A004981 -id:A004981 - cp's OEIS frontend

A298799 Expansion of (1-27*x)^(-1/9).

1, 3, 45, 855, 17955, 398601, 9167823, 216098685, 5186368440, 126201632040, 3104560148184, 77049538223112, 1926238455577800, 48452305767226200, 1225151160114148200, 31118839466899364280, 793530406405933789140, 20305042752151835192700

Offset: 0

Seiichi Manyama, Jun 22 2018

nonn

Conjecture: a(p*n) == a(n) (mod p^2) for prime p == 1 (mod 9) and all positive integers n except those n of the form n = m*p + k for 0 <= m <= (p-1)/9 and 1 <= k <= (p-1)/9. Cf. A034171, A004981 and A004982. - Peter Bala, Dec 23 2019

Seiichi Manyama, Table of n, a(n) for n = 0..700

(1-b*x)^(-1/A003557(b)): A000984 (b=4), A004981 (b=8), A004987 (b=9), A098658 (b=12), A224881 (b=16), A034688 (b=25), this sequence (b=27), A004993 (b=36), A034835 (b=49).

Cf. A305991, A034171, A004981, A004982.

List([0..20],n->(3^n/Factorial(n))*Product([0..n-1],k->9*k+1)); # Muniru A Asiru, Jun 23 2018

seq(coeff(series((1-27*x)^(-1/9), x, n+1), x, n), n=0..20); # Muniru A Asiru, Jun 23 2018
# Alternative:
A298799 := n -> (-27)^n*binomial(-1/9, n):
seq(A298799(n), n=0..17); # Peter Luschny, Dec 26 2019

N=20; x='x+O('x^N); Vec((1-27*x)^(-1/9))

a(n) = 3^n/n! * Product_{k=0..n-1} (9*k + 1) for n > 0.
a(n) ~ 3^(3*n) / (Gamma(1/9) * n^(8/9)). - Vaclav Kotesovec, Jun 23 2018
From Peter Luschny, Dec 26 2019: (Start)
a(n) = (-27)^n*binomial(-1/9, n).
a(n) = n! * [x^n] hypergeom([1/9], [1], 27*x). (End)
D-finite with recurrence: n*a(n) +3*(-9*n+8)*a(n-1)=0. - R. J. Mathar, Jan 20 2020

A186284 Self-convolution square equals A127776.

Original entry on oeis.org

1, 2, 48, 1704, 71490, 3291780, 160844160, 8189867280, 429832053840, 23088359467040, 1263134996327680, 70138971602098560, 3942799810867610280, 223942062435751452240, 12831882367225056387840, 740872398293620831990080

Offset: 0

Paul D. Hanna, Feb 16 2011

nonn

			G.f.: A(x) = 1 + 2*x + 48*x^2 + 1704*x^3 + 71490*x^4 + 3291780*x^5 +...
Related expansions.
The g.f. of A127776 equals A(x)^2:
A(x)^2 = 1 + 4*x + 100*x^2 + 3600*x^3 + 152100*x^4 + 7033104*x^5 +...+ A004981(n)^2*x^n +...
The g.f. of A002897 equals A(x)^4:
A(x)^4 = 1 + 8*x + 216*x^2 + 8000*x^3 + 343000*x^4 + 16003008*x^5 +...+ A000984(n)^3*x^n +...
The g.f. of A004981 begins:
1/(1-8*x)^(1/4) = 1 + 2*x + 10*x^2 + 60*x^3 + 390*x^4 + 2652*x^5 +...
where A004981(n) = (2^n/n!)*Product_{k=0..n-1} (4k + 1).
The g.f. of A000984 begins:
1/(1-4*x)^(1/2) = 1 + 2*x + 6*x^2 + 20*x^3 + 70*x^4 + 252*x^5 +...
where A000984(n) = (2n)!/(n!)^2 forms the central binomial coefficients.

G. C. Greubel, Table of n, a(n) for n = 0..500

Cf. A004981, A000984, A127776, A002897.

nmax = 20; CoefficientList[Series[Sqrt[Hypergeometric2F1[ 1/4, 1/4, 1, 64*x]], {x, 0, nmax}], x] (* Vaclav Kotesovec, Apr 10 2018 *)

{a(n)=local(A004981=1/(1-8*x+x*O(x^n))^(1/4),A=sum(m=0,n,polcoeff(A004981,m)^2*x^m+x*O(x^n))^(1/2));polcoeff(A,n)}

{a(n)=local(A000984=1/(1-4*x+x*O(x^n))^(1/2),A=sum(m=0,n,polcoeff(A000984,m)^3*x^m+x*O(x^n))^(1/4));polcoeff(A,n)}

Self-convolution 4th power equals A002897.
G.f.: sqrt( K(k)/(Pi/2) ) in powers of (kk'/4)^2, where K(k) is complete elliptic integral of first kind evaluated at modulus k. [From a formula by Michael Somos in A002897]
G.f.: sqrt( 1/AGM(1, (1-16x)^(1/2)) ) in powers of x(1-16x) where AGM() is the arithmetic-geometric mean. [From a formula by Michael Somos in A004981]
a(n) ~ Pi^(3/4) * 2^(6*n - 1/2) / (Gamma(1/4)^3 * n^(3/2)). - Vaclav Kotesovec, Apr 10 2018

A209200 G.f.: (1-4x)^(-1/2) (1-8*x)^(-1/4).

Original entry on oeis.org

1, 4, 20, 112, 680, 4384, 29536, 205440, 1462368, 10587520, 77633920, 574845440, 4289409280, 32206976000, 243074083840, 1842511532032, 14018197145088, 106996519311360, 818973463721984, 6284217844736000, 48327723087278080, 372397083591557120

Offset: 0

Paul D. Hanna, Mar 06 2012

nonn

Equals the convolution of sequences A000984 and A004981.
The sequences A000984 and A004981 are related by the identity:
Sum_{n>=0} A000984(n)^3 *x^n = ( Sum_{n>=0} A004981(n)^2 *x^n )^2.

			G.f.: A(x) = 1 + 4*x + 60*x^2 + 1200*x^3 + 27300*x^4 + 668304*x^5 +...
This sequence equals the convolution of the sequences:
A000984 = [1, 2, 6, 20, 70, 252, 924, 3432, 12870, 48620, ...], and
A004981 = [1, 2, 10, 60, 390, 2652, 18564, 132600, 961350, ...].
Related sequences:
A^2: [1, 8, 56, 384, 2656, 18688, 133888, 974848, 7194112, ...],
A^4: [1, 16, 176, 1664, 14592, 122880, 1011712, 8224768, ...].

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

Cf. A000984, A004981.

CoefficientList[Series[(1-4*x)^(-1/2)*(1-8*x)^(-1/4), {x, 0, 20}], x] (* Vaclav Kotesovec, Oct 20 2012 *)

{a(n)=polcoeff((1-4*x +x*O(x^n))^(-1/2)*(1-8*x +x*O(x^n))^(-1/4),n)}

{A000984(n)=polcoeff((1-4*x +x*O(x^n))^(-1/2),n)}
{A004981(n)=polcoeff((1-8*x +x*O(x^n))^(-1/4),n)}
{a(n)=sum(k=0,n,A000984(n-k)*A004981(k))}
for(n=0,20,print1(a(n),", "))

a(n) = Sum_{k=0..n} A000984(n-k)*A004981(k).
Recurrence: n*a(n) = 4*(3*n-2)*a(n-1) - 8*(4*n-5)*a(n-2). - Vaclav Kotesovec, Oct 20 2012
a(n) ~ Gamma(3/4)*8^n/(Pi*n^(3/4)). - Vaclav Kotesovec, Oct 20 2012

A349844 Expansion of -(1 - 16x)^(1/2) / (1 - 8x)^(1/4).

Original entry on oeis.org

-1, 6, 38, 340, 3482, 38740, 457500, 5654440, 72412410, 953696900, 12844323828, 176130113432, 2450987760676, 34524885571400, 491309242342264, 7052495781361488, 101992452504973882, 1484590294804096356, 21732695236734410500, 319745609409940857144

Offset: 0

Jianing Song, Dec 01 2021

sign

Let b(n) = -a(n)/(-8)^n, {b(n)} = {1, 3/4, -19/32, 85/128, -1741/2048, 9685/8192, -114375/65536, ...}, then Sum_{n>=0} b(n) is clearly divergent since Sum_{n>=0} a(n)*x^n has radius of convergence 1/16. Let c(n) = -A349846(n)/(-4)^n, {c(n)} = {1, 3/2, -5/8, 7/16, -45/128, 77/256, -273/1024, ...}, then Sum_{n>=1} c(n) is the Cauchy product of Sum_{n>=0} b(n) with itself. Since |c(n)| ~ 1/sqrt(Pi*n) and |c(n+1)|/|c(n)| = ((2*n-1)*(2*n+3)) / ((2*n+1)*(2*n+2)) < 1, Sum_{n>=0} c(n) is conditionally convergent by Leibniz's criterion. {b(n)} serves as an example such that the Cauchy product of a divergent series with itself can be conditionally convergent.

			Let C(n) denote the Catalan numbers, P = A004981.
a(0) = -P(0) = -1;
a(1) = 2^3 * C(0) * P(0) - P(1) = 6;
a(2) = 2^3 * C(0) * P(1) + 2^5 * C(1) * P(0) - P(2) = 38;
a(3) = 2^3 * C(0) * P(2) + 2^5 * C(1) * P(1) + 2^7 * C(2) * P(0) - P(3) = 340;
a(4) = 2^3 * C(0) * P(3) + 2^5 * C(1) * P(2) + 2^7 * C(2) * P(1) + 2^9 * C(3) * P(0) - P(4) = 3482.

Wikipedia, Cauchy product

Cf. A000108, A004981, A349845, A349846.

C(n) = binomial(2*n,n)/(n+1)
a(n) = sum(k=0, n-1, 2^(2*k+3) * C(k) * A004981(n-1-k)) - A004981(n) \\ See A004981 for its program

a(n) = (Sum_{k=0..n-1} 2^(2*k+3) * CatalanNumber(k) * A004981(n-1-k)) - A004981(n).

A349845 Expansion of -(1 - 16x)^(1/2) / (1 + 8x)^(1/4).

Original entry on oeis.org

-1, 10, 6, 332, 1498, 29964, 269660, 4066456, 48190842, 679524828, 8993585460, 126419889960, 1757062172580, 25004701186680, 356647387079160, 5145713721249072, 74607994412294970, 1089344167433473788, 15981504546211353156, 235635552851036269704

Offset: 0

Jianing Song, Dec 01 2021

sign

Let b(n) = -a(n)/8^n, {b(n)} = {1, -5/4, -3/32, -83/128, -749/2048, -7491/8192, -67415/65536, ...}, then Sum_{n>=0} b(n) is clearly divergent since Sum_{n>=0} a(n)*x^n has radius of convergence 1/16. Let c(n) = A349847(n)/(-4)^n, {c(n)} = {1, -5/2, 11/8, -17/16, 115/128, -203/256, 735/1024, ...}, then Sum_{n>=0} c(n) is the Cauchy product of Sum_{n>=1} b(n) with itself. Since |c(n)| ~ 3/sqrt(Pi*n) and |c(n+1)|/|c(n)| = ((6*n+5)*(2*n-1)) / ((6*n-1)*(2*n+2)) < 1, Sum_{n>=0} c(n) is conditionally convergent by Leibniz's criterion. {b(n)} serves as an example such that the Cauchy product of a divergent series with itself can be conditionally convergent.

			Let C(n) denote the Catalan numbers, P = A004981.
a(0) = -P(0) = -1;
a(1) = 2^3 * C(0) * P(0) + P(1) = 10;
a(2) = -2^3 * C(0) * P(1) + 2^5 * C(1) * P(0) - P(2) = 6;
a(3) = 2^3 * C(0) * P(2) - 2^5 * C(1) * P(1) + 2^7 * C(2) * P(0) + P(3) = 332;
a(4) = -2^3 * C(0) * P(3) + 2^5 * C(1) * P(2) - 2^7 * C(2) * P(1) + 2^9 * C(3) * P(0) - P(4) = 1498.

Wikipedia, Cauchy product

Cf. A000108, A004981, A349844, A349847.

C(n) = binomial(2*n,n)/(n+1)
a(n) = sum(k=0, n-1, (-1)^(n-1-k) * 2^(2*k+3) * C(k) * A004981(n-1-k)) + (-1)^(n-1) * A004981(n) \\ See A004981 for its program

a(n) = (Sum_{k=0..n-1} (-1)^(n-1-k) * 2^(2*k+3) * CatalanNumber(k) * A004981(n-1-k)) + (-1)^(n-1) * A004981(n).

A383600 Expansion of 1/( (1-x)^3 * (1-9*x) )^(1/4).

Original entry on oeis.org

1, 3, 15, 97, 699, 5313, 41689, 334215, 2721411, 22423737, 186497325, 1562826195, 13178010405, 111700773135, 951026829255, 8128169277897, 69701329848051, 599462375836185, 5169038197383789, 44674793959777443, 386916485124220929, 3357265884164614707

Offset: 0

Seiichi Manyama, May 01 2025

Vincenzo Librandi, Table of n, a(n) for n = 0..300

Cf. A084771, A383602.

Cf. A004981, A231482.

R:=PowerSeriesRing(Rationals(), 25); Coefficients(R!( 1/( (1-x)^3 * (1-9*x) )^(1/4))); // Vincenzo Librandi, May 05 2025

Table[Sum[(-8)^(k)* Binomial[-1/4,k]* Binomial[n,k],{k,0,n}],{n,0,22}] (* Vincenzo Librandi, May 05 2025 *)

a(n) = sum(k=0, n, (-8)^k*binomial(-1/4, k)*binomial(n, k));

a(n) = Sum_{k=0..n} (-8)^k * binomial(-1/4,k) * binomial(n,k).
n*a(n) = (10*n-7)*a(n-1) - 9*(n-1)*a(n-2) for n > 1.
a(n) ~ 3^(2*n + 3/2) / (Gamma(1/4) * 2^(9/4) * n^(3/4)). - Vaclav Kotesovec, May 02 2025
a(n) = hypergeom([1/4, -n], [1], -8). - Stefano Spezia, May 05 2025

A122882 Array of T(n,m)=15...(4n-3)37...(4m-1)2^(n+m)/(n+m)! by antidiagonals.

Original entry on oeis.org

1, 2, 6, 10, 6, 42, 60, 20, 28, 308, 390, 90, 70, 154, 2310, 2652, 468, 252, 308, 924, 17556, 18564, 2652, 1092, 924, 1540, 5852, 134596, 132600, 15912, 5304, 3432, 3960, 8360, 38456, 1038312, 961350, 99450, 27846, 14586, 12870, 18810, 48070

Offset: 0

Michael Somos, Sep 16 2006

T(n,m)=2*A(m,n) in Problem A10527 Solution.

			       1        6       42      308     2310    17556 ...
       2        6       28      154      924     5852 ...
      10       20       70      308     1540     8360 ...
      60       90      252      924     3960    18810 ...
     390      468     1092     3432    12870    54340 ...
    2652     2652     5304    14586    48620   184756 ...
   18564    15912    27846    68068   204204   705432 ...
  132600    99450   154700   340340   928200  2939300 ...
  961350   640900   897260  1794520  4486300 13113800 ...
 7049900  4229940  5383560  9869860 22776600 61822200 ...

V. Pasol, Problem 10527, Amer. Math. Monthly, 103 (1996), p. 427; Is it an integer: solution to problem 10527, Amer. Math. Monthly, 104 (1997), 980-981.

Cf. A004981(n)=T(n, 0), A004982(n)=T(0, n), A001448(n)=T(n, n).

A122882 := proc(n,m)
    mul(4*i-3,i=1..n)*mul(4*i-1,i=1..m) ;
    %*2^(n+m)/(n+m)! ;
end proc: # R. J. Mathar, Sep 24 2021

{T(n,m)=if(n<0||m<0, 0, 2^(n+m)/(n+m)!*prod(k=1, m, 4*k-1)*prod(k=1, n, 4*k-3))}

T(n,m) = T(n,m-1)*(8*m-2)/(n+m) = T(n-1,m)*(8*n-6)/(n+m). T(0,0) = 1.

A209358 G.f.: (1-4x)^(-1/4) (1-8*x)^(-1/8).

Original entry on oeis.org

1, 2, 8, 40, 228, 1416, 9312, 63648, 446760, 3195728, 23179840, 169929280, 1256234720, 9350462400, 69993150720, 526455847680, 3976132184160, 30138433333440, 229168000121600, 1747455531216640, 13358199405416320, 102345801274115840, 785740341422453760

Offset: 0

Paul D. Hanna, Mar 06 2012

nonn

			G.f.: A(x) = 1 + 2*x + 8*x^2 + 40*x^3 + 228*x^4 + 1416*x^5 + 9312*x^6 +...
such that the square of the g.f. A(x) equals the g.f. of A209200:
A(x)^2 = 1 + 4*x + 20*x^2 + 112*x^3 + 680*x^4 + 4384*x^5 + 29536*x^6 +...
Sequence A209200 equals the convolution of the sequences:
A000984 = [1, 2, 6, 20, 70, 252, 924, 3432, 12870, 48620, ...],
A004981 = [1, 2, 10, 60, 390, 2652, 18564, 132600, 961350, ...].

Iain Fox, Table of n, a(n) for n = 0..1111 (first 201 terms from Vincenzo Librandi)

Cf. A209200, A000984, A004981.

I:=[2,8]; [1] cat [n le 2 select I[n] else (2*(6*n-5)*Self(n-1) - 4*(8*n-13)*Self(n-2))/n: n in [1..30]]; // G. C. Greubel, Jan 03 2018

CoefficientList[Series[(1-4*x)^(-1/4)*(1-8*x)^(-1/8), {x, 0, 20}], x] (* Vaclav Kotesovec, Oct 20 2012 *)

{a(n)=polcoeff((1-4*x +x*O(x^n))^(-1/4)*(1-8*x +x*O(x^n))^(-1/8), n)}
for(n=0,30,print1(a(n),", "))

Equals the self-convolution square root of A209200, which equals the convolution of sequences A000984 and A004981.
Recurrence: n*a(n) = 2*(6*n-5)*a(n-1) - 4*(8*n-13)*a(n-2). - Vaclav Kotesovec, Oct 20 2012
a(n) ~ Gamma(7/8)*2^(1/4)*8^n*sin(Pi/8)/(n^(7/8)*Pi). - Vaclav Kotesovec, Oct 20 2012

A275607 a(n) = 212^nGamma(n+1/2)(n+1)/(sqrt(Pi)Gamma(n+3)).

Original entry on oeis.org

1, 4, 27, 216, 1890, 17496, 168399, 1667952, 16888014, 173997720, 1818276174, 19225409616, 205299909828, 2210922105840, 23984556773175, 261854925711840, 2874948871877910, 31722346066169880, 351589335566716170, 3912422681494285200, 43694647856506630620, 489597172255515289680

Offset: 0

Karol A. Penson, Nov 14 2016

nonn

In reference of K. Szymanski et al. the function g(x) from the Eq.(4.6) satisfies the equality g(x/4)/4 = W(x) where W(x) is the weight function of the integral representation, see below.

G. C. Greubel, Table of n, a(n) for n = 0..925
Simeon T. Stefanov, Counting fixed points free vector fields on B^2, arXiv:1807.03714 [math.GT], 2018.
K. Szymanski, B. Collins, T. Szarek and K. Zyczkowski, Convex set of quantum states with positive partial transpose analysed by hit and run algorithm, arXiv:1611.01194 [quant-ph], 2016.

Cf. A000108, A002293, A002894, A000168, A000139, A000257, A004987, A005568, A000888, A004981.

a := n -> (2^(2*n+1)*3^n*(n+1)*GAMMA(n+1/2))/(sqrt(Pi)*GAMMA(n+3)):
seq(a(n), n=0..21); # Peter Luschny, Nov 14 2016

g[z_] :=  E^z (BesselI[0,z] - (1-1/z) BesselI[1,z])
Table[CoefficientList[2/3 Series[g[6z], {z,0,21}],z]] Range[0, 21]! //Flatten (* Peter Luschny, Nov 14 2016 *)
Table[ 2*12^n*(n + 1)*Gamma[n + 1/2]/(Sqrt[Pi]*Gamma[n + 3]), {n,0,100}] (* G. C. Greubel, Jan 13 2017 *)

a(n)=2*12^n*gamma(n+1/2)*(n+1)\/(sqrt(Pi)*(n+2)!) \\ Charles R Greathouse IV, Nov 14 2016

a(n)=2*3^n*binomial(2*n+1,n-1)*(n+1)/(2*n+1)/n \\ Charles R Greathouse IV, Nov 14 2016

O.g.f: (1/54)*(1-(6*z+1)*sqrt(1-12*z))/z^2;
E.g.f.(in Maple notation): (1/9)*exp(6*z)*(6*z*(BesselI(0,6*z)-BesselI(1,6*z))+ BesselI(1,6*z))/z;
Recurrence: (-12*n^2-54*n-54)*a(n+1)+(n^2+6*n+8)*a(n+2)=0, n=0,1..., for the initial values a(0)=1, a(1)=4.
Integral representation as the n-th Hausdorff moment of the positive function W(x) on the segment x=(0,12), i.e., a(n) = Integral_{x=0..12} x^n*W(x) dx, where W(x) = (1/27)*sqrt(12-x)*(3+(1/2)*x)/(Pi*sqrt(x)). This representation is unique.
a(n) ~ 2^(2*n+1)*3^n/(sqrt(Pi)*n^(3/2)). - Ilya Gutkovskiy, Nov 14 2016
a(n) = 2*3^n*binomial(2n+1, n-1)*(n+1)/(2n^2+n). - Charles R Greathouse IV, Nov 14 2016

A371927 Expansion of 1/(1 - x/(1 - 8*x^2)^(1/4)).

Original entry on oeis.org

1, 1, 1, 3, 5, 17, 33, 113, 237, 803, 1769, 5915, 13493, 44547, 104337, 340527, 814397, 2630857, 6399865, 20486905, 50548997, 160507953, 400834465, 1263577141, 3188428301, 9985916077, 25426685961, 79168607025, 203193847381, 629311885861, 1626634117809

Offset: 0

Seiichi Manyama, Jun 07 2024

nonn

Cf. A373509, A004981.

A371927 := proc(n)
    add(8^k*binomial((n+2*k)/4-1,k),k=0..floor(n/2)) ;
end proc:
seq(A371927(n),n=0..70) ; # R. J. Mathar, Jun 07 2024

CoefficientList[Series[1/(1-x/(1-8x^2)^(1/4)),{x,0,30}],x] (* Harvey P. Dale, Dec 20 2024 *)

a(n) = sum(k=0, n\2, 8^k*binomial((n+2*k)/4-1, k));

a(n) = Sum_{k=0..floor(n/2)} 8^k * binomial((n+2*k)/4-1,k).

cp's OEIS Frontend

A298799 Expansion of (1-27*x)^(-1/9).

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A186284 Self-convolution square equals A127776.

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A209200 G.f.: (1-4*x)^(-1/2) * (1-8*x)^(-1/4).

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A349844 Expansion of -(1 - 16*x)^(1/2) / (1 - 8*x)^(1/4).

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A349845 Expansion of -(1 - 16*x)^(1/2) / (1 + 8*x)^(1/4).

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A383600 Expansion of 1/( (1-x)^3 * (1-9*x) )^(1/4).

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A122882 Array of T(n,m)=1*5*...*(4n-3)*3*7*...*(4m-1)*2^(n+m)/(n+m)! by antidiagonals.

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A209200 G.f.: (1-4x)^(-1/2) (1-8*x)^(-1/4).

A349844 Expansion of -(1 - 16x)^(1/2) / (1 - 8x)^(1/4).

A349845 Expansion of -(1 - 16x)^(1/2) / (1 + 8x)^(1/4).

A122882 Array of T(n,m)=15...(4n-3)37...(4m-1)2^(n+m)/(n+m)! by antidiagonals.

A209358 G.f.: (1-4x)^(-1/4) (1-8*x)^(-1/8).

A275607 a(n) = 212^nGamma(n+1/2)(n+1)/(sqrt(Pi)Gamma(n+3)).