A002897 -id:A002897 - cp's OEIS frontend

A014137 Partial sums of Catalan numbers (A000108).

1, 2, 4, 9, 23, 65, 197, 626, 2056, 6918, 23714, 82500, 290512, 1033412, 3707852, 13402697, 48760367, 178405157, 656043857, 2423307047, 8987427467, 33453694487, 124936258127, 467995871777, 1757900019101, 6619846420553, 24987199492705, 94520750408709, 358268702159069

Offset: 0

N. J. A. Sloane

This is also the result of applying the transformation on generating functions A -> 1/((1 - x)*(1 - x*A)) to the g.f. for the Catalan numbers.
p divides a(p) - 3 for prime p = 3 and p = {7, 13, 19, 31, 37, 43, ...} = A002476 (Primes of the form 6*n + 1). p^2 divides a(p^2) - 3 for prime p > 3. - Alexander Adamchuk, Jul 11 2006
Prime p divides a(p) for p = {2, 3, 5, 11, 17, 23, 29, 41, 47, 53, 59, 71, 83, 89, 101, ...} = A045309 (Primes congruent to {0, 2} mod 3); and A045309 (Primes p such that x^3 = n (integer) has only one solution mod p). Nonprime numbers n such that n divides a(n) are listed in A128287 = {1, 8, 133, ...}. - Alexander Adamchuk, Feb 23 2007
For p prime >= 5, a(p-1) = 1 or -2 (mod p) according as p = 1 or -1 (mod 3) (see Pan and Sun link). For example, with p=5, a(p-1) = 23 = -2 (mod p). - David Callan, Nov 29 2007
Hankel transform is A010892(n+1). - Paul Barry, Apr 24 2009
Equals INVERTi transform of A000245: (1, 3, 9, 28, ...). - Gary W. Adamson, May 15 2009
The subsequence of prime partial sums of Catalan numbers begins: a(1) = 2, a(4) = 23, a(6) = 197, a(16) = 48760367; see A121852. - Jonathan Vos Post, Feb 10 2010
Number of lattice paths from (0,0) to (n,n) which do not go above the diagonal x=y using steps (1,k), (k,1) with k >= 1 including two kinds of (1,1). - Alois P. Heinz, Oct 14 2015
Binomial transform of A086246(n+1) = [1, 1, 1, 2, 4, 9, ...], or, equivalently, of A001006 (Motzkin numbers) with 1 prepended.

			G.f. = 1 + 2*x + 4*x^2 + 9*x^3 + 23*x^4 + 65*x^5 + 197*x^6 + 626*x^7 + 2056*x^8 + ...

G. C. Greubel, Table of n, a(n) for n = 0..1000 (terms 0 to 200 by T. D. Noe)
G. Alvarez, J. E. Bergner, and R. Lopez, Action Graphs and Catalan Numbers, J. Int. Seq. 18 (2015), 15.7.2.
M. Apagodu, D. Zeilberger, Using the Freshman's dream to prove combinatorial congruences, Am. Math. Mon. 124, No. 7, 597-608 (2017), prop. 2'
Maciej Bendkowski and Pierre Lescanne, Combinatorics of explicit substitutions, arXiv:1804.03862 [cs.LO], 2018.
W. Chammam, F. Marcellán and R. Sfaxi, Orthogonal polynomials, Catalan numbers, and a general Hankel determinant evaluation, Linear Algebra Appl. 436(7) (2012), 2105-2116.
Joel E. Cohen, Variance Functions of Asymptotically Exponentially Increasing Integer Sequences Go Beyond Taylor's Law, J. Int. Seq., Vol. 25 (2022), Article 22.9.3.
Dennis E. Davenport, Lara K. Pudwell, Louis W. Shapiro, and Leon C. Woodson, The Boundary of Ordered Trees, Journal of Integer Sequences, 18 (2015), Article 15.5.8.
Nancy S. S. Gu, Nelson Y. Li, and Toufik Mansour, 2-Binary trees: bijections and related issues, Discr. Math., 308 (2008), 1209-1221.
Guo-Niu Han, Enumeration of Standard Puzzles, 2011. [Cached copy]
Guo-Niu Han, Enumeration of Standard Puzzles, arXiv:2006.14070 [math.CO], 2020.
Ângela Mestre and José Agapito, A Family of Riordan Group Automorphisms, J. Int. Seq., Vol. 22 (2019), Article 19.8.5.
I. Pak, Partition identities and geometric bijections, Proc. Amer. Math. Soc. 132 (2004), 3457-3462.
Hao Pan and Zhi-Wei Sun, A combinatorial identity with application to Catalan numbers, arXiv:math/0509648 [math.CO], 2005-2006.
Murray Tannock, Equivalence classes of mesh patterns with a dominating pattern, MSc Thesis, Reykjavik Univ., May 2016. See Appendix B2.
Kevin Topley, Computationally Efficient Bounds for the Sum of Catalan Numbers, arXiv:1601.04223 [math.CO], 2016.

Cf. A000108, A000245, A000984, A001246, A002476, A002897, A006134, A033536, A045309, A079727, A082894, A094638, A094639, A128287, A358436.

[(&+[Catalan(k): k in [0..n]]): n in [0..40]]; // G. C. Greubel, Jun 30 2024

a:= proc(n) option remember; `if`(n<2, n+1,
      ((5*n-1)*a(n-1)-(4*n-2)*a(n-2))/(n+1))
    end:
seq(a(n), n=0..30);  # Alois P. Heinz, May 18 2013
A014137List := proc(m) local A, P, n; A := [1]; P := [1];
for n from 1 to m - 2 do P := ListTools:-PartialSums([op(P), P[-n]]);
A := [op(A), P[-1]] od; A end: A014137List(30); # Peter Luschny, Mar 26 2022

Table[Sum[(2k)!/(k!)^2/(k+1),{k,0,n}],{n,0,30}] (* Alexander Adamchuk, Jul 11 2006 *)
Accumulate[CatalanNumber[Range[0,30]]] (* Harvey P. Dale, May 08 2012 *)
a[ n_] := SeriesCoefficient[ (1 - (1 - 4 x)^(1/2)) / (2 x (1 - x)), {x, 0, n}]; (* Michael Somos, Oct 24 2015 *)
Table[(1 + CatalanNumber[n] (3 (n + 1) Hypergeometric2F1[1, -n, 1/2 - n, 1/4] - 4 n - 2))/2, {n, 0, 20}] (* Vladimir Reshetnikov, Oct 03 2016 *)

Vec((1-(1-4*x)^(1/2))/(2*x*(1-x))+O(x^99)) \\ Charles R Greathouse IV, Feb 11 2011

sm(v)={my(s=vector(#v)); s[1]=v[1]; for(n=2, #v, s[n]=v[n]+s[n-1]); s; }
C(n)=binomial(2*n, n)/(n+1);
sm(vector(66, n, C(n-1)))
/* Joerg Arndt, May 04 2013 */

from _future_ import division
A014137_list, b, s = [], 1, 0
for n in range(10**2):
    s += b
    A014137_list.append(s)
    b = b*(4*n+2)//(n+2) # Chai Wah Wu, Jan 28 2016

def A014137():
    f, c, n = 1, 1, 1
    while True:
        yield f
        n += 1
        c = c * (4*n - 6) // n
        f = c + f
a = A014137()
print([next(a) for  in range(29)]) # _Peter Luschny, Nov 30 2016

a(n) = A014138(n-1) + 1.
G.f.: (1 - (1 - 4*x)^(1/2))/(2*x*(1 - x)).
a(n) = Sum_{k=0..n} (2k)!/(k!)^2/(k+1). - Alexander Adamchuk, Jul 11 2006
D-finite with recurrence: (n+1)*a(n) + (1-5*n)*a(n-1) + 2*(2*n-1)*a(n-2) = 0. - R. J. Mathar, Dec 14 2011
Mathar's formula reduces to 2*(2*n-1)*C(n-1) = (n+1)*C(n), which is a known recurrence of the Catalan numbers, so the conjecture is true. - Peter J. Taylor, Mar 23 2015
Let C(n+1) = binomial(2*n+2,n+1)/(n+2) and H(n) = hypergeometric([1,n+3/2],[n+3],4) then A014137(n) = -(-1)^(2/3) - C(n+1)*H(n) and A014138(n) = -I^(2/3) - C(n+1)*H(n). - Peter Luschny, Aug 09 2012
G.f. (conjecture): Q(0)/(1-x), where Q(k)= 1 + (4*k + 1)*x/(k + 1 - 2*x*(k + 1)*(4*k + 3)/(2*x*(4*k + 3) + (2*k + 3)/Q(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 14 2013
a(n) ~ 2^(2*n + 2)/(3*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Dec 10 2013
0 =  a(n)*(16*a(n+1) - 26*a(n+2) + 10*a(n+3)) + a(n+1)*(-14*a(n+1) + 23*a(n+2) - 11*a(n+3)) + a(n+2)*(a(n+2) + a(n+3)) if n >= 0. - Michael Somos, Oct 24 2015
a(n) = (1 + A000108(n)*(3*(n+1)*hypergeom([1,-n], [1/2-n], 1/4) - 4*n - 2))/2. - Vladimir Reshetnikov, Oct 03 2016
G.f. A(x) satisfies: A(x) = 1 / (1 - x) + x * (1 - x) * A(x)^2. - Ilya Gutkovskiy, Jul 25 2021
From Peter Luschny, Nov 16 2022: (Start)
a(n) = C(n)*hypergeom([1, -n - 1], [1/2 - n], 1/4) + 1/2.
a(n) = A358436(n) / C(n). (End)
E.g.f.: exp(2*x)*(BesselI(0, 2*x)/2 - BesselI(1, 2*x)) + exp(x)/2*(3*Integral_{x=-oo..oo} BesselI(0,2*x)*exp(x) dx + 1). - Mélika Tebni, Sep 01 2024

A002894 a(n) = binomial(2n, n)^2.

Original entry on oeis.org

1, 4, 36, 400, 4900, 63504, 853776, 11778624, 165636900, 2363904400, 34134779536, 497634306624, 7312459672336, 108172480360000, 1609341595560000, 24061445010950400, 361297635242552100, 5445717990022688400, 82358080713306090000, 1249287673091590440000

Offset: 0

N. J. A. Sloane

a(n) is the number of monotonic paths (only moving N and E) in the lattice [0..2n] X [0..2n] that contain the points (0,0), (n,n) and (2n,2n). - Joe Keane (jgk(AT)jgk.org), Jun 06 2002
This is the Taylor expansion of a special point on a curve described by Beauville. - Matthijs Coster, Apr 28 2004
Expansion of K(k) / (Pi/2) in powers of m/16 = (k/4)^2, where K(k) is the complete elliptic integral of the first kind evaluated at k. - Michael Somos, Mar 04 2003
Square lattice walks that start and end at origin after 2n steps. - Gareth McCaughan and Michael Somos, Jun 12 2004
If A is a random matrix in USp(4) (4 X 4 complex matrices that are unitary and symplectic) then a(n)=E[(tr(A^k))^{2n}] for any k > 4. - Andrew V. Sutherland, Apr 01 2008
From R. H. Hardin, Feb 03 2016 and R. J. Mathar, Feb 18 2016: (Start)
Also, number of 2 X (2n) arrays of permutations of 2n copies of 0 or 1 with row sums equal.
For example, some solutions for n=3:
  0 1 0 1 0 1   0 1 0 1 1 0   0 0 1 0 1 1   1 1 1 0 0 0
  1 0 0 0 1 1   1 1 0 1 0 0   0 0 0 1 1 1   0 0 1 1 0 1
There is a simple combinatorial argument to show that this is a(n): We have 2n copies of 0's and 1's and need equal row sums. Therefore there must be n 1's in each of the two rows. Otherwise there are no constraints, so there are C(2n,n) ways of placing the 1's in the first row and independently C(2n,n) ways of placing the 1's in the second. The product is clearly C(2n,n)^2. (End)
Also the even part of the bisection of A241530. One half of the odd part is given in A000894. - Wolfdieter Lang, Sep 06 2016
From Peter Bala, Jan 26 2018: (Start)
Let S = {[1,0,0], [0,1,0], [1,0,1], [0,1,1]} be a set of four column vectors. Then a(n) equals the number of 3 X k arrays whose columns belong to the set S and whose row sums are all equal to n (apply Eger, Theorem 3). An example is given below. Equivalently, a(n) equals the number of lattice paths from (0,0,0) to (n,n,n) using steps (1,0,0), (0,1,0), (1,0,1) and (0,1,1).
The o.g.f. for the sequence equals the diagonal of the rational function 1/(1 - (x + y + x*z + y*z)).
Row sums of A069466. (End)
Also, the constant term in the expansion of (x + 1/x + y + 1/y)^(2n). - Christopher J. Smyth, Sep 26 2018
Number of ways to place 2n^2 nonattacking pawns on a 2n x 2n board. - Tricia Muldoon Brown, Dec 12 2018
For n>0, a(n) is the number of Littlewood polynomials of degree 4n-1 that have a closed Lill path. A polynomial p(x) has a closed Lill path if and only if p(x) is divisible by x^(2)+1. - Raul Prisacariu, Aug 28 2024

			G.f. = 1 + 4*x + 36*x^2 + 400*x^3 + 4900*x^4 + 63504*x^5 + 853776*x^6 + ... - _Michael Somos_, Aug 06 2014
From _Peter Bala_, Jan 26 2018: (Start)
a(2) = 36: The thirty six 3 x k arrays with columns belonging to the set of column vectors S = {[1,0,0], [0,1,0], [1,0,1], [0,1,1]} and having all row sums equal to 2 are the 6 distinct arrays obtained by permuting the columns of
  /1 1 0 0\
  |0 0 1 1|,
  \0 0 1 1/
the 6 distinct arrays obtained by permuting the columns of
  /0 0 1 1\
  |1 1 0 0|
  \0 0 1 1/
and the 24 arrays obtained by permuting the columns of
  /1 0 1 0\
  |0 1 0 1|. (End)
  \0 0 1 1/

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 591,828.
J. M. Borwein and P. B. Borwein, Pi and the AGM, Wiley, 1987, p. 8.
Matthijs Coster, Over 6 families van krommen [On 6 families of curves], Master's Thesis (unpublished), Aug 26 1983.
Leonard Lipshitz and A. van der Poorten. "Rational functions, diagonals, automata and arithmetic." In Number Theory, Richard A. Mollin, ed., Walter de Gruyter, Berlin (1990): 339-358.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

T. D. Noe, Table of n, a(n) for n = 0..100
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
R. Bacher, Meander algebras, Institut Fourier, 1999.
E. Barcucci, A. Frosini and S. Rinaldi, On directed-convex polyominoes in a rectangle, Discr. Math., 298 (2005). 62-78.
Arnaud Beauville, Les familles stables de courbes elliptiques sur P^1 admettant quatre fibres singulières, Comptes Rendus, Académie Sciences Paris, no. 294, May 24 1982, 657-660. MR0664643 (83h:14008)
Alin Bostan, Armin Straub, and Sergey Yurkevich, On the representability of sequences as constant terms, arXiv:2212.10116 [math.NT], 2022.
Tricia Muldoon Brown, The Problem of Pawns, The Electronic Journal of Combinatorics (2019) Vol. 26, Issue 3, #P3.21. Also arXiv:1811.09606, [math.CO], 2018.
John Maxwell Campbell, New series involving harmonic numbers and squared central binomial coefficients, Rocky Mountain J. Math., 49 (2019), 2513-2544.
C. Domb, On the theory of cooperative phenomena in crystals, Advances in Phys., 9 (1960), 149-361.
Steffen Eger, On the Number of Many-to-Many Alignments of N Sequences, arXiv:1511.00622 [math.CO], 2015.
Murray Elder, Cogrowth, 2011.
M. Elder, A. Rechnitzer, E. J. Janse van Rensburg, and T. Wong, The cogrowth series for BS(N,N) is D-finite, arXiv:1309.4184 [math.GR], 2013.
Philippe Flajolet and Robert Sedgewick, Analytic Combinatorics, 2009; see page 90.
Florian Fürnsinn and Sergey Yurkevich, Algebraicity of hypergeometric functions with arbitrary parameters, arXiv:2308.12855 [math.CA], 2023.
Davidson Noby Joseph and Igor Boettcher, Walking on Archimedean Lattices: Insights from Bloch Band Theory, arXiv:2507.12662 [cond-mat.stat-mech], 2025. See p. 18.
Kiran S. Kedlaya and Andrew V. Sutherland, Hyperelliptic curves, L-polynomials and random matrices, arXiv:0803.4462 [math.NT], 2010.
Markus Kuba and Alois Panholzer, Lattice paths and the diagonal of the cube, arXiv:2411.03930 [math.CO], 2024. See p. 14.
L. Lipshitz and A. J. van der Poorten, Rational functions, diagonals, automata and arithmetic
Raul Prisacariu, Littlewood Polynomials of Degree n with Closed Lill Paths
Eric M. Rains, High powers of random elements of compact Lie groups, Probability Theory and Related Fields 107 (1997), 219-241.
Grzegorz Siudem and Agata Fronczak, Bell polynomials in the series expansions of the Ising model, arXiv:2007.16132 [math-ph], 2020.
Eric Weisstein's World of Mathematics, Lattice Path.
D. Zagier, Integral solutions of Apéry-like recurrence equations. See line G in sporadic solutions table of page 5.

Row sums of A069466.
Row 2 of A268367 (even terms).
Equals 4*A060150.
Cf. A000984, A000515, A010370, A054474 (INVERTi transform), A172390, A000897, A002897, A006480, A008977, A186420, A188662, A000894, A241530, A002898 (walks hex lattice).

[Binomial(2*n, n)^2: n in [0..20]]; // Vincenzo Librandi, Aug 07 2014

Maple
```
A002894 := n-> binomial(2*n,n)^2.
```

CoefficientList[Series[Hypergeometric2F1[1/2, 1/2, 1, 16x], {x, 0, 20}], x]
Table[Binomial[2n,n]^2,{n,0,20}] (* Harvey P. Dale, Jul 06 2011 *)
a[ n_] := SeriesCoefficient[ EllipticK[16 x] / (Pi/2), {x, 0, n}]; (* Michael Somos, Aug 06 2014 *)
a[n_] := 16^n HypergeometricPFQ[{1/2, -2 n, 2 n + 1}, {1, 1}, 1];
Table[a[n], {n, 0, 19}] (* Peter Luschny, Mar 14 2018 *)

PARI
```
{a(n) = binomial(2*n, n)^2};
```

{a(n) = if( n<0, 0, polcoeff( polcoeff( polcoeff( 1 / (1 - x * (y + z + 1/y + 1/z)) + x * O(x^(2*n)), 2*n), 0), 0))}; /* Michael Somos, Jun 12 2004 */

[binomial(2*n, n)**2 for n in range(17)]  # Zerinvary Lajos, Apr 21 2009

D-finite with recurrence: (n+1)^2*a(n+1) = 4*(2*n + 1)^2*a(n). - Matthijs Coster, Apr 28 2004
a(n) ~ Pi^(-1)*n^(-1)*2^(4*n). - Joe Keane (jgk(AT)jgk.org), Jun 06 2002
G.f.: F(1/2, 1/2; 1; 16*x) = 1 / AGM(1, (1 - 16*x)^(1/2)) = K(4*sqrt(x)) / (Pi/2), where AGM(x, y) is the arithmetic-geometric mean of Gauss and Legendre. - Michael Somos, Mar 04 2003
G.f.: 2*EllipticK(4*sqrt(x))/Pi, using Maple's convention for elliptic integrals.
E.g.f.: Sum_{n>=0} a(n)*x^(2*n)/(2*n)! = BesselI(0, 2x)^2.
a(n) = A000984(n)^2. - Jonathan Vos Post, Jun 17 2007
E.g.f.: (BesselI(0, 2*x))^2 = 1+2*x^2/(U(0)-2*x^2); U(k) = 2*x^2*(2*k+1)+(k+1)^3-2*x^2*(2*k+3)*(k+1)^3/U(k+1); (continued fraction). - Sergei N. Gladkovskii, Nov 23 2011
In generally, for (BesselI(b, 2x))^2=((x^(2*b))/(GAMMA(b+1))^2)*(1+(2*x^2)*(2*b+1)/(Q(0)-(2*x^2)*(2*b+1)); Q(k)=(2*x^2)*(2*k+2*b+1)+(k+1)*(k+b+1)*(k+2*b+1)-(2*x^2)*(k+1)*(k+b+1)*(k+2*b+1)*(2*k+2*b+3)/Q(k+1)); (continued fraction). - Sergei N. Gladkovskii, Nov 23 2011
G.f.: G(0)/2, where G(k)= 1 + 1/(1 - 4*(2*k+1)^2*x*(1+4*x)^2/(4*(2*k+1)^2*x*(1+4*x)^2 + (k+1)^2*(1+4*x)^2/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Aug 01 2013
0 = +a(n)*(+393216*a(n+2) -119040*a(n+3) +6860*a(n+4)) +a(n+1)*(-16128*a(n+2) +6928*a(n+3) -465*a(n+4)) +a(n+2)*(+36*a(n+2) -63*a(n+3) +6*a(n+4)) for all n in Z. - Michael Somos, Aug 06 2014
Integral representation as the n-th moment of a positive function W(x) on (0,16), in Maple notation, W(x) = EllipticK(sqrt(1-x/16))/(2*Pi^2*sqrt(x)); a(n) = Integral_{x=0..16} x^n*W(x) dx, n>=0. The function W(x) is singular at x=0 and W(16) = 1/(16*Pi). This representation is unique since W(x) is the solution of the Hausdorff moment problem. - Stanley Smith and Karol A. Penson, Jun 19 2015
a(n) ~ 16^n*(2-2/(8*n+2)^2+21/(8*n+2)^4-671/(8*n+2)^6+45081/(8*n+2)^8)^2/((4*n+1)* Pi). - Peter Luschny, Oct 14 2015
a(n) = binomial(2*n,n)*binomial(2*n,n) = ( [x^n](1 + x)^(2*n) ) *( [x^n](1 + x)^(2*n) ) = [x^n](F(x)^(4*n)), where F(x) = 1 + x + x^2 + 4*x^3 + 20*x^4 + 120*x^5 + 798*x^6 + 5697*x^7 + ... appears to have integer coefficients. For similar results see A000897, A002897, A006480, A008977, A186420 and A188662. - Peter Bala, Jul 14 2016
a(n) = Sum_{k = 0..n} binomial(2*n + k,k)*binomial(n,k)^2. Cf. A005258(n) = Sum_{k = 0..n} binomial(n + k,k)*binomial(n,k)^2. - Peter Bala, Jul 27 2016
a(n) = A241530(2*n), n >= 0. - Wolfdieter Lang, Sep 06 2016
E.g.f.: 2F2(1/2,1/2; 1,1; 16*x). - Ilya Gutkovskiy, Jan 23 2018
a(n) = 16^n*hypergeom([1/2, -2*n, 2*n + 1], [1, 1], 1). - Peter Luschny, Mar 14 2018
The right-hand side of the binomial coefficient identity Sum_{k = 0..n} C(n,k)*C(n+k,k)*C(2*n+2*k,n+k)*(-4)^(n-k) = a(n). - Peter Bala, Mar 16 2018
a(n) = [x^n] (1 - x)^(2*n) * P(2*n,(1 + x)/(1 - x)), where P(n,x) denotes the n-th Legendre polynomial. Compare with A245086(n) = [x^n] (1 - x)^(2*n) * P(n,(1 + x)/(1 - x)). - Peter Bala, Mar 23 2022
a(n) = Sum_{k=0..n} multinomial(2n [k k (n-k) (n-k)]), which is another way to count random walks on Z^2, with steps of (0,+-1) or (+-1,0), that return to the point of origin after 2n steps (not necessarily for the first time), as is C(2n,n)^2. - Shel Kaphan, Jan 12 2023
0 = a(n)*(+393216*a(n+2) -119040*a(n+3) +6860*a(n+4)) +a(n+1)*(-16128*a(n+2) +6928*a(n+3) -465*a(n+4)) +a(n+2)*(+36*a(n+2) -63*a(n+3) +6*a(n+4)) for n>=0. - Michael Somos, May 30 2023
From Peter Bala, Sep 12 2023: (Start)
Right-hand side of the binomial coefficient identities
1) Sum_{k = 0..n} (-1)^(n+k) * C(n,k)*C(n+k,n)*C(2*n+k,n) = a(n).
2) 2*Sum_{k = 0..n} (-1)^(n+k) * C(n,k)*C(n+k-1,n)*C(2*n+k-1,n) = a(n) for n >= 1.
3) (4/3)*Sum_{k = 0..n} (-1)^(n+k) * C(n,k)*C(n+k,n)*C(2*n+k-1,n) = a(n) for n >= 1. (End)

Edited by N. J. A. Sloane, Feb 18 2016

A094638 Triangle read by rows: T(n,k) = |s(n,n+1-k)|, where s(n,k) are the signed Stirling numbers of the first kind A008276 (1 <= k <= n; in other words, the unsigned Stirling numbers of the first kind in reverse order).

Original entry on oeis.org

1, 1, 1, 1, 3, 2, 1, 6, 11, 6, 1, 10, 35, 50, 24, 1, 15, 85, 225, 274, 120, 1, 21, 175, 735, 1624, 1764, 720, 1, 28, 322, 1960, 6769, 13132, 13068, 5040, 1, 36, 546, 4536, 22449, 67284, 118124, 109584, 40320, 1, 45, 870, 9450, 63273, 269325, 723680, 1172700, 1026576, 362880

Offset: 1

André F. Labossière, May 17 2004

Triangle of coefficients of the polynomial (x+1)(x+2)...(x+n), expanded in decreasing powers of x. - T. D. Noe, Feb 22 2008
Row n also gives the number of permutation of 1..n with complexity 0,1,...,n-1. See the comments in A008275. - N. J. A. Sloane, Feb 08 2019
T(n,k) is the number of deco polyominoes of height n and having k columns. A deco polyomino is a directed column-convex polyomino in which the height, measured along the diagonal, is attained only in the last column. Example: T(2,1)=1 and T(2,2)=1 because the deco polyominoes of height 2 are the vertical and horizontal dominoes, having, respectively, 1 and 2 columns. - Emeric Deutsch, Aug 14 2006
Sum_{k=1..n} k*T(n,k) = A121586. - Emeric Deutsch, Aug 14 2006
Let the triangle U(n,k), 0 <= k <= n, read by rows, be given by [1,0,1,0,1,0,1,0,1,0,1,...] DELTA [1,1,2,2,3,3,4,4,5,5,6,...] where DELTA is the operator defined in A084938; then T(n,k) = U(n-1,k-1). - Philippe Deléham, Jan 06 2007
From Tom Copeland, Dec 15 2007: (Start)
Consider c(t) = column vector(1, t, t^2, t^3, t^4, t^5, ...).
Starting at 1 and sampling every integer to the right, we obtain (1,2,3,4,5,...). And T * c(1) = (1, 1*2, 1*2*3, 1*2*3*4,...), giving n! for n > 0. Call this sequence the right factorial (n+)!.
Starting at 1 and sampling every integer to the left, we obtain (1,0,-1,-2,-3,-4,-5,...). And T * c(-1) = (1, 1*0, 1*0*-1, 1*0*-1*-2,...) = (1, 0, 0, 0, ...), the left factorial (n-)!.
Sampling every other integer to the right, we obtain (1,3,5,7,9,...). T * c(2) = (1, 1*3, 1*3*5, ...) = (1,3,15,105,945,...), giving A001147 for n > 0, the right double factorial, (n+)!!.
Sampling every other integer to the left, we obtain (1,-1,-3,-5,-7,...). T * c(-2) = (1, 1*-1, 1*-1*-3, 1*-1*-3*-5,...) = (1,-1,3,-15,105,-945,...) = signed A001147, the left double factorial, (n-)!!.
Sampling every 3 steps to the right, we obtain (1,4,7,10,...). T * c(3) = (1, 1*4, 1*4*7,...) = (1,4,28,280,...), giving A007559 for n > 0, the right triple factorial, (n+)!!!.
Sampling every 3 steps to the left, we obtain (1,-2,-5,-8,-11,...), giving T * c(-3) = (1, 1*-2, 1*-2*-5, 1*-2*-5*-8,...) = (1,-2,10,-80,880,...) = signed A008544, the left triple factorial, (n-)!!!.
The list partition transform A133314 of [1,T * c(t)] gives [1,T * c(-t)] with all odd terms negated; e.g., LPT[1,T*c(2)] = (1,-1,-1,-3,-15,-105,-945,...) = (1,-A001147). And e.g.f. for [1,T * c(t)] = (1-xt)^(-1/t).
The above results hold for t any real or complex number. (End)
Let R_n(x) be the real and I_n(x) the imaginary part of Product_{k=0..n} (x + I*k). Then, for n=1,2,..., we have R_n(x) = Sum_{k=0..floor((n+1)/2)}(-1)^k*Stirling1(n+1,n+1-2*k)*x^(n+1-2*k), I_n(x) = Sum_{k=0..floor(n/2)}(-1)^(k+1)*Stirling1(n+1,n-2*k)*x^(n-2*k). - Milan Janjic, May 11 2008
T(n,k) is also the number of permutations of n with "reflection length" k (i.e., obtained from 12..n by k not necessarily adjacent transpositions). For example, when n=3, 132, 213, 321 are obtained by one transposition, while 231 and 312 require two transpositions. - Kyle Petersen, Oct 15 2008
From Tom Copeland, Nov 02 2010: (Start)
[x^(y+1) D]^n = x^(n*y) [T(n,1)(xD)^n + T(n,2)y (xD)^(n-1) + ... + T(n,n)y^(n-1)(xD)], with D the derivative w.r.t. x.
E.g., [x^(y+1) D]^4 = x^(4*y) [(xD)^4 + 6 y(xD)^3 + 11 y^2(xD)^2 + 6 y^3(xD)].
(xD)^m can be further expanded in terms of the Stirling numbers of the second kind and operators of the form x^j D^j. (End)
With offset 0, 0 <= k <= n: T(n,k) is the sum of products of each size k subset of {1,2,...,n}. For example, T(3,2) = 11 because there are three subsets of size two: {1,2},{1,3},{2,3}. 1*2 + 1*3 + 2*3 = 11. - Geoffrey Critzer, Feb 04 2011
The Kn11, Fi1 and Fi2 triangle sums link this triangle with two sequences, see the crossrefs. For the definitions of these triangle sums see A180662. The mirror image of this triangle is A130534. - Johannes W. Meijer, Apr 20 2011
T(n+1,k+1) is the elementary symmetric function a_k(1,2,...,n), n >= 0, k >= 0, (a_0(0):=1). See the T. D. Noe and Geoffrey Critzer comments given above. For a proof see the Stanley reference, p. 19, Second Proof. - Wolfdieter Lang, Oct 24 2011
Let g(t) = 1/d(log(P(j+1,-t)))/dt (see Tom Copeland's 2007 formulas). The Mellin transform (t to s) of t*Dirac[g(t)] gives Sum_{n=1..j} n^(-s), which as j tends to infinity gives the Riemann zeta function for Re(s) > 1. Dirac(x) is the Dirac delta function. The complex contour integral along a circle of radius 1 centered at z=1 of z^s/g(z) gives the same result. - Tom Copeland, Dec 02 2011
Rows are coefficients of the polynomial expansions of the Pochhammer symbol, or rising factorial, Pch(n,x) = (x+n-1)!/(x-1)!. Expansion of Pch(n,xD) = Pch(n,Bell(.,:xD:)) in a polynomial with terms :xD:^k=x^k*D^k gives the Lah numbers A008297. Bell(n,x) are the unsigned Bell polynomials or Stirling polynomials of the second kind A008277. - Tom Copeland, Mar 01 2014
From Tom Copeland, Dec 09 2016: (Start)
The Betti numbers, or dimension, of the pure braid group cohomology. See pp. 12 and 13 of the Hyde and Lagarias link.
Row polynomials and their products appear in presentation of the Jack symmetric functions of R. Stanley. See Copeland link on the Witt differential generator.
(End)
From Tom Copeland, Dec 16 2019: (Start)
The e.g.f. given by Copeland in the formula section appears in a combinatorial Dyson-Schwinger equation of quantum field theory in Yeats in Thm. 2 on p. 62 related to a Hopf algebra of rooted trees. See also the Green function on p. 70.
Per comments above, this array contains the coefficients in the expansion in polynomials of the Euler, or state number, operator xD of the rising factorials Pch(n,xD) = (xD+n-1)!/(xD-1)! = x [:Dx:^n/n!]x^{-1} = L_n^{-1}(-:xD:), where :Dx:^n = D^n x^n and :xD:^n = x^n D^n. The polynomials L_n^{-1} are the Laguerre polynomials of order -1, i.e., normalized Lah polynomials.
The Witt differential operators L_n = x^(n+1) D and the row e.g.f.s appear in Hopf and dual Hopf algebra relations presented by Foissy. The Witt operators satisfy L_n L_k - L_k L_n = (k-n) L_(n+k), as for the dual Hopf algebra. (End)

			Triangle starts:
  1;
  1,  1;
  1,  3,  2;
  1,  6, 11,  6;
  1, 10, 35, 50, 24;
...

M. Miyata and J. W. Son, On the complexity of permutations and the metric space of bijections, Tensor, 60 (1998), No. 1, 109-116 (MR1768839).
Jerome Spanier and Keith B. Oldham, "Atlas of Functions", Hemisphere Publishing Corp., 1987, chapter 18, equations 18:4:2 - 18:4:8 at page 151.
R. P. Stanley, Enumerative Combinatorics, Vol. 1, Cambridge University Press, 1997.

T. D. Noe, Rows n = 1..51 of triangle, flattened
E. Barcucci, A. Del Lungo and R. Pinzani, "Deco" polyominoes, permutations and random generation, Theoretical Computer Science, 159, 1996, 29-42.
F. Bergeron, Ph. Flajolet and B. Salvy, Varieties of Increasing Trees, Lecture Notes in Computer Science vol. 581, ed. J.-C. Raoult, Springer 1992, pp. 24-48. Added Mar 01 2014
F. Bergeron, Philippe Flajolet and Bruno Salvy, Varieties of increasing trees, HAL, Rapport De Recherche Inria. Added Mar 01 2014
Tom Copeland, Addendum to Mathemagical Forests
Tom Copeland, A Class of Differential Operators and the Stirling Numbers
Tom Copeland, Mellin Interpolation of Differential Ops and Associated Infinigens and Appell Polynomials: The Ordered, Laguerre, and Scherk-Witt-Lie Diff Ops
Tom Copeland, Witt differential generator for special Jack symmetric functions / polynomials
FindStat - Combinatorial Statistic Finder, The absolute length of a permutation
L. Foissy, Faa di Bruno subalgebras of the Hopf algebra of planar trees from combinatorial Dyson-Schwinger equations, arXiv:0707.1204 [math.RA], (2007).
O. Furdui, T. Trif, On the Summation of Certain Iterated Series, J. Int. Seq. 14 (2011) #11.6.1
F. Hivert, J.-C. Novelli and J.-Y. Thibon, The Algebra of Binary Search Trees, Theoretical Computer Science, 339 (2005), 129-165.
T. Hyde and J. Lagarias Polynomial splitting measures and cohomology of the pure braid group, arXiv preprint arXiv:1604.05359 [math.RT], 2016.
MathOverflow, Motivation of Virasoro algebra, an answer by Tom Copeland to an MO question posed in 2012.
Romeo Mestrovic, Lucas' theorem: its generalizations, extensions and applications (1878--2014), arXiv preprint arXiv:1409.3820 [math.NT], 2014.
Robert E. Moritz, On the sum of products of n consecutive integers, Univ. Washington Publications in Math., 1 (No. 3, 1926), 44-49 [Annotated scanned copy]
M. D. Schmidt, Generalized j-Factorial Functions, Polynomials, and Applications , J. Int. Seq. 13 (2010), 10.6.7.
M. Z. Spivey, On Solutions to a General Combinatorial Recurrence, J. Int. Seq. 14 (2011) # 11.9.7.
K. Yeats, A Combinatorial Perspective on Quantum Field Theory, SpringerBriefs in Mathematical Physics, Vol. 15, 2017.

A008276 gives the (signed) Stirling numbers of the first kind.
Cf. A000108, A014137, A001246, A033536, A000984, A094639, A006134, A082894, A002897, A079727, A000217 (2nd column), A000914 (3rd column), A001303 (4th column), A000915 (5th column), A053567 (6th column), A000142 (row sums).
Triangle sums (see the comments): A124380 (Kn11), A001710 (Fi1, Fi2). - Johannes W. Meijer, Apr 20 2011
Cf. A121586, A130534, A143491.

Flat(List([1..10], n-> List([1..n], k-> Stirling1(n,n-k+1) ))); # G. C. Greubel, Dec 29 2019

a094638 n k = a094638_tabl !! (n-1) !! (k-1)
a094638_row n = a094638_tabl !! (n-1)
a094638_tabl = map reverse a130534_tabl
-- Reinhard Zumkeller, Aug 01 2014

[(-1)^(k+1)*StirlingFirst(n,n-k+1): k in [1..n], n in [1..10]]; // G. C. Greubel, Dec 29 2019

T:=(n,k)->abs(Stirling1(n,n+1-k)): for n from 1 to 10 do seq(T(n,k),k=1..n) od; # yields sequence in triangular form. # Emeric Deutsch, Aug 14 2006

Table[CoefficientList[Series[Product[1 + i x, {i,n}], {x,0,20}], x], {n,0,6}] (* Geoffrey Critzer, Feb 04 2011 *)
Table[Abs@StirlingS1[n, n-k+1], {n, 10}, {k, n}]//Flatten (* Michael De Vlieger, Aug 29 2015 *)

create_list(abs(stirling1(n+1,n-k+1)),n,0,10,k,0,n); /* Emanuele Munarini, Jun 01 2012 */

{T(n,k)=if(n<1 || k>n,0,(n-1)!*polcoeff(polcoeff(x*y/(1 - x*y+x*O(x^n))^(1 + 1/y),n,x),k,y))} /* Paul D. Hanna, Jul 21 2011 */

[[stirling_number1(n, n-k+1) for k in (1..n)] for n in (1..10)] # G. C. Greubel, Dec 29 2019

With P(n,t) = Sum_{k=0..n-1} T(n,k+1) * t^k = 1*(1+t)*(1+2t)...(1+(n-1)*t) and P(0,t)=1, exp[P(.,t)*x] = (1-tx)^(-1/t). T(n,k+1) = (1/k!) (D_t)^k (D_x)^n [ (1-tx)^(-1/t) - 1 ] evaluated at t=x=0. (1-tx)^(-1/t) - 1 is the e.g.f. for a plane m-ary tree when t=m-1. See Bergeron et al. in "Varieties of Increasing Trees". - Tom Copeland, Dec 09 2007
First comment and formula above rephrased as o.g.f. for row n: Product_{i=0...n} (1+i*x). - Geoffrey Critzer, Feb 04 2011
n-th row polynomials with alternate signs are the characteristic polynomials of the (n-1)x(n-1) matrices with 1's in the superdiagonal, (1,2,3,...) in the main diagonal, and the rest zeros. For example, the characteristic polynomial of [1,1,0; 0,2,1; 0,0,3] is x^3 - 6*x^2 + 11*x - 6. - Gary W. Adamson, Jun 28 2011
E.g.f.: A(x,y) = x*y/(1 - x*y)^(1 + 1/y) = Sum_{n>=1, k=1..n} T(n,k)*x^n*y^k/(n-1)!. - Paul D. Hanna, Jul 21 2011
With F(x,t) = (1-t*x)^(-1/t) - 1 an e.g.f. for the row polynomials P(n,t) of A094638 with P(0,t)=0, G(x,t)= [1-(1+x)^(-t)]/t is the comp. inverse in x. Consequently, with H(x,t) = 1/(dG(x,t)/dx) = (1+x)^(t+1),
  P(n,t) = [(H(x,t)*d/dx)^n] x evaluated at x=0; i.e.,
  F(x,t) = exp[x*P(.,t)] = exp[x*H(u,t)*d/du] u, evaluated at u = 0.
  Also, dF(x,t)/dx = H(F(x,t),t). - Tom Copeland, Sep 20 2011
T(n,k) = |A008276(n,k)|. - R. J. Mathar, May 19 2016
The row polynomials of this entry are the reversed row polynomials of A143491 multiplied by (1+x). E.g., (1+x)(1 + 5x + 6x^2) = (1 + 6x + 11x^2 + 6x^3). - Tom Copeland, Dec 11 2016
Regarding the row e.g.f.s in Copeland's 2007 formulas, e.g.f.s for A001710, A001715, and A001720 give the compositional inverses of the e.g.f. here for t = 2, 3, and 4 respectively. - Tom Copeland, Dec 28 2019

Edited by Emeric Deutsch, Aug 14 2006

A008977 a(n) = (4*n)!/(n!)^4.

Original entry on oeis.org

1, 24, 2520, 369600, 63063000, 11732745024, 2308743493056, 472518347558400, 99561092450391000, 21452752266265320000, 4705360871073570227520, 1047071828879079131681280, 235809301462142612780721600, 53644737765488792839237440000, 12309355935372581458927646400000

Offset: 0

N. J. A. Sloane

Number of paths of length 4*n in an n X n X n X n grid from (0,0,0,0) to (n,n,n,n).
a(n) occurs in Ramanujan's formula 1/Pi = (sqrt(8)/9801) * Sum_{n>=0} (4*n)!/(n!)^4 * (1103 + 26390*n)/396^(4*n). - Susanne Wienand, Jan 05 2013
a(n) is the number of ballot results that lead to a 4-way tie when 4*n voters each cast three votes for three out of four candidates vying for 3 slots on a county commission; each of these ballot results give 3*n votes to each of the four candidates. - Dennis P. Walsh, May 02 2013
a(n) is the constant term of (X + Y + Z + 1/(X*Y*Z))^(4*n). - Mark van Hoeij, May 07 2013
In Narumiya and Shiga on page 158 the g.f. is given as a hypergeometric function. - Michael Somos, Aug 12 2014
Diagonal of the rational function R(x,y,z,w) = 1/(1-(w+x+y+z)). - Gheorghe Coserea, Jul 15 2016

			a(13)=52!/(13!)^4=53644737765488792839237440000 is the number of ways of dealing the four hands in Bridge or Whist. - _Henry Bottomley_, Oct 06 2000
a(1)=24 since, in a 4-voter 3-vote election that ends in a four-way tie for candidates A, B, C, and D, there are 4! ways to arrange the needed vote sets {A,B,C}, {A,B,D}, {A,C,D}, and {B,C,D} among the 4 voters. - _Dennis P. Walsh_, May 02 2013
G.f. = 1 + 24*x + 2520*x^2 + 369600*x^3 + 63063000*x^4 + 11732745024*x^5 + ...

T. D. Noe, Table of n, a(n) for n=0..100
R. M. Dickau, Paths through a 4-D lattice.
S. Hassani, J.-M. Maillard, and N. Zenine, On the diagonals of rational functions: the minimal number of variables (unabridged version), arXiv:2502.05543 [math-ph], 2025. See pp. 41.
Timothy Huber, Daniel Schultz, and Dongxi Ye, Ramanujan-Sato series for 1/pi, Acta Arith. (2023) Vol. 207, 121-160. See p. 11.
Markus Kuba and Alois Panholzer, Lattice paths and the diagonal of the cube, arXiv:2411.03930 [math.CO], 2024.
R. Mestrovic, Wolstenholme's theorem: Its Generalizations and Extensions in the last hundred and fifty years (1862-2011), arXiv:1111.3057 [math.NT], 2011.
Michaël Moortgat, The Tamari order for D^3 and derivability in semi-associative Lambek-Grishin Calculus, 15th Workshop: Computational Logic and Applications (CLA 2020).
N. Narumiya and H. Shiga, The mirror map for a family of K3 surfaces induced from the simplest 3-dimensional reflexive polytope, Proceedings on Moonshine and related topics (Montréal, QC, 1999), 139-161, CRM Proc. Lecture Notes, 30, Amer. Math. Soc., Providence, RI, 2001. MR1877764 (2002m:14030).

Cf. A000984, A006480, A008978, A178529, A000897, A002894, A002897, A006480, A008979, A186420, A188662.
Row 4 of A187783.
Related to diagonal of rational functions: A268545-A268555.

[Factorial(4*n)/Factorial(n)^4: n in [0..20]]; // Vincenzo Librandi, Aug 13 2014

Maple
```
A008977 := n->(4*n)!/(n!)^4;
```

Table[(4n)!/(n!)^4,{n,0,16}] (* Harvey P. Dale, Oct 24 2011 *)
a[ n_] := If[ n < 0, 0, (4 n)! / n!^4]; (* Michael Somos, Aug 12 2014 *)
a[ n_] := SeriesCoefficient[ HypergeometricPFQ[ {1/4, 2/4, 3/4}, {1, 1}, 256 x], {x, 0, n}]; (* Michael Somos, Aug 12 2014 *)

A008977(n):=(4*n)!/(n!)^4$ makelist(A008977(n),n,0,20); /* Martin Ettl, Nov 15 2012 */

a(n) = (4*n)!/n!^4; \\ Gheorghe Coserea, Jul 15 2016

from math import factorial
def A008977(n): return factorial(n<<2)//factorial(n)**4 # Chai Wah Wu, Mar 15 2023

a(n) = A139541(n)*(A001316(n)/A049606(n))^3. - Reinhard Zumkeller, Apr 28 2008
Self-convolution of A178529, where A178529(n) = (4^n/n!^2) * Product_{k=0..n-1} (8*k + 1)*(8*k + 3).
G.f.: hypergeom([1/8, 3/8], [1], 256*x)^2. - Mark van Hoeij, Nov 16 2011
a(n) ~ 2^(8*n - 1/2) / (Pi*n)^(3/2). - Vaclav Kotesovec, Mar 07 2014
G.f.: hypergeom([1/4, 2/4, 3/4], [1, 1], 256*x). - Michael Somos, Aug 12 2014
From Peter Bala, Jul 12 2016: (Start)
a(n) = binomial(2*n,n)*binomial(3*n,n)*binomial(4*n,n) = ( [x^n](1 + x)^(2*n) ) * ( [x^n](1 + x)^(3*n) ) * ( [x^n](1 + x)^(4*n) ) = [x^n](F(x)^(24*n)), where F(x) = 1 + x + 29*x^2 + 2246*x^3 + 239500*x^4 + 30318701*x^5 + 4271201506*x^6 + ... appears to have integer coefficients. For similar results see A000897, A002894, A002897, A006480, A008978, A008979, A186420 and A188662. (End)
0 = (x^2-256*x^3)*y''' + (3*x-1152*x^2)*y'' + (1-816*x)*y' - 24*y, where y is the g.f. - Gheorghe Coserea, Jul 15 2016
From Peter Bala, Jul 17 2016: (Start)
a(n) = Sum_{k = 0..3*n} (-1)^(n+k)*binomial(4*n,n + k)* binomial(n + k,k)^4.
a(n) = Sum_{k = 0..4*n} (-1)^k*binomial(4*n,k)*binomial(n + k,k)^4. (End)
E.g.f.: 3F3(1/4,1/2,3/4; 1,1,1; 256*x). - Ilya Gutkovskiy, Jan 23 2018
From Peter Bala, Feb 16 2020: (Start)
a(m*p^k) == a(m*p^(k-1)) ( mod p^(3*k) ) for prime p >= 5 and positive integers m and k - apply Mestrovic, equation 39, p. 12.
a(n) = [(x*y*z)^n] (1 + x + y + z)^(4*n). (End)
D-finite with recurrence n^3*a(n) -8*(4*n-3)*(2*n-1)*(4*n-1)*a(n-1)=0. - R. J. Mathar, Aug 01 2022
a(n) = 24*A082368(n). - R. J. Mathar, Jun 21 2023

A000897 a(n) = (4n)! / ((2n)!*n!^2).

Original entry on oeis.org

1, 12, 420, 18480, 900900, 46558512, 2498640144, 137680171200, 7735904619300, 441233078286000, 25467973278667920, 1484298740174927040, 87202550985276963600, 5157850293780050462400, 306839461354466267304000, 18344908596179023234548480

Offset: 0

N. J. A. Sloane

Appears in Ramanujan's theory of elliptic functions of signature 4.
H. A. Verrill proves that a(n) = Sum_{p + q + r = 3n} w^(p-q) * {(3n)!/(p! q! r!)}^2, with p, q, r >= 0 and w = primitive 3rd root of unity.
The family of elliptic curves "x=2*H1=p^2+q^2-(1/4)*q^4, 0sqrt(-1)*q" to H1 produces "x=2*H2=p^2-q^2-(1/4)*q^4, 0Bradley Klee, Feb 25 2018
Even-order terms in the diagonal of rational function 1/(1 - (x^2 + y^2 + z)). - Gheorghe Coserea, Aug 09 2018

			G.f.: 1 + 12*x + 420*x^2 + 18480*x^3 + 900900*x^4 + 46558512*x^5 + 2498640144*x^6 + ...

E. R. Hansen, A Table of Series and Products, Prentice-Hall, Englewood Cliffs, NJ, 1975, p. 96.

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Alin Bostan, Armin Straub, and Sergey Yurkevich, On the representability of sequences as constant terms, arXiv:2212.10116 [math.NT], 2022.
H. J. Brothers, Pascal's Prism: Supplementary Material
B. Klee, Geometric G.F. for Ramanujan Periods, seqfan mailing list, 2017.
S. Ramanujan, Modular Equations and Approximations to Pi, Quarterly Journal of Mathematics, XLV (1914), 350-372.
L. C. Shen, A note on Ramanujan's identities involving the hypergeometric function 2F1(1/6,5/6;1;z), The Ramanujan Journal, 30.2 (2013), 211-222.
H. A. Verrill, Sums of squares of binomial coefficients, with applications to Picard-Fuchs equations, arXiv:math/0407327v1 [math.CO], 2004.

Cf. A002897, A008977, A186420, A188662. Elliptic Integrals: A002894, A113424, A006480. Factors: A005809, A005810, A000984, A001448.

a:=n->Sum([0..3*n],k->(-1)^k*Binomial(3*n,k)*Binomial(6*n-k,3*n)*
Binomial(2*k,k));;
A000897:=List([0..14],n->a(n)); # Muniru A Asiru, Feb 11 2018

seq((4*n)!/(n!)^4/binomial(2*n,n), n=0..14); # Zerinvary Lajos, Jun 28 2007

Table[(4n)!/((2n)! n!^2), {n, 0, 30}] (* Stefan Steinerberger, Apr 14 2006 *)
a[ n_] := Binomial[ 4 n, 2 n] Binomial[ 2 n, n]; (* Michael Somos, Mar 24 2013 *)
a[ n_] := SeriesCoefficient[ Hypergeometric2F1[ 1/4, 3/4, 1, 64 x], {x, 0, n}]; (* Michael Somos, Mar 24 2013 *)
a[ n_] := If[ n < 0, 0, With[{m = 4 n}, (-1)^n m! SeriesCoefficient[ BesselI[ 0, 2 x] BesselJ[ 0, 2 x], {x, 0, m}]]]; (* Michael Somos, Aug 12 2014 *)
a[ n_] := 64^n Pochhammer[1/4, n] Pochhammer[3/4, n] / n!^2; (* Michael Somos, Aug 12 2014 *)

{a(n) = if( n<0, 0, (4*n)! / ((2*n)! * n!^2))}; /* Michael Somos, Oct 31 2005 */

E.g.f.: Sum_{k>=0} (-1)^k * a(k) * x^(4*k) / (4*k)! = BesselI(0, 2x) * BesselJ(0, 2x).
G.f.: F(1/4, 3/4; 1; 64*x). - Michael Somos, Oct 31 2005
a(n) = A008977(n)/A000984(n) - Zerinvary Lajos, Jun 28 2007
Sum_{k>=0} a(k) * x^(3k)/(3k)!^2 = f(x)*f(x*w)*f(x/w) where f(x) = BesselI(0, 2*sqrt(x)) and w = primitive 3rd root of unity. - Michael Somos, Jul 25 2007
In general, for (BesselI(b, 2x))*(BesselJ(b, 2x))=((x^(2*b))/((GAMMA(b+1))^2)*(1-(x^4)/(Q(0)+(x^4))); Q(k)=(k+1)*(k+b+1)*(2*k+b+1)*(2*k+b+2)-(x^4)+(x^4)*(k+1)*(k+b+1)*(2*k+b+1)*(2*k+b+2)/Q(k+1)) ; (continued fraction). - Sergei N. Gladkovskii, Nov 24 2011
D-finite with recurrence 0 = a(n)*4*(4*n + 1)*(4*n + 3) - a(n+1)*(n + 1)^2 for all n in Z. - Michael Somos, Aug 12 2014
0 = a(n)*(-4026531840*a(n+2) +2005401600*a(n+3) -103896576*a(n+4) +1251948*a(n+5)) + a(n+1)*(+41418752*a(n+2) -30435328*a(n+3) +1863228*a(n+4) -24604*a(n+5)) + a(n+2)*(-16896*a(n+2) +75608*a(n+3) -6740*a(n+4) +105*a(n+5)) for all n in Z. - Michael Somos, Aug 12 2014
From Peter Bala, Jul 12 2016: (Start)
a(n) = binomial(3*n,n)*binomial(4*n,n) = A005809(n)*A005810(n) = ( [x^n](1 + x)^(3*n) ) * ( [x^n](1 + x)^(4*n) ) = [x^n](F(x)^(12*n)), where F(x) = 1 + x + 6*x^2 + 105*x^3 + 2448*x^4 + 67043*x^5 + 2028307*x^6 + ... appears to have integer coefficients. Cf. A002894, A002897, A006480, A008977, A186420 and A188662. (End)
a(n) ~ 2^(6*n-1/2)/(Pi*n). - Ilya Gutkovskiy, Jul 12 2016
G.f.: 2*EllipticK(sqrt((sqrt(1-64*x)-1)/(2*sqrt(1-64*x))))/(Pi*(1-64*x)^(1/4)) where EllipticK is the complete elliptic integral of the first kind (in Maple's notation). - Robert Israel, Jul 12 2016
a(n) = Sum_{k = 0..3*n} (-1)^k*C(3*n,k)*C(6*n-k,3*n)*C(2*k,k). - Peter Bala, Feb 10 2018
From Bradley Klee, Feb 27 2018: (Start)
a(n) = A000984(n)*A001448(n).
G.f.: (1/(sqrt(2)*Pi))*Integral_{q=-oo..oo} 1/sqrt(q^2+(1/4)*q^4+(1-64*x)) dq.
G.f.: (1/(2*Pi))*Integral_{phi=0..2*Pi} 1/sqrt(1-64*x*sin^4(phi)) dphi. (End)
From Peter Bala, Mar 20 2022: (Start)
Right-hand side of the following identities valid for n >= 1:
Sum_{k = 0..2*n} 2*n*(2*n+k-1)!/(k!*n!^2) = (4*n)!/((2*n)!*n!^2);
(3/2)*Sum_{k = 0..n} 2*n*(3*n+k-1)!/(k!*n!*(2*n)!) = (4*n)!/((2*n)!*n!^2).
Cf. A001451. (End)
a(n) = (4^n/n!^2)*Product_{k = 0..2*n-1} (2*k + 1). - Peter Bala, Feb 26 2023
a(n) = Sum_{k = 0..n} (-1)^(n+k) * binomial(n, k) * A108625(3*n, k) (verified using the MulZeil procedure in Doron Zeilberger's MultiZeilberger package). - Peter Bala, Oct 15 2024

A008978 a(n) = (5*n)!/(n!)^5.

Original entry on oeis.org

1, 120, 113400, 168168000, 305540235000, 623360743125120, 1370874167589326400, 3177459078523411968000, 7656714453153197981835000, 19010638202652030712978200000, 48334775757901219912115629238400, 125285878026462826569986857692288000

Offset: 0

N. J. A. Sloane

Number of paths of length 5n in Z^5 from (0,0,0,0,0) to (n,n,n,n,n).

Vincenzo Librandi, Table of n, a(n) for n = 0..100
V. Batyrev, Review of "Mirror Symmetry and Algebraic Geometry", by D. A. Cox and S. Katz, Bull. Amer. Math. Soc., 37 (No. 4, 2000), 473-476.
R. M. Dickau, 5-D shortest path diagrams
R. Mestrovic, Wolstenholme's theorem: Its Generalizations and Extensions in the last hundred and fifty years (1862-2011), arXiv:1111.3057 [math.NT], 2011.

Cf. A000984, A006480, A008977, A002894, A002897, A186420, A188662, A001460.

Row 5 of A187783.

[Factorial(5*n)/Factorial(n)^5: n in [0..10]]; // Vincenzo Librandi, Mar 08 2014

Table[(5 n)!/(n)!^5, {n, 0, 20}] (* Vincenzo Librandi, Mar 08 2014 *)

a(n) = (5*n)!/(n!)^5; \\ Michel Marcus, Mar 08 2014

a(n) ~ 5^(5*n+1/2) / (4 * Pi^2 * n^2). - Vaclav Kotesovec, Mar 07 2014
From Peter Bala, Jul 12 2016: (Start)
a(n) = binomial(2*n,n)*binomial(3*n,n)*binomial(4*n,n)*binomial(5*n,n) = ( [x^n](1 + x)^(2*n) ) * ( [x^n](1 + x)^(3*n) ) * ( [x^n](1 + x)^(4*n) ) * ( [x^n](1 + x)^(5*n) ) = [x^n]( F(x)^(120*n) ), where F(x) = 1 + x + 353*x^2 + 318986*x^3 + 408941594*x^4 + 633438203535*x^5 + 1105336091531052*x^6 + ... appears to have integer coefficients. For similar results see A000897, A002894, A002897, A006480, A008977, A186420 and A188662. (End)
From Peter Bala, Jul 17 2016: (Start)
a(n) = Sum_{k = 0..4*n} (-1)^k*binomial(5*n,n + k)*binomial(n + k,k)^5.
a(n) = Sum_{k = 0..5*n} (-1)^(n+k)*binomial(5*n,k)*binomial(n + k,k)^5. (End)
From Ilya Gutkovskiy, Nov 23 2017: (Start)
O.g.f.: 4F3(1/5,2/5,3/5,4/5; 1,1,1; 3125*x).
E.g.f.: 4F4(1/5,2/5,3/5,4/5; 1,1,1,1; 3125*x). (End)
From Peter Bala, Feb 16 2020: (Start)
a(m*p^k) == a(m*p^(k-1)) ( mod p^(3*k) ) for prime p >= 5 and positive integers m and k - apply Mestrovic, equation 39, p. 12.
a(n) = [(x*y*z*u)^n] (1 + x + y + z + u )^(5*n). (End)
a(n) = 120*A322252(n). - R. J. Mathar, Jun 21 2023
a(n) = a(n-1)*5*(5*n - 1)*(5*n - 2)*(5*n - 3)*(5*n - 4)/n^4. - Neven Sajko, Jul 21 2023

A004981 a(n) = (2^n/n!) * Product_{k=0..n-1} (4*k + 1).

Original entry on oeis.org

1, 2, 10, 60, 390, 2652, 18564, 132600, 961350, 7049900, 52169260, 388898120, 2916735900, 21987701400, 166478310600, 1265235160560, 9647418099270, 73774373700300, 565603531702300, 4346216612028200, 33465867912617140, 258165266754475080, 1994913424920943800

Offset: 0

Joe Keane (jgk(AT)jgk.org)

The convolution of this sequence with itself yields A059304. - T. D. Noe, Jun 11 2002

Conjecture: a(p*n) = a(n) (mod p^2) for prime p = 1 (mod 4) and all positive integers n. Cf. A004982 and A298799. - Peter Bala, Dec 22 2019

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
A. Vogt, Resummation of small-x double logarithms in QCD: semi-inclusive electron-positron annihilation, arXiv preprint arXiv:1108.2993 [hep-ph], 2011.

Cf. A002897, A004982, A059304, A127776, A298799.

List([0..25], n-> 2^n*Product([0..n-1], k-> 4*k+1)/Factorial(n) ); # G. C. Greubel, Aug 22 2019

[1] cat [2^n*&*[4*k+1: k in [0..n-1]]/Factorial(n): n in [1..25]]; // G. C. Greubel, Aug 22 2019

A004981 := n -> (-8)^n*binomial(-1/4, n):
seq(A004981(n), n=0..25); # Peter Luschny, Oct 23 2018

CoefficientList[Series[(1-8x)^(-1/4), {x, 0, 25}], x] (* Vincenzo Librandi, Mar 16 2014 *)
Table[8^n*Pochhammer[1/4, n]/n!, {n,0,25}] (* G. C. Greubel, Aug 22 2019 *)

a(n):=if n=0 then 1 else (sum(m*sum(binomial(-m+2*k-1,k-1) *2^(n+m-k)*binomial(2*n-k-1,n-1),k,m,n),m,1,n))/(n); /* Vladimir Kruchinin, Dec 26 2011 */

PARI
```
a(n)=if(n<0,0,prod(k=1,n,(8*k-6)/k))
```

{a(n)=if(n<0, 0, polcoeff( (1-8*x+x*O(x^n))^(-1/4), n))} /* Michael Somos, Jan 31 2007 */

[8^n*rising_factorial(1/4, n)/factorial(n) for n in (0..25)] # G. C. Greubel, Aug 22 2019

a(n) ~ Gamma(1/4)^-1*n^(-3/4)*2^(3*n)*{1 - 3/32*n^-1 - ...}
G.f.: (1-8*x)^(-1/4).
A002897(n) = Sum_{k=0..n} a(k)^2*a(n-k)^2. - Michael Somos, Jan 31 2007
a(n) = (Sum_{m=1..n} m*Sum_{k=m..n} binomial(-m+2*k-1,k-1)*2^(n+m-k)*binomial(2*n-k-1,n-1))/n, n>0, a(0)=1. - Vladimir Kruchinin, Dec 26 2011
D-finite with recurrence: n*a(n) = 2*(4*n-3)*a(n-1). - R. J. Mathar, Mar 14 2014
From Karol A. Penson, Dec 19 2015: (Start)
a(n) = (-8)^n*binomial(-1/4,n).
E.g.f.: is the hypergeometric function of type 1F1, in Maple notation hypergeom([1/4], [1], 8*x).
Representation as n-th moment of a positive function on (0, 8): a(n)=int(x^n*(sqrt(2)/(16*Pi*(x/8)^(3/4)*(1-x/8)^(1/4))), x=0..8), n=0, 1, ... . This function is the solution of the Hausdorff moment problem on (0, 8) with moments equal to a(n). As a consequence this representation is unique. (End)

More terms from James Sellers, May 01 2000

A188662 Binomial coefficients: a(n) = binomial(3*n,n)^2.

Original entry on oeis.org

1, 9, 225, 7056, 245025, 9018009, 344622096, 13521038400, 540917591841, 21966328580625, 902702926350225, 37456461988358400, 1566697064677290000, 65973795093338597136, 2794203818390077646400, 118933541228935777741056, 5084343623375056062840609

Offset: 0

Emanuele Munarini, Apr 07 2011

Even-order terms in the diagonal of rational function 1/(1 - (x^2 + y^2 + z^2 - x*y - y*z - x*z)). - Gheorghe Coserea, Aug 09 2018

Seiichi Manyama, Table of n, a(n) for n = 0..604

Cf. A005809, A000897, A002894, A002897, A006480, A008977, A186420.

[Binomial(3*n,n)^2: n in [0..100]]; // Vincenzo Librandi, Apr 08 2011

Mathematica
```
Table[Binomial[3 n, n]^2, {n, 0, 16}]
```
Maxima
```
makelist(binomial(3*n,n)^2,n,0,16);
```

a(n) = binomial(3*n,n)^2; \\ Michel Marcus, Nov 01 2016

from math import comb
def A188662(n): return comb(3*n,n)**2 # Chai Wah Wu, Mar 15 2023

a(n) = A005809(n)^2.
a(n) = binomial(3*n,n)^2 = ( [x^n](1 + x)^(3*n) )^2 = [x^n](F(x)^(9*n)), where F(x) = 1 + x + 4*x^2 + 49*x^3 + 795*x^4 + 15180*x^5 + 320422*x^6 + ... appears to have integer coefficients. For similar results see A000897, A002894, A002897, A006480, A008977 and A186420. - Peter Bala, Jul 12 2016
a(n) ~ 3^(6*n+1)*4^(-2*n-1)/(Pi*n). - Ilya Gutkovskiy, Jul 13 2016
a(n) = Sum_{k=0..n} binomial(n, k)^2*binomial(3*n+k, 2*n). - Seiichi Manyama, Jan 09 2017

A181543 Triangle of cubed binomial coefficients, T(n,k) = C(n,k)^3, read by rows.

Original entry on oeis.org

1, 1, 1, 1, 8, 1, 1, 27, 27, 1, 1, 64, 216, 64, 1, 1, 125, 1000, 1000, 125, 1, 1, 216, 3375, 8000, 3375, 216, 1, 1, 343, 9261, 42875, 42875, 9261, 343, 1, 1, 512, 21952, 175616, 343000, 175616, 21952, 512, 1, 1, 729, 46656, 592704, 2000376, 2000376, 592704, 46656, 729, 1

Offset: 0

Paul D. Hanna, Oct 30 2010

Diagonal of rational function R(x,y,z,t) = 1/(1 + y + z + x*y + y*z + t*x*z + (t+1)*x*y*z) with respect to x, y, z, i.e., T(n,k) = [(xyz)^n*t^k] R(x,y,z,t). - Gheorghe Coserea, Jul 01 2018

			Triangle begins:
  1;
  1,   1;
  1,   8,     1;
  1,  27,    27,      1;
  1,  64,   216,     64,       1;
  1, 125,  1000,   1000,     125,       1;
  1, 216,  3375,   8000,    3375,     216,      1;
  1, 343,  9261,  42875,   42875,    9261,    343,     1;
  1, 512, 21952, 175616,  343000,  175616,  21952,   512,   1;
  1, 729, 46656, 592704, 2000376, 2000376, 592704, 46656, 729, 1;
  ...

Indranil Ghosh, Rows 0..120 of triangle, flattened
Jeffrey S. Geronimo, Hugo J. Woerdeman, and Chung Y. Wong, The autoregressive filter problem for multivariable degree one symmetric polynomials, arXiv:2101.00525 [math.CA], 2021.

Cf. A000172 (row sums), A181545 (antidiagonal sums), A002897, A181544, A248658.

Variants: A008459, A007318.

T:= (n, k)-> binomial(n, k)^3:
seq(seq(T(n, k), k=0..n), n=0..10);  # Alois P. Heinz, Jan 06 2021

Flatten[Table[Binomial[n,k]^3,{n,0,10},{k,0,n}]] (* Harvey P. Dale, May 23 2011 *)

T(n,k)=binomial(n,k)^3
for(n=0,10,for(k=0,n,print1(T(n,k),", "));print())

T(n,k)=polcoeff(polcoeff(sum(m=0,n,(3*m)!/m!^3*x^(2*m)*y^m/(1-x-x*y+x*O(x^n))^(3*m+1)),n,x),k,y)
for(n=0,10,for(k=0,n,print1(T(n,k),", "));print()) \\ Paul D. Hanna, Nov 04 2010

diag(expr, N=22, var=variables(expr)) = {
  my(a = vector(N));
  for (k = 1, #var, expr = taylor(expr, var[#var - k + 1], N));
  for (n = 1, N, a[n] = expr;
    for (k = 1, #var, a[n] = polcoeff(a[n], n-1)));
  return(a);
};
x='x; y='y; z='z; t='t;
concat(apply(Vec, diag(1/(1 + y + z + x*y + y*z + t*x*z + (t+1)*x*y*z), 10, [x, y, z]))) \\ Gheorghe Coserea, Jul 01 2018

Row sums equal A000172, the Franel numbers.
Central terms are A002897(n) = C(2n,n)^3.
Antidiagonal sums equal A181545;
The g.f. of the antidiagonal sums is Sum_{n>=0} (3n)!/(n!)^3 * x^(3n)/(1-x-x^2)^(3n+1).
G.f. for column k: [Sum_{j=0..2k} A181544(k,j)*x^j]/(1-x)^(3k+1), where the row sums of A181544 equals De Bruijn's s(3,n) = (3n)!/(n!)^3.
G.f.: A(x,y) = Sum_{n>=0} (3n)!/n!^3 * x^(2n)*y^n/(1-x-x*y)^(3n+1). - Paul D. Hanna, Nov 04 2010

cp's OEIS Frontend

A014137 Partial sums of Catalan numbers (A000108).

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A002894 a(n) = binomial(2n, n)^2.

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A094638 Triangle read by rows: T(n,k) = |s(n,n+1-k)|, where s(n,k) are the signed Stirling numbers of the first kind A008276 (1 <= k <= n; in other words, the unsigned Stirling numbers of the first kind in reverse order).

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A008977 a(n) = (4*n)!/(n!)^4.

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A000897 a(n) = (4*n)! / ((2*n)!*n!^2).

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A000897 a(n) = (4n)! / ((2n)!*n!^2).