A002457 -id:A002457 - cp's OEIS frontend

A329913 The fifth moments of the squared binomial coefficients; a(n) = Sum_{m=0..n} m^5*binomial(n, m)^2.

0, 1, 36, 540, 6080, 56250, 455112, 3342192, 22809600, 146988270, 904475000, 5358254616, 30750385536, 171773279860, 937514244240, 5014575000000, 26351064760320, 136319273714070, 695429503781400, 3503580441563400, 17452918098000000, 86055711108818220

Offset: 0

Nikita D. Gogin, Nov 24 2019

nonn

H. W. Gould, Combinatorial Identities, 1972. (See formulas 3.77, 3.78, and 3.79 on page 31.)

Robert Israel, Table of n, a(n) for n = 0..1638

Cf. A074334, A294486.

Cf. A000984, A002457, A037966, A037972, A074334, A329444.

[(&+[Binomial(n,k)^2*k^5: k in [0..n]]): n in [0..30]]; // G. C. Greubel, Jun 23 2022

seq( binomial(2*n,n)*n^4*(n^3 + 3*n^2 - 3*n - 5)/((16*n-8)*(2*n-3)),n=0..30); # Robert Israel, Jan 26 2020

Table[Sum[m^5*(Binomial[n, m])^2, {m, 0, n}], {n, 21}]

a(n) = sum(k=0, n, k^5*binomial(n, k)^2); \\ Michel Marcus, Nov 24 2019

[n^4*(n+1)*(n^3+3*n^2-3*n-5)/(8*(2*n-1)*(2*n-3))*catalan_number(n) for n in (0..30)] # G. C. Greubel, Jun 23 2022

a(n) = binomial(2*n,n)*n^4*(n^3 + 3*n^2 - 3*n - 5)/(8*(2*n-1)*(2*n-3)).
G.f.: x*(1 + 14*x - 54*x^2 + 404*x^3 - 1544*x^4 + 2880*x^5 - 2160*x^6)/(1-4*x)^(11/2). - Stefano Spezia, Jan 03 2020
(-12960 + 8640*n)*a(n) + (7200 - 13680*n)*a(n + 1) + (3920 + 9056*n)*a(n + 2) + (-4184 - 3160*n)*a(n + 3) + (1404 + 620*n)*a(n + 4) + (-584 - 110*n)*a(n + 5) + (14 + 10*n)*a(n + 6) + (n + 6)*a(n + 7) = 0. - Robert Israel, Jan 26 2020

A368846 Triangle read by rows: T(n, k) = (-1)^(n + k)2binomial(2k - 1, n) binomial(2n + 1, 2k) for k > 0, and k^n for k = 0.

Original entry on oeis.org

1, 0, 6, 0, 0, 30, 0, 0, -70, 140, 0, 0, 0, -840, 630, 0, 0, 0, 924, -6930, 2772, 0, 0, 0, 0, 18018, -48048, 12012, 0, 0, 0, 0, -12870, 216216, -300300, 51480, 0, 0, 0, 0, 0, -350064, 2042040, -1750320, 218790, 0, 0, 0, 0, 0, 184756, -5542680, 16628040, -9699690, 923780

Offset: 0

Peter Luschny, Jan 07 2024

The row sums of the inverse triangle (A368847/A368848) are the unsigned Bernoulli numbers |B(2n)|. To get the signed Bernoulli numbers B(2n), one only needs to change the sign factor in the definition from (-1)^(n + k) to (-1)^(n + 1).

Conjecture: |Sum_{j=0..k} T(k + j, k)| = A229580(k + 1) for k >= 0.

			[0] [1]
[1] [0, 6]
[2] [0, 0,  30]
[3] [0, 0, -70,  140]
[4] [0, 0,   0, -840,    630]
[5] [0, 0,   0,  924,  -6930,   2772]
[6] [0, 0,   0,    0,  18018,  -48048,   12012]
[7] [0, 0,   0,    0, -12870,  216216, -300300,    51480]
[8] [0, 0,   0,    0,      0, -350064, 2042040, -1750320, 218790]

Paolo Xausa, Table of n, a(n) for n = 0..11475 (rows 0..150 of the triangle, flattened).
Thomas Curtright, Scale Invariant Scattering and the Bernoulli Numbers, arXiv:2401.00586 [math-ph], Jan 2024.
Thomas L. Curtright, Bernoulli Partitions, arXiv:2502.09633 [math.CO], 2025.

Cf. A368847/A368848 (inverse), A369134, A369135, A002457 (main diagonal), A000367/A002445 (Bernoulli(2n)), A229580.

A368846[n_,k_] := If[k==0, Boole[n==0], (-1)^(n+k) 2 Binomial[2k-1, n] Binomial[2n+1, 2k]];
Table[A368846[n, k], {n,0,10}, {k,0,n}] (* Paolo Xausa, Jan 08 2024 *)

def A368846(n, k):
    if k == 0: return k^n
    if k  > n: return 0
    return (-1)^(n + k)*2*binomial(2*k - 1, n)*binomial(2*n + 1, 2*k)
for n in range(10): print([A368846(n, k) for k in range(n+1)])

A377197 Expansion of 1/(1 - 4*x/(1-x))^(3/2).

Original entry on oeis.org

1, 6, 36, 206, 1146, 6258, 33728, 180018, 953628, 5021698, 26315676, 137350746, 714455826, 3705635646, 19171860336, 98973407550, 509963556330, 2623133951730, 13472299015580, 69098721151530, 353966981339070, 1811212435206070, 9258333786967920, 47281424213258070

Offset: 0

Seiichi Manyama, Oct 19 2024

nonn

Vincenzo Librandi, Table of n, a(n) for n = 0..1000

Cf. A085362, A377199, A377200.
Cf. A002457, A377198.
Cf. A374497.

R:=PowerSeriesRing(Rationals(), 35); Coefficients(R!( 1/(1 - 4*x/(1-x))^(3/2))); // Vincenzo Librandi, May 11 2025

Table[Sum[(2*k+1)*Binomial[2*k,k]*Binomial[n-1,n-k],{k,0,n}],{n,0,30}] (* Vincenzo Librandi, May 11 2025 *)

a(n) = sum(k=0, n, (2*k+1)*binomial(2*k, k)*binomial(n-1, n-k));

a(0) = 1; a(n) = 2 * Sum_{k=0..n-1} (3-k/n) * a(k).
a(n) = (6*n*a(n-1) - 5*(n-2)*a(n-2))/n for n > 1.
a(n) = Sum_{k=0..n} (2*k+1) * binomial(2*k,k) * binomial(n-1,n-k).
a(n) ~ 16 * sqrt(n) * 5^(n - 3/2) / sqrt(Pi). - Vaclav Kotesovec, Oct 26 2024
a(n) = 6*hypergeom([5/2, 1-n], [2], -4) for n > 0. - Stefano Spezia, May 08 2025

A053124 Triangle of coefficients of Chebyshev's U(n,2*x-1) polynomials (exponents of x in increasing order).

Original entry on oeis.org

1, -2, 4, 3, -16, 16, -4, 40, -96, 64, 5, -80, 336, -512, 256, -6, 140, -896, 2304, -2560, 1024, 7, -224, 2016, -7680, 14080, -12288, 4096, -8, 336, -4032, 21120, -56320, 79872, -57344, 16384, 9, -480, 7392, -50688, 183040, -372736, 430080, -262144, 65536, -10, 660, -12672, 109824, -512512, 1397760, -2293760, 2228224

Offset: 0

Wolfdieter Lang

a(n,m) = (4^m)*A053122(n,m).
G.f. for row polynomials U^{*}(n,x) = U(n,2*x-1) (signed triangle): 1/(1+2*z*(1-2*x) + z^2). Unsigned triangle |a(n,m)| has g.f. 1/(1-2*z*(1+2*x)+z^2) for the row polynomials.
Row sums (signed triangle) A000027(n+1) (natural numbers). Row sums (unsigned triangle) A001109(n+1).
In the language of Shapiro et al. (see A053121 for the reference) such a lower triangular (ordinary) convolution array, considered as a matrix, belongs to a Riordan group.

			{1}; {-2,4}; {3,-16,16}; {-4,40,-96,64}; {5,-80,336,-512,256};... E.g., fourth row (n=3) {-4,40,-96,64} corresponds to polynomial U(3,2*x-1)= -4+40*x-96*x^2+64*x^3.

C. Lanczos, Applied Analysis. Prentice-Hall, Englewood Cliffs, NJ, 1956, p. 518.
G. Pólya and G. Szegő, Problems and Theorems in Analysis I (Springer 1924, reprinted 1972), Part One, Chap. 1, problem 39, page 7.
Theodore J. Rivlin, Chebyshev polynomials: from approximation theory to algebra and number theory, 2. ed., Wiley, New York, 1990.

T. D. Noe, Rows n = 0..50 of triangle, flattened
C. Lanczos, Applied Analysis (Annotated scans of selected pages). See page 518.
Index entries for sequences related to Chebyshev polynomials.

Cf. A053122, A053125.

Table[ CoefficientList[ ChebyshevU[n, 2x - 1], x], {n, 0, 9}] // Flatten (* Jean-François Alcover, Dec 05 2012 *)

a(n, m) := 0 if n < m, otherwise (4^m)*((-1)^(n-m))*binomial(n+m+1, 2*m+1);
a(n, m) = -2*a(n-1, m) + 4*a(n-1, m-1) - a(n-2, m), a(n, m) := 0 if n=-1 or m=-1 or n < m, a(0, 0)=1;
g.f. for m-th column (signed triangle): ((4*x/(1+x)^2)^m)/(1+x)^2.
In other words, Riordan array (1/(1+x)^2, 4x/(1+x)^2). - Ralf Stephan, Jan 21 2014

A078817 Table by antidiagonals giving variants on Catalan sequence: T(n,k)=C(2n,n)C(2k,k)(2k+1)/(n+k+1).

Original entry on oeis.org

1, 3, 1, 10, 4, 2, 35, 15, 9, 5, 126, 56, 36, 24, 14, 462, 210, 140, 100, 70, 42, 1716, 792, 540, 400, 300, 216, 132, 6435, 3003, 2079, 1575, 1225, 945, 693, 429, 24310, 11440, 8008, 6160, 4900, 3920, 3080, 2288, 1430, 92378, 43758, 30888, 24024, 19404

Offset: 0

Henry Bottomley, Dec 07 2002

			Rows start:
     1,     3,    10,    35,   126,   462,  1716,
     1,     4,    15,    56,   210,   792,  3003,
     2,     9,    36,   140,   540,  2079,  8008,
     5,    24,   100,   400,  1575,  6160, 24024,
    14,    70,   300,  1225,  4900, 19404, 76440,
    42,   216,   945,  3920, 15876, 63504,252252,
   132,   693,  3080, 12936, 52920,213444,853776,
etc.

Ira Gessel, Super ballot numbers, J. Symbolic Computation 14 (1992), 179-194.
Isabel Cação, Helmuth R. Malonek, Maria Irene Falcão, and Graça Tomaz, Combinatorial Identities Associated with a Multidimensional Polynomial Sequence, J. Int. Seq., Vol. 21 (2018), Article 18.7.4.
Jovan Mikic, A Note on the Gessel Numbers, arXiv:2203.12931 [math.CO], 2022.

Columns include A000108 (catalan), A038629, A078818 and A078819. Rows include A001700, A001791, A007946 and A078820. Diagonals include A002894 and A060150.

Essentially a reflected version of A033820.

A078817 := proc(n,k)
    binomial(2*n,n)*binomial(2*k,k)*(2*k+1)/(n+k+1) ;
end proc: # R. J. Mathar, Dec 06 2018

T(n, k) = A000984(n)*A002457(k)/(n+k+1) = T(k, n)*(2k+1)/(2n+1).

A107254 a(n) = SF(2n-1)/SF(n-1)^2 where SF = A000178.

Original entry on oeis.org

1, 1, 12, 8640, 870912000, 22122558259200000, 222531556847250309120000000, 1280394777025250130271722799104000000000, 5746332926632566442385615219551212618645504000000000000

Offset: 0

Henry Bottomley, May 14 2005

nonn

Inverse product of all matrix elements of n X n Hilbert matrix M(i,j) = 1/(i+j-1) (i,j = 1..n). - Alexander Adamchuk, Apr 12 2006

The n X n matrix with A(i,j) = 1/(i+j-1)! (i,j = 1..n) has determinant (-1)^floor(n/2)/a(n). - Mikhail Lavrov, Nov 01 2022

			a(3) = 1!*2!*3!*4!*5!/(1!*2!*1!*2!) = 34560/4 = 8640.
n = 2: HilbertMatrix[n,n]
  1/1 1/2
  1/2 1/3
so a(2) = 1 / (1 * 1/2 * 1/2 * 1/3) = 12.
The n X n Hilbert matrix begins:
  1/1 1/2 1/3 1/4  1/5  1/6  1/7  1/8 ...
  1/2 1/3 1/4 1/5  1/6  1/7  1/8  1/9 ...
  1/3 1/4 1/5 1/6  1/7  1/8  1/9 1/10 ...
  1/4 1/5 1/6 1/7  1/8  1/9 1/10 1/11 ...
  1/5 1/6 1/7 1/8  1/9 1/10 1/11 1/12 ...
  1/6 1/7 1/8 1/9 1/10 1/11 1/12 1/13 ...

Alois P. Heinz, Table of n, a(n) for n = 0..20
Mathematics Stack Exchange, Determinant of a matrix involving factorials.
Eric Weisstein's World of Mathematics, Hilbert Matrix.

Cf. A000178, A002457, A005249, A098118.

A107254:= func< n | n eq 0 select 1 else (&*[Factorial(n+j)/Factorial(j): j in [0..n-1]]) >;
[A107254(n): n in [0..12]]; // G. C. Greubel, Apr 21 2021

a:= n-> mul((n+i)!/i!, i=0..n-1):
seq(a(n), n=0..10);  # Alois P. Heinz, Jul 23 2012

Table[Product[(i+j-1),{i,1,n},{j,1,n}], {n,1,10}] (* Alexander Adamchuk, Apr 12 2006 *)
Table[n!*BarnesG[2n+1]/(BarnesG[n+2]*BarnesG[n+1]), {n,0,12}] (* G. C. Greubel, Apr 21 2021 *)

a = lambda n: prod(rising_factorial(k,n) for k in (1..n))
print([a(n) for n in (0..10)]) # Peter Luschny, Nov 29 2015

a(n) = n!*(n+1)!*(n+2)!*...*(2n-1)!/(0!*1!*2!*3!*...*(n-1)!) = A000178(2n-1)/A000178(n-1)^2 = A079478(n)/A000984(n) = A079478(n-1)*A009445(n-1) = A107252(n)*A000142(n) = A088020(n)/A039622(n).
a(n) = 1/Product_{j=1..n} ( Product_{i=1..n} 1/(i+j-1) ). - Alexander Adamchuk, Apr 12 2006
a(n) = 2^(n*(n-1)) * A136411(n) for n > 0 . - Robert Coquereaux, Apr 06 2013
a(n) = A136411(n) * A053763(n) for n > 0. [Following remark from Robert Coquereaux] - M. F. Hasler, Apr 06 2013
a(n) ~ A * 2^(2*n^2-1/12) * n^(n^2+1/12) / exp(3*n^2/2+1/12), where A = 1.28242712910062263687534256886979... is the Glaisher-Kinkelin constant (see A074962). - Vaclav Kotesovec, Feb 10 2015
a(n) = Product_{k=1..n} rf(k,n) where rf denotes the rising factorial. - Peter Luschny, Nov 29 2015
a(n) = (n! * G(2*n+1))/(G(n+1)*G(n+2)), where G(n) is the Barnes G - function. - G. C. Greubel, Apr 21 2021

A116666 Triangle, row sums = number of edges in n-dimensional hypercubes.

Original entry on oeis.org

1, 1, 3, 1, 6, 5, 1, 9, 15, 7, 1, 12, 30, 28, 9, 1, 15, 50, 70, 45, 11, 1, 18, 75, 140, 135, 66, 13, 1, 21, 105, 245, 315, 231, 91, 15, 1, 24, 140, 392, 630, 616, 364, 120, 17, 1, 27, 180, 588, 1134, 1386, 1092, 540, 153, 19, 1, 30, 225, 840, 1890, 2772

Offset: 1

Gary W. Adamson, Feb 22 2006

Terms in the array rows tend to A001787, number of edges in n-dimensional hypercubes: 1, 4, 12, 32, 80, 192, 448... Row sums of the sequence also = A001787.

			First few rows of the array are:
1 1 1 1 1...
1 4 7 10 13...
1 4 12 25 43...
1 4 12 32 71...
1 4 12 32 80...
...
Then take differences of columns which become rows of the triangle:
1;
1, 3;
1, 6, 5;
1, 9, 15, 7;
1, 12, 30, 28, 9;
1, 15, 50, 70, 45, 11;
1, 18, 75, 140, 135, 66, 13;
1, 21, 105, 245, 315, 231, 91, 15;
...

Reinhard Zumkeller, Rows n = 1..125 of table, flattened

Cf. A001787.

Cf. A007318, A005408, A002457 (central terms).

Flat(List([0..100],n->List([1..n+1],k->Binomial(n,k-1)*(2*k-1)))); # Muniru A Asiru, Jan 30 2018

a116666 n k = a116666_tabl !! (n-1) !! (k-1)
a116666_row n = a116666_tabl !! (n-1)
a116666_tabl = zipWith (zipWith (*)) a007318_tabl a158405_tabl
-- Reinhard Zumkeller, Nov 02 2013

/* As triangle */ [[(2*k+1)*Binomial(n,k): k in [0..n]]: n in [0..10]]; // G. C. Greubel, Jan 29 2018

seq(seq(binomial(n,k-1)*(2*k-1), k=1..n+1),n=0..100); # Muniru A Asiru, Jan 30 2018

Table[Binomial[n,k]*(2*k+1), {n,0,10}, {k,0,n}] (* G. C. Greubel, Jan 29 2018 *)

for(n=0,10, for(k=0,n, print1(binomial(n,k)*(2*k+1), ", "))) \\ G. C. Greubel, Jan 29 2018

From an array, rows = binomial transforms of (1,0,0,0...); (1,3,0,0,0...); (1,3,5,0,0,0...); difference rows of the columns become rows of the triangle.

T(n,k) = binomial(n,k-1) * (2*k - 1), 1 <= k <= n. - Reinhard Zumkeller, Nov 02 2013

A167867 a(n) = 2^n * Sum_{k=0..n} binomial(2*k,k)^3 / 2^k.

Original entry on oeis.org

1, 10, 236, 8472, 359944, 16722896, 822334816, 42068907200, 2215884717400, 119364801362800, 6545334930678816, 364137834051739200, 20502307365808906816, 1166063313963833813632, 66893439680369963627264

Offset: 0

Alexander Adamchuk, Nov 14 2009

nonn

The expression a(n) = B^n*Sum_{ k=0..n } binomial(2*k,k)/B^k gives A006134 for B=1, A082590 (B=2), A132310 (B=3), A002457 (B=4), A144635 (B=5), A167713 (B=16).

The expression a(n) = B^n*Sum_{ k=0..n } binomial(2*k,k)^3/B^k gives A079727 for B=1, A167867 (B=2), A167868 (B=3), A167869 (B=4), A167870 (B=16), A167871 (B=64).

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

Cf. A079727, A167867, A167868, A167869, A167870, A167872.

Cf. A000984, A066796, A006134, A082590, A132310, A002457, A144635, A167713, A167859.

Table[2^n Sum[Binomial[2k,k]^3/2^k,{k,0,n}],{n,0,30}] (* Vincenzo Librandi, Mar 26 2012 *)

a(n) = 2^n * Sum_{k=0..n} binomial(2*k,k)^3 / 2^k.
Recurrence: n^3*a(n) = 2*(33*n^3 - 48*n^2 + 24*n - 4)*a(n-1) - 16*(2*n-1)^3*a(n-2). - Vaclav Kotesovec, Aug 13 2013
a(n) ~ 2^(6*n+5)/(31*(Pi*n)^(3/2)). - Vaclav Kotesovec, Aug 13 2013

More terms from Sean A. Irvine, Apr 27 2010

A167868 a(n) = 3^n * Sum_{k=0..n} binomial(2*k,k)^3 / 3^k.

Original entry on oeis.org

1, 11, 249, 8747, 369241, 17110731, 840221217, 42944901219, 2260581606657, 121714776747971, 6671749658197129, 371062413164972955, 20887218937200347281, 1187720356043817041843, 68124474120573747125529

Offset: 0

Alexander Adamchuk, Nov 14 2009

nonn

The expression a(n) = B^n*Sum_{k=0..n} binomial(2*k,k)/B^k gives A006134 for B=1, A082590 (B=2), A132310 (B=3), A002457 (B=4), A144635 (B=5), A167713 (B=16).

The expression a(n) = B^n*Sum_{k=0..n} binomial(2*k,k)^3/B^k gives A079727 for B=1, A167867 (B=2), A167868 (B=3), A167869 (B=4), A167870 (B=16), A167871 (B=64).

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

Cf. A079727, A167867, A167868, A167869, A167870, A167872.

Cf. A000984, A066796, A006134, A082590, A132310, A002457, A144635, A167713, A167859.

Table[3^n Sum[Binomial[2k,k]^3/3^k,{k,0,n}],{n,0,20}] (* Vincenzo Librandi, Mar 26 2012 *)

a(n) = 3^n * Sum_{k=0..n} binomial(2*k,k)^3 / 3^k.
Recurrence: n^3*a(n) = (67*n^3 - 96*n^2 + 48*n - 8)*a(n-1) - 24*(2*n-1)^3*a(n-2). - Vaclav Kotesovec, Aug 13 2013
a(n) ~ 2^(6*n+6)/(61*(Pi*n)^(3/2)). - Vaclav Kotesovec, Aug 13 2013

More terms from Sean A. Irvine, Apr 27 2010

A167869 a(n) = 4^n * Sum_{k=0..n} binomial(2*k,k)^3 / 4^k.

Original entry on oeis.org

1, 12, 264, 9056, 379224, 17519904, 858968640, 43860112128, 2307187351512, 124161781334048, 6803252453289408, 378260174003539200, 21287072393719585216, 1210206988807094340864, 69402141007670673363456

Offset: 0

Alexander Adamchuk, Nov 14 2009

nonn

The expression a(n) = B^n*Sum_{k=0..n} binomial(2*k,k)/B^k gives A006134 for B=1, A082590 (B=2), A132310 (B=3), A002457 (B=4), A144635 (B=5), A167713 (B=16).

The expression a(n) = B^n*Sum_{k=0..n} binomial(2*k,k)^3/B^k gives A079727 for B=1, A167867 (B=2), A167868 (B=3), A167869 (B=4), A167870 (B=16), A167871 (B=64).

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

Cf. A079727, A167867, A167868, A167869, A167870, A167872.

Cf. A000984, A066796, A006134, A082590, A132310, A002457, A144635, A167713, A167859.

Table[4^n Sum[Binomial[2k,k]^3/4^k,{k,0,n}],{n,0,30}] (* Vincenzo Librandi, Mar 26 2012 *)

a(n) = 4^n * Sum_{k=0..n} binomial(2*k,k)^3 / 4^k.
Recurrence: n^3*a(n) = 4*(17*n^3 - 24*n^2 + 12*n - 2)*a(n-1) - 32*(2*n-1)^3*a(n-2). - Vaclav Kotesovec, Aug 13 2013
a(n) ~ 2^(6*n+4)/(15*(Pi*n)^(3/2)). - Vaclav Kotesovec, Aug 13 2013

More terms from Sean A. Irvine, Apr 27 2010

cp's OEIS Frontend

A329913 The fifth moments of the squared binomial coefficients; a(n) = Sum_{m=0..n} m^5*binomial(n, m)^2.

Views

Author

Keywords

References

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

SageMath

Formula

A368846 Triangle read by rows: T(n, k) = (-1)^(n + k)*2*binomial(2*k - 1, n)* binomial(2*n + 1, 2*k) for k > 0, and k^n for k = 0.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

SageMath

A377197 Expansion of 1/(1 - 4*x/(1-x))^(3/2).

Views

Author

Keywords

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Formula

A053124 Triangle of coefficients of Chebyshev's U(n,2*x-1) polynomials (exponents of x in increasing order).

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Mathematica

Formula

A078817 Table by antidiagonals giving variants on Catalan sequence: T(n,k)=C(2n,n)*C(2k,k)*(2k+1)/(n+k+1).

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Maple

Formula

A107254 a(n) = SF(2n-1)/SF(n-1)^2 where SF = A000178.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Maple

Mathematica

Sage

Formula

A116666 Triangle, row sums = number of edges in n-dimensional hypercubes.

Views

Author

Keywords

Comments

Examples

Links

A368846 Triangle read by rows: T(n, k) = (-1)^(n + k)2binomial(2k - 1, n) binomial(2n + 1, 2k) for k > 0, and k^n for k = 0.

A078817 Table by antidiagonals giving variants on Catalan sequence: T(n,k)=C(2n,n)C(2k,k)(2k+1)/(n+k+1).