input T(n,n) = 1, for n >= 0. Here b(n) = [x^n]*(d/dx)log(G(x)) = A001608(n+1), for n >= 0.
Recurrences from the A- and Z- sequences (see the W. Lang link under A006232 with references), which are A(n) = A319202(n) and Z(n) = A(n+1).
  T(0, 0) = 1, T(n, k) = 0 for n < k, and
  T(n, 0) = Sum_{j=0..n-1} Z(j)*T(n-1, j), for n >= 1, and
  T(n, k) = Sum_{j=0..n-k} A(j)*T(n-1, k-1+j), for n >= m >= 1.
(End)

A020924 Expansion of 1/(1-4*x)^(13/2).

Original entry on oeis.org

1, 26, 390, 4420, 41990, 352716, 2704156, 19315400, 130378950, 840219900, 5209363380, 31256180280, 182327718300, 1037865473400, 5782393351800, 31610416989840, 169905991320390, 899502306990300, 4697400936504900, 24228699567235800, 123566367792902580, 623715951716555880

Offset: 0

N. J. A. Sloane

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

Cf. A000984, A001622, A020922, A046521 (seventh column).

List([0..30], n-> Binomial(n+6,n)*Binomial(2*n+12,n+6)/924); # G. C. Greubel, Jul 20 2019

[&*[2*n+i: i in [1..11 by 2]]*Binomial(2*n, n)/10395: n in [0..20]]; // Vincenzo Librandi, Jul 05 2013

[Binomial(n+6,n)*Binomial(2*n+12,n+6)/924: n in [0..30]]; // G. C. Greubel, Jul 20 2019

CoefficientList[Series[1/(1-4x)^(13/2), {x,0,30}], x] (* Vincenzo Librandi, Jul 05 2013 *)

vector(30, n, n--; m=n+6; binomial(m,n)*binomial(2*m,m)/924)

[binomial(n+6,n)*binomial(2*n+12,n+6)/924 for n in (0..30)] # G. C. Greubel, Jul 20 2019

a(n) = binomial(n+6, 6)*A000984(n+6)/A000984(6), where A000984 are the central binomial coefficients. - Wolfdieter Lang
a(n) = (2*n+11)*(2*n+9)*(2*n+7)*(2*n+5)*(2*n+3)*(2*n+1)*Binomial(2*n, n)/10395. - Vincenzo Librandi, Jul 05 2013
n*a(n) - 2*(2*n+11)*a(n-1) = 0. - Bruno Berselli, Jul 05 2013
Boas-Buck recurrence: a(n) = (26/n)*Sum_{k=0..n-1} 4^(n-k-1)*a(k), n >= 1, a(0) = 1. Proof from a(n) = A046521(n+6, 6). See a comment there. - Wolfdieter Lang, Aug 10 2017
a(n) = 45*binomial(n+6,n)*binomial(2*n+12,n+6)/(4*binomial(2*n,n)). - G. C. Greubel, Jul 20 2019
From Amiram Eldar, Mar 25 2022: (Start)
Sum_{n>=0} 1/a(n) = 1018468/315 - 594*sqrt(3)*Pi.
Sum_{n>=0} (-1)^n/a(n) = 27500*sqrt(5)*log(phi) - 1864148/63, where phi is the golden ratio (A001622). (End)

A020926 Expansion of 1/(1-4*x)^(15/2).

Original entry on oeis.org

1, 30, 510, 6460, 67830, 624036, 5200300, 40116600, 290845350, 2003601300, 13223768580, 84151254600, 518932736700, 3113596420200, 18236779032600, 104557533120240, 588136123801350, 3252046802195700, 17705588145287700, 95051052148386600, 503770576386448980

Offset: 0

N. J. A. Sloane

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

Cf. A000984, A001622, A020924, A046521 (eighth column).

List([0..20], n-> Binomial(2*(n+7), n+7)*Binomial(n+7,7)/3432); # G. C. Greubel, Jul 21 2019

[&*[2*n+i: i in [1..13 by 2]]*Binomial(2*n, n)/135135: n in [0..20]]; // Vincenzo Librandi, Jul 05 2013

CoefficientList[Series[1/(1-4x)^(15/2), {x,0,20}], x] (* Vincenzo Librandi, Jul 05 2013 *)

vector(20, n, n--; m=n+7; binomial(2*m,m)*binomial(m,7)/3432) \\ G. C. Greubel, Jul 21 2019

[binomial(2*(n+7), n+7)*binomial(n+7,7)/3432 for n in (0..20)] # G. C. Greubel, Jul 21 2019

a(n) = binomial(n+7, 7)*A000984(n+7)/A000984(7), A000984: central binomial coefficients. - Wolfdieter Lang
a(n) = ((2*n+13)*(2*n+11)*(2*n+9)*(2*n+7)*(2*n+5)*(2*n+3)*(2*n+1)/135135) * binomial(2*n, n). - Vincenzo Librandi, Jul 05 2013
Boas-Buck recurrence: a(n) = (30/n)*Sum_{k=0..n-1} 4^(n-k-1)*a(k), n >= 1, a(0) = 1. Proof from a(n) = A046521(n+7, 7). See a comment there. - Wolfdieter Lang, Aug 10 2017
From Amiram Eldar, Mar 25 2022: (Start)
Sum_{n>=0} 1/a(n) = 2106*sqrt(3)*Pi - 13234624/1155.
Sum_{n>=0} (-1)^n/a(n) = 162500*sqrt(5)*log(phi) - 121172896/693, where phi is the golden ratio (A001622). (End)

A020928 Expansion of 1/(1-4*x)^(17/2).

Original entry on oeis.org

1, 34, 646, 9044, 104006, 1040060, 9360540, 77558760, 601080390, 4407922860, 30855460020, 207573094680, 1349225115420, 8510496881880, 52278766560120, 313672599360720, 1842826521244230, 10623352887172620, 60198999693978180, 335847050924299320, 1847158780083646260

Offset: 0

N. J. A. Sloane

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

Cf. A000984, A001622, A020926, A046521 (ninth column).

List([0..20], n-> Binomial(2*(n+8),n+8)*Binomial(n+8,8)/12870); # G. C. Greubel, Jul 21 2019

[&*[2*n+i: i in [1..15 by 2]]*Binomial(2*n, n)/2027025: n in [0..20]]; // Vincenzo Librandi, Jul 05 2013

CoefficientList[Series[1/(1-4x)^(17/2), {x,0,20}], x] (* Vincenzo Librandi, Jul 05 2013 *)

vector(20, n, n--; m=n+8; binomial(2*m,m)*binomial(m,8)/12870) \\ G. C. Greubel, Jul 21 2019

[binomial(2*(n+8),n+8)*binomial(n+8,8)/12870 for n in (0..20)] # G. C. Greubel, Jul 21 2019

a(n) = binomial(n+8, 8)*A000984(n+8)/A000984(8), A000984: central binomial coefficients. - Wolfdieter Lang
a(n) = ((2*n+15)*(2*n+13)*(2*n+11)*(2*n+9)*(2*n+7)*(2*n+5)*(2*n+3)*(2*n+1)/2027025)*binomial(2*n, n). - Vincenzo Librandi, Jul 05 2013
Boas-Buck recurrence: a(n) = (34/n)*Sum_{k=0..n-1} 4^(n-k-1)*a(k), n >= 1, a(0) = 1. Proof from a(n) = A046521(n+8, 8). See a comment there. - Wolfdieter Lang, Aug 10 2017
From Amiram Eldar, Mar 27 2022: (Start)
Sum_{n>=0} 1/a(n) = 39708492/1001 - 7290*sqrt(3)*Pi.
Sum_{n>=0} (-1)^n/a(n) = 937500*sqrt(5)*log(phi) - 3029336260/3003, where phi is the golden ratio (A001622). (End)

A020930 Expansion of 1/(1-4*x)^(19/2).

Original entry on oeis.org

1, 38, 798, 12236, 152950, 1651860, 15967980, 141430680, 1166803110, 9075135300, 67156001220, 476197099560, 3254013513660, 21526550936520, 138384970306200, 867212480585520, 5311676443586310, 31870058661517860, 187679234340049620, 1086563988284497800

Offset: 0

N. J. A. Sloane

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

Cf. A000984, A001622, A020928, A046521 (tenth column).

List([0..20], n-> Binomial(2*(n+9), n+9)*Binomial(n+9, 9)/Binomial(18, 9)); # G. C. Greubel, Jul 21 2019

[&*[2*n+i: i in [1..17 by 2]]*Binomial(2*n, n)/34459425: n in [0..20]]; // Vincenzo Librandi, Jul 05 2013

CoefficientList[Series[1/(1-4x)^(19/2), {x,0,20}], x] (* Vincenzo Librandi, Jul 05 2013 *)

vector(20, n, n--; m=n+9; binomial(2*m,m)*binomial(m, 9)/binomial(18,9) ) \\ G. C. Greubel, Jul 21 2019

[binomial(2*(n+9),n+9)*binomial(n+9, 9)/binomial(18,9) for n in (0..20)] # G. C. Greubel, Jul 21 2019

a(n) = binomial(n+9, 9)*A000984(n+9)/A000984(9), where A000984 are the central binomial coefficients. - Wolfdieter Lang
a(n) = ((2*n+17)*(2*n+15)*(2*n+13)*(2*n+11)*(2*n+9)*(2*n+7)*(2*n+5)*(2*n+3)*(2*n+1)/34459425)*binomial(2*n, n). - Vincenzo Librandi, Jul 05 2013
Boas-Buck recurrence: a(n) = (38/n)*Sum_{k=0..n-1} 4^(n-k-1)*a(k), n >= 1, a(0) = 1. Proof from a(n) = A046521(n+9, 9). See a comment there. - Wolfdieter Lang, Aug 10 2017
a(n) = binomial(2*(n+9),n+9)*binomial(n+9, 9)/binomial(18,9). - G. C. Greubel, Jul 21 2019
From Amiram Eldar, Mar 27 2022: (Start)
Sum_{n>=0} 1/a(n) = 24786*sqrt(3)*Pi - 2025065024/15015.
Sum_{n>=0} (-1)^n/a(n) = 5312500*sqrt(5)*log(phi) - 257493786304/45045, where phi is the golden ratio (A001622). (End)

cp's OEIS Frontend

A129818 Riordan array (1/(1+x), x/(1+x)^2), inverse array is A039599.

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

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

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A001780 Expansion of 1/((1+x)(1-x)^9).

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A001786 Expansion of 1/((1+x)*(1-x)^11).

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A104578 A Padovan convolution triangle.

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A020924 Expansion of 1/(1-4*x)^(13/2).

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