A038845 -id:A038845 - cp's OEIS frontend

A020918 Expansion of 1/(1-4*x)^(7/2).

1, 14, 126, 924, 6006, 36036, 204204, 1108536, 5819814, 29745716, 148728580, 730122120, 3528923580, 16830250920, 79342611480, 370265520240, 1712478031110, 7857252142740, 35794148650260

Offset: 0

N. J. A. Sloane

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

5*a(n) is the number of (n+3) X 2 Young tableaux with a four horizontal walls between the first and second column. If there is a wall between two cells, the entries may be decreasing; see [Banderier, Wallner 2021], A000984 for one horizontal wall, A002457 for two, and A002802 for three. - Michael Wallner, Mar 09 2022

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Cyril Banderier and Michael Wallner, Young Tableaux with Periodic Walls: Counting with the Density Method, Séminaire Lotharingien de Combinatoire, 85B (2021), Art. 47, 12 pp.

Cf. A000302, A000984, A001622, A002457, A002697, A002802, A000292, A038845, A046521 (fourth column).

List([0..30], n-> Binomial(2*n+6, n+3)*Binomial(n+3, 3)/20); # G. C. Greubel, Jul 20 2019

[Binomial(2*n+6, n+3)*Binomial(n+3, 3)/20: n in [0..30]]; // G. C. Greubel, Jul 20 2019

seq(binomial(2*n,n)*binomial(n,(n-3))/20, n=2..21); # Zerinvary Lajos, May 05 2007
seq(simplify(4^n*hypergeom([-n,-5/2], [1], 1)),n=0..18); # Peter Luschny, Apr 26 2016

CoefficientList[Series[1/(1-4x)^(7/2), {x, 0, 30}], x] (* Vincenzo Librandi, Jul 04 2013 *)

vector(30, n, n--; binomial(2*n+6, n+3)*binomial(n+3, 3)/20 ) \\ G. C. Greubel, Jul 20 2019

[binomial(2*n+6, n+3)*binomial(n+3, 3)/20 for n in (0..30)] # G. C. Greubel, Jul 20 2019

a(n) = binomial(n+3, 3)*A000984(n+3)/A000984(3), where A000984 are the central binomial coefficients. - Wolfdieter Lang
a(n) ~ 8/15*Pi^(-1/2)*n^(5/2)*2^(2*n)*{1 + 35/8*n^-1 + ...}. - Joe Keane (jgk(AT)jgk.org), Nov 22 2001
a(n) = Sum_{a+b+c+d+e+f+g=n} f(a)*f(b)*f(c)*f(d)*f(e)*f(f)*f(g) with f(n)=A000984(n). - Philippe Deléham, Jan 22 2004
a(n) = A000292(n)*A000984(n+2)/20. - Zerinvary Lajos, May 05 2007
From Rui Duarte, Oct 08 2011: (Start)
a(n) = ((2n+5)(2n+3)(2n+1)/(5*3*1)) * binomial(2n, n).
a(n) = binomial(2n+6, 6) * binomial(2n, n) / binomial(n+3, 3).
a(n) = binomial(n+3, 3) * binomial(2n+6, n+3) / binomial(6, 3). (End)
a(n) = 4^n*hypergeom([-n,-5/2], [1], 1). - Peter Luschny, Apr 26 2016
Boas-Buck recurrence: a(n) = (14/n)*Sum_{k=0..n-1} 4^(n-k-1)*a(k), n >= 1, a(0) = 1. Proof from a(n) = A046521(n+3, 3). See a comment there. - Wolfdieter Lang, Aug 10 2017
From Amiram Eldar, Apr 07 2022: (Start)
Sum_{n>=0} 1/a(n) = 10*sqrt(3)*Pi - 160/3.
Sum_{n>=0} (-1)^n/a(n) = 10*sqrt(5)*log(phi) - 320/3, where phi is the golden ratio (A001622). (End)
D-finite with recurrence n*a(n) +2*(-2*n-5)*a(n-1)=0. - R. J. Mathar, Aug 01 2022

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

Original entry on oeis.org

1, 20, 240, 2240, 17920, 129024, 860160, 5406720, 32440320, 187432960, 1049624576, 5725224960, 30534533120, 159719096320, 821412495360, 4161823309824, 20809116549120, 102821517066240, 502682972323840, 2434043865989120, 11683410556747776, 55635288365465600

Offset: 0

Wolfdieter Lang

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]

Vincenzo Librandi, Table of n, a(n) for n = 0..400
Eric Weisstein's World of Mathematics, Idempotent Number.
Index entries for linear recurrences with constant coefficients, signature (20, -160, 640, -1280, 1024).

Cf. A000302, A020920, A000984.

Cf. A038231.

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

[4^n*Binomial(n+4, 4): n in [0..30]]; // Vincenzo Librandi, Oct 15 2011

seq(seq(binomial(i, j)*4^(i-4), j =i-4), i=4..22); # Zerinvary Lajos, Dec 03 2007
seq(binomial(n+4,4)*4^n,n=0..30); # Zerinvary Lajos, May 19 2008
spec := [S, {B=Set(Z, 0 <= card), S=Prod(Z, Z, Z, Z, B, B, B, B)}, labeled]: seq(combstruct[count](spec, size=n)/24, n=4..34); # Zerinvary Lajos, Apr 05 2009

Table[Binomial[n+4,4]*4^n, {n,0,30}] (* Michael De Vlieger, Aug 21 2015 *)

vector(30, n, n--; 4^n*binomial(n+4, 4)) \\ G. C. Greubel, Jul 20 2019

[lucas_number2(n, 4, 0)*binomial(n,4)/2^8 for n in range(4, 34)] # Zerinvary Lajos, Mar 11 2009

a(n) = binomial(n+4, 4)*4^n.
G.f.: 1/(1-4*x)^5.
a(n) = Sum_{ i_1+i_2+i_3+i_4+i_5+i_6+i_7+i_8+i_9+i_10 = 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) with f(k)=A000984(k). - Rui Duarte, Oct 08 2011
E.g.f.: (3 + 48*x + 144*x^2 + 128*x^3 + 32*x^4)*exp(4*x)/3. - G. C. Greubel, Jul 20 2019
From Amiram Eldar, Mar 25 2022: (Start)
Sum_{n>=0} 1/a(n) = 376/3 - 432*log(4/3).
Sum_{n>=0} (-1)^n/a(n) = 2000*log(5/4) - 1336/3. (End)

A081140 10th binomial transform of (0,0,1,0,0,0,...).

Original entry on oeis.org

0, 0, 1, 30, 600, 10000, 150000, 2100000, 28000000, 360000000, 4500000000, 55000000000, 660000000000, 7800000000000, 91000000000000, 1050000000000000, 12000000000000000, 136000000000000000, 1530000000000000000

Offset: 0

Paul Barry, Mar 08 2003

Starting at 1, the three-fold convolution of A011557 (powers of 10).

Vincenzo Librandi, Table of n, a(n) for n = 0..400
Index entries for linear recurrences with constant coefficients, signature (30,-300,1000).

Sequences similar to the form q^(n-2)*binomial(n, 2): A000217 (q=1), A001788 (q=2), A027472 (q=3), A038845 (q=4), A081135 (q=5), A081136 (q=6), A027474 (q=7), A081138 (q=8), A081139 (q=9), this sequence (q=10), A081141 (q=11), A081142 (q=12), A027476 (q=15).

[10^n* Binomial(n+2, 2): n in [-2..20]]; // Vincenzo Librandi, Oct 16 2011

Table[10^(n-2)*Binomial[n, 2], {n, 0, 30}] (* G. C. Greubel, May 13 2021 *)

a(n) = 30*a(n-1) - 300*a(n-2) + 1000*a(n-3), a(0)=a(1)=0, a(2)=1.
a(n) = 10^(n-2)*binomial(n, 2).
G.f.: x^2/(1-10*x)^3.
E.g.f.: (x^2/2)*exp(10*x). - G. C. Greubel, May 13 2021
From Amiram Eldar, Jan 06 2022: (Start)
Sum_{n>=2} 1/a(n) = 20 - 180*log(10/9).
Sum_{n>=2} (-1)^n/a(n) = 220*log(11/10) - 20. (End)

A081141 11th binomial transform of (0,0,1,0,0,0,...).

Original entry on oeis.org

0, 0, 1, 33, 726, 13310, 219615, 3382071, 49603708, 701538156, 9646149645, 129687123005, 1711870023666, 22254310307658, 285596982281611, 3624884775112755, 45569980029988920, 568105751040528536

Offset: 0

Paul Barry, Mar 08 2003

Starting at 1, the three-fold convolution of A001020 (powers of 11).

Vincenzo Librandi, Table of n, a(n) for n = 0..400
Index entries for linear recurrences with constant coefficients, signature (33,-363,1331).

Cf. A001020.

Sequences similar to the form q^(n-2)*binomial(n, 2): A000217 (q=1), A001788 (q=2), A027472 (q=3), A038845 (q=4), A081135 (q=5), A081136 (q=6), A027474 (q=7), A081138 (q=8), A081139 (q=9), A081140 (q=10), this sequence (q=11), A081142 (q=12), A027476 (q=15).

[11^(n-2)*Binomial(n, 2): n in [0..20]]; // Vincenzo Librandi, Oct 16 2011

seq((11)^(n-2)*binomial(n,2), n=0..30); # G. C. Greubel, May 13 2021

LinearRecurrence[{33,-363,1331},{0,0,1},30] (* Harvey P. Dale, Dec 15 2014 *)

vector(20, n, n--; 11^(n-2)*binomial(n, 2)) \\ G. C. Greubel, Nov 23 2018

[11^(n-2)*binomial(n, 2) for n in range(20)] # G. C. Greubel, Nov 23 2018

a(n) = 33*a(n-1) - 363*a(n-2) + 1331*a(n-3), a(0) = a(1) = 0, a(2) = 1.
a(n) = 11^(n-2)*binomial(n, 2).
G.f.: x^2/(1 - 11*x)^3.
E.g.f.: (1/2)*exp(11*x)*x^2. - Franck Maminirina Ramaharo, Nov 23 2018
From Amiram Eldar, Jan 06 2022: (Start)
Sum_{n>=2} 1/a(n) = 22 - 220*log(11/10).
Sum_{n>=2} (-1)^n/a(n) = 264*log(12/11) - 22. (End)

A027476 Third column of A027467.

Original entry on oeis.org

1, 45, 1350, 33750, 759375, 15946875, 318937500, 6150937500, 115330078125, 2114384765625, 38058925781250, 674680957031250, 11806916748046875, 204350482177734375, 3503151123046875000, 59553569091796875000

Offset: 3

Olivier Gérard

nonn

Vincenzo Librandi, Table of n, a(n) for n = 3..200
Index entries for linear recurrences with constant coefficients, signature (45,-675,3375).

Sequences similar to the form q^(n-2)*binomial(n, 2): A000217 (q=1), A001788 (q=2), A027472 (q=3), A038845 (q=4), A081135 (q=5), A081136 (q=6), A027474 (q=7), A081138 (q=8), A081139 (q=9), A081140 (q=10), A081141 (q=11), A081142 (q=12), this sequence (q=15).

[(n-1)*(n-2)/2 * 15^(n-3): n in [3..20]]; // Vincenzo Librandi, Dec 29 2012

seq((15)^(n-3)*binomial(n-1, 2), n=3..30) # G. C. Greubel, May 13 2021

Table[(n-1)*(n-2)/2 * 15^(n-3), {n, 3, 30}] (* Vincenzo Librandi, Dec 29 2012 *)

[(15)^(n-3)*binomial(n-1,2) for n in (3..30)] # G. C. Greubel, May 13 2021

Numerators of sequence a[3,n] in (a[i,j])^4 where a[i,j] = binomial(i-1, j-1)/2^(i-1) if j<=i, 0 if j>i.
a(n) = 15^(n-3)*binomial(n-1, 2).
From G. C. Greubel, May 13 2021: (Start)
a(n) = 45*a(n-1) - 675*a(n-2) + 3375*a(n-3).
G.f.: x^3/(1 - 15*x)^3.
E.g.f.: (-2 + (2 - 30*x + 225*x^2)*exp(15*x))/6750. (End)
From Amiram Eldar, Jan 06 2022: (Start)
Sum_{n>=3} 1/a(n) = 30 - 420*log(15/14).
Sum_{n>=3} (-1)^(n+1)/a(n) = 480*log(16/15) - 30. (End)

A081142 12th binomial transform of (0,0,1,0,0,0,...).

Original entry on oeis.org

0, 0, 1, 36, 864, 17280, 311040, 5225472, 83607552, 1289945088, 19349176320, 283787919360, 4086546038784, 57954652913664, 811365140791296, 11234286564802560, 154070215745863680, 2095354934143746048

Offset: 0

Paul Barry, Mar 08 2003

Starting at 1, the three-fold convolution of A001021 (powers of 12).

Vincenzo Librandi, Table of n, a(n) for n = 0..400
Index entries for linear recurrences with constant coefficients, signature (36,-432,1728).

Cf. A001021.

Sequences similar to the form q^(n-2)*binomial(n, 2): A000217 (q=1), A001788 (q=2), A027472 (q=3), A038845 (q=4), A081135 (q=5), A081136 (q=6), A027474 (q=7), A081138 (q=8), A081139 (q=9), A081140 (q=10), A081141 (q=11), this sequence (q=12), A027476 (q=15).

List([0..20],n->12^(n-2)*Binomial(n,2)); # Muniru A Asiru, Nov 24 2018

[12^(n-2)* Binomial(n, 2): n in [0..20]]; // Vincenzo Librandi, Oct 16 2011

seq(coeff(series(x^2/(1-12*x)^3,x,n+1), x, n), n = 0 .. 20); # Muniru A Asiru, Nov 24 2018

LinearRecurrence[{36,-432,1728},{0,0,1},30] (* or *) Table[(n-1) (n-2) 3^(n-3) 2^(2n-7),{n,20}] (* Harvey P. Dale, Jul 25 2013 *)

vector(20, n, n--; 2^(2*n-5)*3^(n-2)*n*(n-1)) \\ G. C. Greubel, Nov 23 2018

[2^(2*n-5)*3^(n-2)*n*(n-1) for n in range(20)] # G. C. Greubel, Nov 23 2018

a(n) = 36*a(n-1) - 432*a(n-2) + 1728*a(n-3), a(0) = a(1) = 0, a(2) = 1.
a(n) = 12^(n-2)*binomial(n, 2).
G.f.: x^2/(1 - 12*x)^3.
a(n) = 2^(2*n-5)*3^(n-2)*n*(n-1). - Harvey P. Dale, Jul 25 2013
E.g.f.: (1/2)*exp(12*x)*x^2. - Franck Maminirina Ramaharo, Nov 23 2018
From Amiram Eldar, Jan 06 2022: (Start)
Sum_{n>=2} 1/a(n) = 24 - 264*log(12/11).
Sum_{n>=2} (-1)^n/a(n) = 312*log(13/12) - 24. (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)

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)

A081130 Square array of binomial transforms of (0,0,1,0,0,0,...), read by antidiagonals.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 3, 0, 0, 0, 1, 6, 6, 0, 0, 0, 1, 9, 24, 10, 0, 0, 0, 1, 12, 54, 80, 15, 0, 0, 0, 1, 15, 96, 270, 240, 21, 0, 0, 0, 1, 18, 150, 640, 1215, 672, 28, 0, 0, 0, 1, 21, 216, 1250, 3840, 5103, 1792, 36, 0, 0, 0, 1, 24, 294, 2160, 9375, 21504, 20412, 4608, 45, 0

Offset: 0

Paul Barry, Mar 08 2003

Rows, of the square array, are three-fold convolutions of sequences of powers.

			The array begins as:
  0,  0,  0,   0,   0,    0, ...
  0,  0,  0,   0,   0,    0, ...
  0,  1,  1,   1,   1,    1, ... A000012
  0,  3,  6,   9,  12,   15, ... A008585
  0,  6, 24,  54,  96,  150, ... A033581
  0, 10, 80, 270, 640, 1250, ... A244729
The antidiagonal triangle begins as:
  0;
  0, 0;
  0, 0, 0;
  0, 0, 1, 0;
  0, 0, 1, 3,  0;
  0, 0, 1, 6,  6,  0;
  0, 0, 1, 9, 24, 10, 0;

G. C. Greubel, Antidiadoganal rows n = 0..50, flattened

Main diagonal: A081131.
Rows: A000012 (n=2), A008585 (n=3), A033581 (n=4), A244729 (n=5).
Columns: A000217 (k=1), A001788 (k=2), A027472 (k=3), A038845 (k=4), A081135 (k=5), A081136 (k=6), A027474 (k=7), A081138 (k=8), A081139 (k=9), A081140 (k=10), A081141 (k=11), A081142 (k=12), A027476 (k=15).

[k eq n select 0 else (n-k)^(k-2)*Binomial(k,2): k in [0..n], n in [0..12]]; // G. C. Greubel, May 14 2021

Table[If[k==n, 0, (n-k)^(k-2)*Binomial[k, 2]], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, May 14 2021 *)

T(n, k)=if (k==0, 0, k^(n-2)*binomial(n, 2));
seq(nn) = for (n=0, nn, for (k=0, n, print1(T(k, n-k), ", ")); );
seq(12) \\ Michel Marcus, May 14 2021

flatten([[0 if (k==n) else (n-k)^(k-2)*binomial(k,2) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, May 14 2021

T(n, k) = k^(n-2)*binomial(n, 2), with T(n, 0) = 0 (square array).
T(n, n) = A081131(n).
Rows have g.f. x^3/(1-k*x)^n.
From G. C. Greubel, May 14 2021: (Start)
T(k, n-k) = (n-k)^(k-2)*binomial(k,2) with T(n, n) = 0 (antidiagonal triangle).
Sum_{k=0..n} T(n, n-k) = A081197(n). (End)

Term a(5) corrected by G. C. Greubel, May 14 2021

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A020918 Expansion of 1/(1-4*x)^(7/2).

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

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A081140 10th binomial transform of (0,0,1,0,0,0,...).

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A081141 11th binomial transform of (0,0,1,0,0,0,...).

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A027476 Third column of A027467.

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A081142 12th binomial transform of (0,0,1,0,0,0,...).

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