A006134 -id:A006134 - cp's OEIS frontend

A293490 a(n) = Sum_{k=0..n} binomial(2k, k)binomial(2*n-k, n).

1, 4, 18, 84, 400, 1932, 9436, 46512, 231066, 1155660, 5813808, 29396952, 149305884, 761282032, 3894953640, 19987999696, 102847396416, 530446714812, 2741576339716, 14196136939600, 73631851898220, 382483602131400, 1989514312826400, 10361255764532400, 54020655931542300, 281933439875693424

Offset: 0

Ilya Gutkovskiy, Oct 10 2017

nonn

Main diagonal of iterated partial sums array of central binomial coefficients (starting with the first partial sums).

Cf. A000984, A002426, A006134, A026375, A270447.

A293490 := Concatenation([1], List([1..3*10^2],n -> Sum([0..n],k -> Binomial(2*k, k)*(Binomial(2*n - k, n))))); # Muniru A Asiru, Oct 15 2017

Table[Sum[Binomial[2 k, k] Binomial[2 n - k, n], {k, 0, n}], {n, 0, 25}]
Table[SeriesCoefficient[1/(Sqrt[1 - 4 x] (1 - x)^(n + 1)), {x, 0, n}], {n, 0, 25}]
Table[SeriesCoefficient[(1/(1 - x)^(n + 1)) 1/(1 - 2 x/(1 + ContinuedFractionK[-x, 1, {k, 1, n}])), {x, 0, n}], {n, 0, 25}]
CoefficientList[Series[1/(Sqrt[2 Sqrt[1-4 x]-1] Sqrt[1-4 x]),{x,0,25}],x] (* Alexander M. Haupt, Jun 24 2018 *)

a(n) = sum(k=0, n, binomial(2*k, k)*binomial(2*n-k, n)); \\ Michel Marcus, Oct 15 2017

a(n) = [x^n] 1/(sqrt(1 - 4*x)*(1 - x)^(n+1)).
a(n) = [x^n] 1/((1 - x)^(n+1)*(1 - 2*x/(1 - x/(1 - x/(1 - x/(1 - ...)))))), a continued fraction.
a(n) = 4^n*Gamma(n+1/2)*2F1(-n,n+1; 1/2-n; 1/4)/(sqrt(Pi)*Gamma(n+1)).
From Vaclav Kotesovec, Oct 16 2017: (Start)
D-finite with recurrence: 3*(n-1)*n*a(n) = 14*(n-1)*(2*n-1)*a(n-1) - 4*(4*n-5)*(4*n-3)*a(n-2).
a(n) ~ 2^(4*n + 3/2) / (3^(n + 1/2) * sqrt(Pi*n)).
(End)
G.f.: 1/(sqrt(2*sqrt(1-4*x)-1)*sqrt(1-4*x)). - Alexander M. Haupt, Jun 24 2018

A307354 a(n) = Sum_{0<=i<=j<=n} (-1)^(i+j) * (i+j)!/(i!*j!).

Original entry on oeis.org

1, 2, 6, 19, 65, 231, 841, 3110, 11628, 43834, 166298, 634140, 2428336, 9331688, 35967462, 138987715, 538287881, 2088842463, 8119916647, 31613327405, 123251518641, 481125828853, 1880262896537, 7355767408395, 28803717914791, 112887697489907, 442784607413427

Offset: 0

Seiichi Manyama, Apr 03 2019

nonn

Seiichi Manyama, Table of n, a(n) for n = 0..1664

Partial sums of A026641. - Seiichi Manyama, Jan 30 2023
Cf. A000108, A006134, A079309, A105872, A120305, A306409.
Cf. A191993, A360186, A360212.

Table[Sum[Sum[(-1)^(i + j)*(i + j)!/(i!*j!), {i, 0, j}], {j, 0, n}], {n, 0, 25}] (* Vaclav Kotesovec, Apr 04 2019 *)

a(n) = sum(i=0, n, sum(j=i, n, (-1)^(i+j)*(i+j)!/(i!*j!)));

a(n) = sum(k=0, n\3, (-1)^k*binomial(2*n-3*k, n)); \\ Seiichi Manyama, Jan 29 2023

my(N=30, x='x+O('x^N)); Vec(1/(sqrt(1-4*x)*(1+x^3*(2/(1+sqrt(1-4*x)))^3))) \\ Seiichi Manyama, Jan 29 2023

a(n) = (A006134(n) + A120305(n))/2.
From Vaclav Kotesovec, Apr 04 2019: (Start)
Recurrence: 2*n*a(n) = (9*n-4)*a(n-1) - (3*n-2)*a(n-2) - 2*(2*n-1)*a(n-3).
a(n) ~ 2^(2*n+3) / (9*sqrt(Pi*n)). (End)
From Seiichi Manyama, Jan 29 2023: (Start)
a(n) = Sum_{k=0..floor(n/3)} (-1)^k * binomial(2*n-3*k,n).
G.f.: 1 / ( sqrt(1-4*x) * (1 + x^3 * c(x)^3) ), where c(x) is the g.f. of A000108. (End)
a(n) = [x^n] 1/((1+x^3) * (1-x)^(n+1)). - Seiichi Manyama, Apr 08 2024

A323222 A(n, k) = [x^k] (1 - 4x)^(-n/2)x/(1 - x), square array read by ascending antidiagonals with n >= 0 and k >= 0.

Original entry on oeis.org

0, 0, 1, 0, 1, 1, 0, 1, 3, 1, 0, 1, 5, 9, 1, 0, 1, 7, 21, 29, 1, 0, 1, 9, 37, 85, 99, 1, 0, 1, 11, 57, 177, 341, 351, 1, 0, 1, 13, 81, 313, 807, 1365, 1275, 1, 0, 1, 15, 109, 501, 1593, 3579, 5461, 4707, 1, 0, 1, 17, 141, 749, 2811, 7737, 15591, 21845, 17577, 1

Offset: 0

Peter Luschny, Jan 24 2019

General asymptotic formula for g.f. (1 - 4*x)^(-j/2)*x/(1 - x) and fixed j>0 is a(n) ~ n^(j/2 - 1) * 4^n / (3*Gamma(j/2)). - Vaclav Kotesovec, Jan 29 2019

			[n\k] 0  1   2    3     4      5       6       7        8         9
-------------------------------------------------------------------
[0]   0, 1,  1,   1,    1,     1,      1,      1,       1,        1, ... A057427
[1]   0, 1,  3,   9,   29,    99,    351,   1275,    4707,    17577, ... A006134
[2]   0, 1,  5,  21,   85,   341,   1365,   5461,   21845,    87381, ... A002450
[3]   0, 1,  7,  37,  177,   807,   3579,  15591,   67071,   285861, ... A277178
[4]   0, 1,  9,  57,  313,  1593,   7737,  36409,  167481,   757305, ... A014916
[5]   0, 1, 11,  81,  501,  2811,  14823,  74883,  366603,  1752273, ... A323223
[6]   0, 1, 13, 109,  749,  4589,  26093, 140781,  730605,  3679725, ...
[7]   0, 1, 15, 141, 1065,  7071,  43107, 247311, 1355847,  7175661, ...
[8]   0, 1, 17, 177, 1457, 10417,  67761, 411825, 2377905, 13191345, ...
[9]   0, 1, 19, 217, 1933, 14803, 102319, 656587, 3982195, 23104441, ...
Triangle given by antidiagonals:
0;
0, 1;
0, 1,  1;
0, 1,  3,   1;
0, 1,  5,   9,   1;
0, 1,  7,  21,  29,    1;
0, 1,  9,  37,  85,   99,    1;
0, 1, 11,  57, 177,  341,  351,    1;
0, 1, 13,  81, 313,  807, 1365, 1275,    1;
0, 1, 15, 109, 501, 1593, 3579, 5461, 4707, 1;

Sums of antidiagonals are A323217. Main diagonal is A323219.
Rows: A057427 (n=0), A006134 (n=1), A002450 (n=2), A277178 (n=3), A014916 (n=4), A323223 (n=5).
Columns: A005408 (k=2), A059993 (k=3), A323218 (k=4).
Similar array based on Catalan numbers is A323224.

Row := proc(n, len) local ogf, ser; ogf := (1 - 4*x)^(-n/2)*x/(1 - x);
ser := series(ogf, x, (n+1)*len+1); seq(coeff(ser, x, j), j=0..len) end:
for n from 0 to 9 do Row(n, 9) od;

BF[N_, K_] := Module[{}, r[n_, k_] := FrobeniusSolve[ConstantArray[1, n], k];
X[n_] := Flatten[Table[r[N, j], {j, 0, n - 1}], 1];
CentralBinomial[n_] := Binomial[2 n, n];
Sum[Product[CentralBinomial[m[[i]]], {i, 1, N}], {m , X[K]}]];
Trow[n_] := Table[BF[n, k], {k, 0, 9}]; Table[Trow[n], {n, 1, 9}]

For n>0 and k>0 let X(n, k) denote the set of all tuples of length n with elements from {0, ..., k-1} with sum < k. Let b(m) = binomial(2*m, m). Then A(n, k) = Sum_{(j1,...,jn) in X(n, k)} b(j1)*b(j2)*...*b(jn).

A006135 T(n+2,2) from table A045912 of characteristic polynomial of negative Pascal matrix.

Original entry on oeis.org

1, 9, 72, 626, 6084, 64974, 744193, 8965323, 112088583, 1441465015, 18952951005, 253712542005, 3447133563343, 47425573790397, 659506609478472, 9256644358552742, 130981854694547790, 1866712391002772586

Offset: 0

N. J. A. Sloane

			1 + 9*x + 72*x^2 + 626*x^3 + 6084*x^4 + 64974*x^5 + 744193*x^6 + 8965323*x^7 + ...

N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Robert Israel, Table of n, a(n) for n = 0..200
W. F. Lunnon, The Pascal matrix, Fib. Quart. vol. 15 (1977) pp. 201-204.

Cf. A045912, A006134, A006136.

f:= n -> coeff(LinearAlgebra:-CharacteristicPolynomial(Matrix(n+2,n+2,(i,j) -> -binomial(i+j-2,i-1)),lambda),lambda,2):
map(f, [$0..20]); # Robert Israel, Jul 09 2018

{a(n) = if( n<0, 0, polcoeff( charpoly( matrix( n+2, n+2, i, j, -binomial( i+j-2, i-1))), 2))} /* Michael Somos, Jul 10 2002 */

Edited by Michael Somos, Jul 19 2002

A096465 Triangle (read by rows) formed by setting all entries in the first column and in the main diagonal ((i,i) entries) to 1 and the rest of the entries by the recursion T(n, k) = T(n-1, k) + T(n, k-1).

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 4, 1, 1, 4, 8, 9, 1, 1, 5, 13, 22, 23, 1, 1, 6, 19, 41, 64, 65, 1, 1, 7, 26, 67, 131, 196, 197, 1, 1, 8, 34, 101, 232, 428, 625, 626, 1, 1, 9, 43, 144, 376, 804, 1429, 2055, 2056, 1, 1, 10, 53, 197, 573, 1377, 2806, 4861, 6917, 6918, 1, 1, 11, 64, 261, 834, 2211, 5017, 9878, 16795, 23713, 23714, 1

Offset: 0

Gerald McGarvey, Aug 12 2004

The third column is A034856 (binomial(n+1, 2) + n-1).
The row sums are A014137 (partial sums of Catalan numbers (A000108)).
The "1st subdiagonal" ((i+1,i) entries) are also A014137.
The "2nd subdiagonal" ((i+2,i) entries) is A014138 ( Partial sums of Catalan numbers (starting 1,2,5,...)).
The "3rd subdiagonal" ((i+3,i) entries) is A001453 (Catalan numbers - 1.)
This is the reverse of A091491 - see A091491 for more information. The sequence of antidiagonal sums gives A124642. - Gerald McGarvey, Dec 09 2006

			Triangle begins as:
  1;
  1, 1;
  1, 2,  1;
  1, 3,  4,  1;
  1, 4,  8,  9,   1;
  1, 5, 13, 22,  23,   1;
  1, 6, 19, 41,  64,  65,   1;
  1, 7, 26, 67, 131, 196, 197, 1;

Reinhard Zumkeller, Rows n=0..150 of triangle, flattened

Cf. A000108, A001453, A014137, A014138, A034856.

Cf. A006134, A024718, A030237, A078478, A091491, A100066, A105848, A124642.

a096465 n k = a096465_tabl !! n !! k
a096465_row n = a096465_tabl !! n
a096465_tabl = map reverse a091491_tabl
-- Reinhard Zumkeller, Jul 12 2012

A096465:= func< n,k | k eq n select 1 else (n-k)*(&+[Binomial(n+k-2*j, n-j)/(n+k-2*j): j in [0..k]]) >;
[A096465(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Apr 30 2021

A096465:= (n,k)-> `if`(k=n, 1, (n-k)*add(binomial(n+k-2*j, n-j)/(n+k-2*j), j=0..k));
seq(seq(A096465(n,k), k=0..n), n=0..12) # G. C. Greubel, Apr 30 2021

T[, 0]= 1; T[n, n_]= 1; T[n_, m_]:= T[n, m]= T[n-1, m] + T[n, m-1]; T[n_, m_] /; n < 0 || m > n = 0; Table[T[n, m], {n, 0, 12}, {m, 0, n}]//Flatten (* Jean-François Alcover, Dec 17 2012 *)

def A096465(n,k): return 1 if (k==n) else (n-k)*sum( binomial(n+k-2*j, n-j)/(n+k-2*j) for j in (0..k))
flatten([[A096465(n,k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Apr 30 2021

From G. C. Greubel, Apr 30 2021: (Start)
T(n, k) = (n-k) * Sum_{j=0..k} binomial(n+k-2*j, n-j)/(n+k-2*j) with T(n,n) = 1.
T(n, k) = A091491(n, n-k).
Sum_{k=0..n} T(n,k) = Sum_{j=0..n} A000108(j) = A014137(n). (End)

Offset changed by Reinhard Zumkeller, Jul 12 2012

A106187 Sequence array for central binomial numbers A000984.

Original entry on oeis.org

1, 2, 1, 6, 2, 1, 20, 6, 2, 1, 70, 20, 6, 2, 1, 252, 70, 20, 6, 2, 1, 924, 252, 70, 20, 6, 2, 1, 3432, 924, 252, 70, 20, 6, 2, 1, 12870, 3432, 924, 252, 70, 20, 6, 2, 1, 48620, 12870, 3432, 924, 252, 70, 20, 6, 2, 1, 184756, 48620, 12870, 3432, 924, 252, 70, 20, 6, 2, 1

Offset: 0

Paul Barry, Apr 24 2005

			Triangle begins:
    1;
    2,  1;
    6,  2, 1;
   20,  6, 2, 1;
   70, 20, 6, 2, 1;
  252, 70, 20, 6, 2, 1;
  ...
The matrix inverse starts:
   1;
  -2,1;
  -2,-2,1;
  -4,-2,-2,1;
  -10,-4,-2,-2,1;
  -28,-10,-4,-2,-2,1;
  -84,-28,-10,-4,-2,-2,1;
  -264,-84,-28,-10,-4,-2,-2,1;
apparently related to A002420. - _R. J. Mathar_, Apr 08 2013

Row sums are A006134.
Antidiagonal sums are A106188.
Cf. A000984.

A106187 := proc(n,k)
    binomial(2*(n-k),n-k) ;
end proc: # R. J. Mathar, Apr 08 2013

T[n_, k_] := (((2*n - 2*k)!)/((n - k)!)^2); Table[T[n, k], {n, 0, 9}, {k, 0, n}] // Flatten (* Detlef Meya, Aug 11 2024 *)

T(n, k) = binomial(2*(n-k), n-k).

Riordan array (1/sqrt(1-4x), x).

A106188 Expansion of 1/((1-x^2)sqrt(1-4x)).

Original entry on oeis.org

1, 2, 7, 22, 77, 274, 1001, 3706, 13871, 52326, 198627, 757758, 2902783, 11158358, 43019383, 166275878, 644099773, 2499882098, 9719235073, 37845145898, 147565763893, 576103020338, 2251664727613, 8809533747938, 34499268410713

Offset: 0

Paul Barry, Apr 24 2005

Diagonal sums of number triangle A106187.

			1 + 2*x + 7*x^2 + 22*x^3 + 77*x^4 + 274*x^5 + 1001*x^6 + 3706*x^7 + 13871*x^8 + ...

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

Cf. A006134, A054108. Convolution of A000984 and A059841.

CoefficientList[Series[1/((1-x^2)*Sqrt[1-4*x]), {x, 0, 20}], x] (* Vaclav Kotesovec, Feb 03 2014 *)

x='x+O('x^50); Vec(1/((1-x^2)*sqrt(1-4*x))) \\ G. C. Greubel, Mar 16 2017

G.f. 1 / ((1 - x^2) * sqrt(1 - 4*x)).
a(n)=sum{k=0..floor(n/2), binomial(2(n-2k), n-2k)}.
PSUMSIGN transform of A006134. a(n+1) + a(n) = A006134(n). a(n) = Sum_{k=0..n} (-1)^k * binomial(2 * (n-k), n-k). - Michael Somos, Jun 20 2012
First difference is A054108. a(n+1) - a(n) = A054108(n). - Michael Somos, Jun 20 2012
D-finite with recurrence: n*a(n)+2*(1-2*n)*a(n-1) -n*a(n-2) +2*(2*n-1)*a(n-3)=0. - R. J. Mathar, Nov 09 2012
a(n) ~ 2^(2*n+4) / (15*sqrt(Pi*n)). - Vaclav Kotesovec, Feb 03 2014

A120580 Hankel transform of Sum_{k=0..n} C(2k,k).

Original entry on oeis.org

1, 0, -4, -8, 0, 32, 64, 0, -256, -512, 0, 2048, 4096, 0, -16384, -32768, 0, 131072, 262144, 0, -1048576, -2097152, 0, 8388608, 16777216, 0, -67108864, -134217728, 0, 536870912, 1073741824, 0, -4294967296, -8589934592, 0, 34359738368, 68719476736, 0, -274877906944, -549755813888, 0

Offset: 0

Paul Barry, Jun 15 2006

Hankel transform of A006134.
Hankel transform of A098479. - Paul Barry, Sep 19 2008
Hankel transform of A025565. - Paul Barry, Mar 26 2010

Paul Barry, Jacobsthal Decompositions of Pascal's Triangle, Ternary Trees, and Alternating Sign Matrices, Journal of Integer Sequences, 19, 2016, #16.3.5.
Index entries for linear recurrences with constant coefficients, signature (2,-4).

Cf. A006134, A025565, A098479, A104538.

LinearRecurrence[{2,-4},{1,0},50] (* Harvey P. Dale, Feb 13 2023 *)

G.f.: (1-2x)/(1-2x+4x^2).
a(n) = 2^n*(cos(Pi*n/3)-sin(Pi*n/3)/sqrt(3)).
E.g.f.: exp(x)*(cos(sqrt(3)*x) - sin(sqrt(3)*x)/sqrt(3)). - Stefano Spezia, Jul 15 2024

A167860 Primes p dividing every A167859(m) from m=(p-1)/2 to m=(p-1).

Original entry on oeis.org

7, 47, 191, 383, 439, 1151, 1399, 2351, 2879, 3119, 3511, 3559, 4127, 5087, 5431, 6911, 8887, 9127, 9791, 9887, 12391, 13151, 14407, 15551, 16607, 19543, 20399, 21031, 21319, 21839, 23039, 25391, 26399, 28087, 28463, 28711, 29287, 33223, 39551, 43103, 44879, 46271

Offset: 1

Alexander Adamchuk, Nov 13 2009

nonn

Apparently A167860 is a subset of primes of the form 8*k + 7 (A007522).
Every A167859(m) from m=(p-1)/2 to m=(p-1) is divisible by prime p belonging to A167860.
7^3 divides A167859(13) and 7^2 divides A167859(10)-A167859(13).
Every A167859(m) from m=(kp-1 - (p-1)/2) to m=(kp-1) is divisible by prime p from A167860.
Every A167859(m) from m=((p^2-1)/2) to m=(p^2-1) is divisible by prime p from A167860. For p=7 every A167859(m) from m=((p^3-1)/2) to m=(p^3-1) and from m=((p^4-1)/2) to m(p^4-1)is divisible by p^2.

R. J. Mathar, Table of n, a(n) for n = 1..57

Cf. A000984, A006134, A007522, A066796, A082590, A132310, A144635, A167713, A167859, A276649.

A167859 := proc(n)
    option remember;
    if n <= 1 then
        add( (binomial(2*k, k)/2^k)^2, k=0..n) ;
        4^n*% ;
    else
        4*(5*n^2 - 4*n + 1)*procname(n-1) - 16*(2*n - 1)^2*procname(n-2) ;
        %/n^2 ;
    end if;
end proc:
isA167860 := proc(p)
    local m ;
    for m from (p-1)/2 to p-1 do
        if modp(A167859(m),p) > 0 then
            return false;
        end if;
    end do:
    true ;
end proc:
A167860 := proc(n)
    option remember ;
    if n = 0 then
        2;
    else
        p := nextprime(procname(n-1)) ;
        while not isA167860(p) do
            p := nextprime(p) ;
        end do ;
        return p;
    end if;
end proc:
seq(A167860(n),n=1..10) ; # R. J. Mathar, Jan 22 2025

is(p) = if(isprime(p)&&p%2, my(m=Mod(1, p), s=m); for(k=1, p\2, s+=(m*=(2*k-1)/k)^2); !s, 0); \\ Jinyuan Wang, Jul 24 2022

More terms from Jinyuan Wang, Jul 24 2022

A006136 T(n+3,3) from table A045912 of characteristic polynomial of negative Pascal matrix.

Original entry on oeis.org

1, 29, 626, 13869, 347020, 9952274, 321541977, 11416400590, 435869304863, 17605464402686, 743624059688891, 32572923621373010, 1470621027107356485, 68120063089374617281, 3225635202844511176442, 155695310201341829770911

Offset: 0

N. J. A. Sloane

nonn

			1 + 29*x + 626*x^2 + 13869*x^3 + 347020*x^4 + 9952274*x^5 + 321541977*x^6 + ...

N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

W. F. Lunnon, The Pascal matrix, Fib. Quart. vol. 15 (1977) pp. 201-204.

Cf. A045912, A006134, A006135.

{a(n) = if( n<0, 0, polcoeff( charpoly( matrix( n+3, n+3, i, j, -binomial( i+j-2, i-1))), 3))} /* Michael Somos, Jul 10 2002 */

Edited by Michael Somos, Jul 19 2002

cp's OEIS Frontend

A293490 a(n) = Sum_{k=0..n} binomial(2*k, k)*binomial(2*n-k, n).

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A307354 a(n) = Sum_{0<=i<=j<=n} (-1)^(i+j) * (i+j)!/(i!*j!).

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A323222 A(n, k) = [x^k] (1 - 4*x)^(-n/2)*x/(1 - x), square array read by ascending antidiagonals with n >= 0 and k >= 0.

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A006135 T(n+2,2) from table A045912 of characteristic polynomial of negative Pascal matrix.

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A096465 Triangle (read by rows) formed by setting all entries in the first column and in the main diagonal ((i,i) entries) to 1 and the rest of the entries by the recursion T(n, k) = T(n-1, k) + T(n, k-1).

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A106187 Sequence array for central binomial numbers A000984.

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A106188 Expansion of 1/((1-x^2)*sqrt(1-4*x)).

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A293490 a(n) = Sum_{k=0..n} binomial(2k, k)binomial(2*n-k, n).

A323222 A(n, k) = [x^k] (1 - 4x)^(-n/2)x/(1 - x), square array read by ascending antidiagonals with n >= 0 and k >= 0.

A106188 Expansion of 1/((1-x^2)sqrt(1-4x)).