A112029 -id:A112029 - cp's OEIS frontend

A045721 a(n) = binomial(3*n+1,n).

1, 4, 21, 120, 715, 4368, 27132, 170544, 1081575, 6906900, 44352165, 286097760, 1852482996, 12033222880, 78378960360, 511738760544, 3348108992991, 21945588357420, 144079707346575, 947309492837400, 6236646703759395, 41107996877935680, 271250494550621040, 1791608261879217600

Offset: 0

Emeric Deutsch

Number of leaves in all noncrossing rooted trees on n nodes on a circle.
Number of standard tableaux of shape (n-1,1^(2n-3)). - Emeric Deutsch, May 25 2004
a(n) = number of Dyck (2n-3)-paths with exactly one descent of odd length. For example, a(3) counts all 5 Dyck 3-paths except UDUDUD. - David Callan, Jul 25 2005
a(n+2) gives row sums of A119301. - Paul Barry, May 13 2006
a(n) is the number of paths avoiding UU from (0,0) to (3n,n) and taking steps from {U,H}. - Shanzhen Gao, Apr 15 2010
Central coefficients of triangle A078812. - Vladimir Kruchinin, May 10 2012
Row sums of A252501. - L. Edson Jeffery, Dec 18 2014

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Milan Janjić, Two Enumerative Functions.
Dmitry Kruchinin and Vladimir Kruchinin, A Method for Obtaining Generating Function for Central Coefficients of Triangles, Journal of Integer Sequence, Vol. 15 (2012), Article 12.9.3.
Wojciech Mlotkowski and Karol A. Penson, Probability distributions with binomial moments, arXiv preprint arXiv:1309.0595 [math.PR], 2013.
Index entries for sequences related to rooted trees.

Cf. A252501, A263134 (partial sums), A001764, A004319, A005809, A006013, A013698, A025174, A117671, A165817, A236194.

[Binomial(3*n+1, n): n in [0..20]]; // Vincenzo Librandi, Aug 07 2014

[seq( binomial(3*n+1,n),n=0..40)]; # N. J. A. Sloane, Jun 09 2007

Table[Binomial[3 n + 1, n], {n, 0, 20}] (* Vincenzo Librandi, Aug 07 2014 *)

a(n)=binomial(3*n+1,n) \\ Charles R Greathouse IV, Mar 18 2014

a(n) is asymptotic to c/sqrt(n)*(27/4)^n with c=0.73... - Benoit Cloitre, Jan 27 2003
G.f.: gz^2/(1-3zg^2), where g=g(z) is given by g=1+zg^3, g(0)=1, i.e. (in Maple command) g := 2*sin(arcsin(3*sqrt(3*z)/2)/3)/sqrt(3*z). - Emeric Deutsch, May 22 2003
a(n+2) = C(3n+1,n) = Sum_{k=0..n} C(3n-k,n-k). - Paul Barry, May 13 2006
a(n+2) = C(3n+1,2n+1) = A078812(2n,n). - Paul Barry, Nov 09 2006
G.f.: A(x)=(2*cos(asin((3^(3/2)*sqrt(x))/2)/3)* sin(asin((3^(3/2)* sqrt(x))/2)/3))/(sqrt(3)*sqrt(1-(27*x)/4)*sqrt(x)). - Vladimir Kruchinin, Jun 10 2012
From Peter Luschny, Sep 04 2012: (Start)
O.g.f.: hypergeometric2F1([2/3, 4/3], [3/2], x*27/4).
a(n) = (n+1)*hypergeometric2F1([-2*n, -n], [2], 1).  (End)
D-finite with recurrence 2*n*(2*n+1)*a(n) - 3*(3*n-1)*(3*n+1)*a(n-1) = 0. - R. J. Mathar, Feb 05 2013
a(n) = Sum_{r=0..n} C(n,r) * C(2*n+1,r). - J. M. Bergot, Mar 18 2014
From Peter Bala, Nov 04 2015: (Start)
a(n) = Sum_{k = 0..n} 1/(2*k + 1)*binomial(3*n - 3*k,n - k)*binomial(3*k, k).
O.g.f. equals f(x)*g(x), where f(x) is the o.g.f. for A005809 and g(x) is the o.g.f. for A001764. More generally, f(x)*g(x)^k is the o.g.f. for the sequence binomial(3*n + k,n). Cf. A025174 (k = 2), A004319 (k = 3), A236194 (k = 4), A013698 (k = 5), A165817 (k = -1), A117671 (k = -2). (End)
a(n) = [x^n] 1/(1 - x)^(2*(n+1)). - Ilya Gutkovskiy, Oct 10 2017
O.g.f.: (i/24)*((4*sqrt(4 - 27*z) + 12*i*sqrt(3)*sqrt(z))^(2/3) - (4*sqrt(4 - 27*z) - 12*i*sqrt(3)*sqrt(z))^(2/3))*sqrt(3)*8^(1/3)*sqrt(4 - 27*z)/(sqrt(z)*(-4 + 27*z)), where i = sqrt(-1). - Karol A. Penson, Dec 13 2023
a(n-1) = (1/(4*n))*binomial(2*n, n)^2 * (1 - 3*((n - 1)/(n + 1))^3 + 5*((n - 1)*(n - 2)/((n + 1)*(n + 2)))^3 - 7*((n - 1)*(n - 2)*(n - 3)/((n + 1)*(n + 2)*(n + 3)))^3 + - ...) for n >= 1. Cf. A112029. - Peter Bala, Aug 08 2024
a(n) = Sum_{k = 0..n} (-1)^(n+k)*binomial(4*n+2, k)*binomial(2*n-k, n-k). - Peter Bala, Sep 04 2025
a(n) ~ 3^(3*n+3/2) / (4^(n+1) * sqrt(Pi*n)). - Amiram Eldar, Sep 05 2025

Simpler definition from Ira M. Gessel, May 26 2007. This change means that most of the offsets in the comments will now need to be changed too.

A173774 The arithmetic mean of (21k + 8)binomial(2*k,k)^3 (k=0..n-1).

Original entry on oeis.org

8, 120, 3680, 144760, 6427008, 306745824, 15364514880, 796663553400, 42395640372800, 2302336805317120, 127078484504270208, 7108177964254013920, 402042028998035350400, 22954860061516225396800

Offset: 1

Zhi-Wei Sun, Feb 24 2010

nonn

On Feb 10 2010, Zhi-Wei Sun introduced the sequence and conjectured that each term a(n) is an integer divisible by 4*binomial(2*n,n). On Feb 11 2011, Kasper Andersen confirmed this conjecture by noting that the sequence b(n) = a(n)/(4*binomial(2*n,n)), for n > 0, coincides with A112029. It was proved that for every prime p and positive integer a we have a(p^a) == 8 + 16*p^3*B_(p-3) (mod p^4), where B_0, B_1, B_2, ... are Bernoulli numbers. Given a prime p, it has been conjectured that Sum_{k=0..(p-1)/2} (21*k + 8)*binomial(2*k,k)^3 == 8*p + (-1)^((p-1)/2)*32*p^3*E_(p-3) (mod p^4) if p > 3 (where E_0, E_1, E_2, ... are Euler numbers), and that Sum_{k=0..floor(2p^a/3)} (21*k + 8)*binomial(2*k,k)^3 == 8*p^a (mod p^(a + 5 + (-1)^p)) if a is a positive integer with p^a == 1 (mod 3). He also observed that b(n) = a(n)/(4*binomial(2*n,n)) is odd if and only if n is a power of two.

			For n=2 we have a(2)=120 since (8*binomial(0,0)^3 + (21+8)*binomial(2,1)^3)/2 = 120.

G. C. Greubel, Table of n, a(n) for n = 1..500
Kasper Andersen, Re: A somewhat surprising conjecture
Zhi-Wei Sun, A somewhat surprising conjecture
Zhi-Wei Sun, Re: A somewhat surprising conjecture
Zhi-Wei Sun, Open conjectures on congruences, preprint, arXiv:0911.5665 [math.NT], 2009-2011.
Zhi-Wei Sun, Super congruences and Euler numbers, preprint, arXiv:1001.4453 [math.NT], 2010-2011.

Cf. A000984, A112029, A122045.

[(&+[(21*j+8)*(j+1)^3*Catalan(j)^2: j in [0..n-1]])/n: n in [1..30]]; // G. C. Greubel, Jul 06 2021

a[n_]:= Sum[(21*k+8)*Binomial[2*k,k]^3, {k,0,n-1}]/n; Table[a[n], {n, 25}]

[(1/n)*sum((21*j+8)*binomial(2*j,j)^3 for j in (0..n-1)) for n in (1..30)] # G. C. Greubel, Jul 06 2021

a(n) = (1/n)*Sum_{k=0..n-1} (21*k + 8)*binomial(2*k,k)^3.
(n+1)*a(n+1) = n*a(n) + 8*(21*n + 8)*binomial(2*n-1, n)^3, n > 0, with a(1) = 8.
a(n) ~ 2^(6*n) / (3 * (Pi*n)^(3/2)). - Vaclav Kotesovec, Jan 24 2019
a(n) = (1/n)*Sum_{j=0..n-1} (21*j + 8)*(j+1)^3*Catalan(j)^3. - G. C. Greubel, Jul 06 2021

A112028 a(n) = Sum_{k=0..n} binomial(n+k,k)^3.

Original entry on oeis.org

1, 9, 244, 9065, 389376, 18188478, 897376152, 46011772521, 2427553965160, 130930630643384, 7186614533569296, 400132290102421214, 22543708920891189136, 1282873288801683197250, 73628947696550668509744, 4257138240245923453355625, 247733479854085081062353400

Offset: 0

N. J. A. Sloane, Nov 28 2005

M.J. Coster: Supercongruences, [Thesis] Univ. of Leiden, the Netherlands, 1988.

Vincenzo Librandi, Table of n, a(n) for n = 0..200
F. Baldassarri, S. Bosch, B. Dwork, (eds), p-adic Analysis. Lecture Notes in Mathematics, vol. 1454, pp. 194 - 204, Springer, Berlin, Heidelberg.
Matthijs Coster, Supercongruences.
C. Elsner, On recurrence formulas for sums involving binomial coefficients, Fib. Q., 43,1 (2005), 31-45.
Vaclav Kotesovec, Asymptotic of generalized Apery sequences with powers of binomial coefficients, Nov 04 2012

Cf. A001700, A112029, A219562, A219563, A219564.

A375178 is an essentially identical sequence.

y2 := hypergeom([2/3, 2/3],[4/3],-x^2/27)*x^(1/3);
h := hypergeom([1/4, 1/4],[1],64*x)^2;
H := (243+247*x)*x*diff(h,x,x) + (463*x+234)*diff(h,x) + (48-24/x)*h;
ogf := y2*Int(Int(y2*H,x)/(8*x*(x^2+27)*y2^2),x);
# Check ogf by computing a series expansion:
SER := proc(a, x) series(a, x, 20) end:
INT := proc(a, x) int(SER(a, x), x) end:
SER(eval(ogf, Int = INT), x); # Mark van Hoeij, Apr 04 2013

Table[Sum[Binomial[n+k,k]^3,{k,0,n}],{n,0,20}] (* Harvey P. Dale, Sep 24 2012 *)

a(n) = sum(k=0, n, binomial(n+k, k)^3); \\ Michel Marcus, Mar 09 2016

a(n) ~ 2^(6*n+3)/(7*(Pi*n)^(3/2)). - Vaclav Kotesovec, Nov 23 2012
Recurrence: 3*(3*n-1)*(3*n+1)*(15799*n^5 - 103177*n^4 + 265789*n^3 - 336367*n^2 + 208000*n - 49852)*n^3*a(n) = 24*(2*n-3)^3*(3*n-4)*(3*n-2)*(15799*n^5 - 24182*n^4 + 11071*n^3 - 72*n^2 - 1080*n + 192)*a(n-1) - (n-1)^2*(15799*n^5 - 103177*n^4 + 265789*n^3 - 336367*n^2 + 208000*n - 49852)*n^3*a(n-2) + 8*(n-2)^2*(2*n-3)^3*(15799*n^5 - 24182*n^4 + 11071*n^3 - 72*n^2 - 1080*n + 192)*a(n-3). - Vaclav Kotesovec, Nov 23 2012
O.g.f. can be expressed in terms of hypergeometric functions (see Maple program). - Mark van Hoeij, Apr 01 2013
From Peter Bala, Mar 29 2023: (Start)
The supercongruence a(p-1) == 1 (mod p^5) appears to hold for all primes p >= 7 (checked up to p = 199). Coster, Theorem 4, proves that a(p-1) == 1 (mod p^3) for primes p >= 5.
For r >= 2, the supercongruence a(p^r - 1) == a(p^(r-1) - 1) (mod p^(3*r+3)) may hold for all primes p >= 7. (End)

A219562 a(n) = Sum_{k=0..n} binomial(n+k,k)^4.

Original entry on oeis.org

1, 17, 1378, 170257, 25561876, 4294835666, 776487013506, 147812510671121, 29234435383857304, 5955068493838815892, 1241820686691538181636, 263946916625793118532050, 56996643356459050103185444, 12473214064899644269110156626, 2760963661677614009262282769378

Offset: 0

Vaclav Kotesovec, Nov 23 2012

Vincenzo Librandi, Table of n, a(n) for n = 0..200
F. Baldassarri, S. Bosch, B. Dwork, (eds), p-adic Analysis. Lecture Notes in Mathematics, vol. 1454, pp. 194 - 204, Springer, Berlin, Heidelberg.
Matthijs Coster, Supercongruences.
Vaclav Kotesovec, Asymptotic of generalized Apery sequences with powers of binomial coefficients, Nov 04 2012

Cf. A001700, A112029, A112028, A219563, A219564.

q := x-4+I*((x+4)*(16-x))^(1/2);
f := x*(q/8)^4;
s := ((q-2)/(8*I-6))^(1/4);
y1 := hypergeom([1/8, 1/8], [3/4], f) * s / x^(1/8);
r := 2/(x*((x+4)*(16-x))^(1/2)*y1^2);
h := hypergeom([1/2, 1/2, 1/2, 1/2],[1, 1, 1],256*x);
u := (15*(223*x+72)*x^2*diff(h,x,x,x)+(14579*x+3226)*x*diff(h,x,x)
+(9969*x+1002)*diff(h,x)+320*h)/(16*(16-x)*(x+4)*x^2);
ogf := y1^2*Int(r*(1+Int(r*Int(u/(r*y1)^2,x),x)),x) ;
# Check o.g.f. by computing a series expansion:
SER := proc(a,x) series(a,x,20) end:
INT := proc(a,x) int(SER(a,x),x) end:
SER(eval(ogf, Int = INT),x); # Mark van Hoeij, Apr 02 2013

Table[Sum[Binomial[n+k,k]^4, {k,0,n}], {n,0,20}]

a(n) = sum(k=0, n, binomial(n+k,k)^4); \\ Michel Marcus, Jul 15 2022

a(n) ~ 2^(8*n+4)/(15*Pi^2*n^2).
Recurrence: 4*(n-1)*(4*n-1)*(4*n+1)*(279825*n^6 - 2240985*n^5 + 7416081*n^4 - 12962383*n^3 + 12597634*n^2 - 6438500*n + 1347304)*n^4*a(n) = 2*(n-1)*(2290647450*n^12 - 22926837585*n^11 + 100717526436*n^10 - 254986993727*n^9 + 410380920831*n^8 - 435959897978*n^7 + 305660392723*n^6 - 134977315842*n^5 + 31413259700*n^4 + 2833672*n^3 - 2076143616*n^2 + 500898816*n - 39813120)*a(n-1) + (859902225*n^13 - 10755967005*n^12 + 60090860763*n^11 - 197381561581*n^10 + 422055067481*n^9 - 613861172995*n^8 + 615013106513*n^7 - 418396400175*n^6 + 182810864162*n^5 - 42759392772*n^4 + 146171272*n^3 + 2813432832*n^2 - 691172352*n + 55738368)*a(n-2) - 16*(n-2)^3*(2*n-3)^4*(279825*n^6 - 562035*n^5 + 408531*n^4 - 111409*n^3 - 5504*n^2 + 7968*n - 1024)*a(n-3).
G.f. as an expression in terms of 2F1 and 4F3 functions is given in the Maple program below. - Mark van Hoeij, Apr 02 2013
From Peter Bala, Nov 29 2024: (Start)
Conjecture: a(p-1) == 1 (mod p^5) for prime p >= 7 (checked up to p = 499). Coster, Theorem 4, proves that a(p-1) == 1 (mod p^3) for primes p >= 5.
Conjecture: for r >= 2, the supercongruence a(p^r - 1) == a(p^(r-1) - 1) (mod p^(3*r+3)) may hold for all primes p >= 5. Coster, Theorem 4, proves that a(p^r -1) == a(p^(r-1) - 1) (mod p^(3*r)) for r >= 2 and all primes p >= 5. (End)

A219563 Sum(binomial(n+k,k)^5, k=0..n).

Original entry on oeis.org

1, 33, 8020, 3301025, 1733984376, 1048567813062, 694995078406056, 491336887915201185, 364377975224032162000, 280380150421755638519408, 222165159124597435189467696, 180288439972217748901049985158, 149230751849318301857448761484400, 125602423480863080624602495191566250

Offset: 0

Vaclav Kotesovec, Nov 23 2012

nonn

Vincenzo Librandi, Table of n, a(n) for n = 0..200
V. Kotesovec, Asymptotic of generalized Apery sequences with powers of binomial coefficients, Nov 04 2012

Cf. A001700, A112029, A112028, A219562, A219564.

Table[Sum[Binomial[n+k,k]^5, {k,0,n}], {n,0,20}]

a(n) ~ 2^(10*n+5)/(31*(Pi*n)^(5/2)).

A219564 Sum(binomial(n+k,k)^6, k=0..n).

Original entry on oeis.org

1, 65, 47386, 65004097, 119498671876, 260128695981674, 632156164654144530, 1659900189891175027265, 4616088190888638302435080, 13418259230056806455830305940, 40401802613222456104862752944356, 125182282922559710456869140648653290, 397195659937314116991934285462527257236

Offset: 0

Vaclav Kotesovec, Nov 23 2012

nonn

Vincenzo Librandi, Table of n, a(n) for n = 0..200
V. Kotesovec, Asymptotic of generalized Apery sequences with powers of binomial coefficients, Nov 04 2012

Cf. A001700 (q=1), A112029 (q=2), A112028 (q=3), A219562 (q=4), A219563 (q=5).

Table[Sum[Binomial[n+k,k]^6, {k,0,n}], {n,0,20}]

a(n) ~ 2^(12*n+6)/(63*Pi^3*n^3)

Generally (for q > 0), Sum_{k=0..n} C(n + k,k)^q is asymptotic to 2^((2*n+1)*q)/((2^q-1)*(Pi*n)^(q/2)) * (1 - q/(2*n)*(1/4+1/(2^q-1)^2) + O(1/n^2))

A176335 Central coefficients T(2n,n) of number triangle A176331.

Original entry on oeis.org

1, 3, 28, 315, 3876, 50358, 678112, 9365499, 131809060, 1882294128, 27193657008, 396600597198, 5829739893264, 86262567856650, 1283677784658528, 19196304797150715, 288295493121264420, 4346056823245242420

Offset: 0

Paul Barry, Apr 15 2010

G. C. Greubel, Table of n, a(n) for n = 0..825
F. Baldassarri, S. Bosch, B. Dwork, (eds), p-adic Analysis. Lecture Notes in Mathematics, vol. 1454, pp. 194 - 204, Springer, Berlin, Heidelberg.
Matthijs J. Coster, Supercongruences.

Cf. A112029, A176331, A176332, A176334.

T:= function(n,k)
    return Sum([0..n], j-> (-1)^(n-j)*Binomial(j,k)*Binomial(j,n-k) );
  end;
List([0..30], n-> T(2*n,n) ); # G. C. Greubel, Dec 07 2019

T:= func< n,k | &+[(-1)^(n-j)*Binomial(j,n-k)*Binomial(j,k): j in [0..n]] >;
[T(2*n,n): n in [0..30]]; // G. C. Greubel, Dec 07 2019

A176335 := proc(n)
    add((-1)^k*binomial(k,n)^2,k=0..2*n);
end proc: # R. J. Mathar, Feb 10 2015

T[n_, k_]:= Sum[(-1)^(n-j)*Binomial[j, k]*Binomial[j, n-k], {j,0,n}]; Table[T[2*n, n], {n,0,30}] (* G. C. Greubel, Dec 07 2019 *)

T(n,k) = sum(j=0, n, (-1)^(n-j)*binomial(j, n-k)*binomial(j, k));
vector(31, n, T(2*(n-1), n-1) ) \\ G. C. Greubel, Dec 07 2019

@CachedFunction
def T(n, k): return sum( (-1)^(n-j)*binomial(j, n-k)*binomial(j, k) for j in (0..n))
[T(2*n, n) for n in (0..30)] # G. C. Greubel, Dec 07 2019

a(n) = Sum_{k=0..2n} C(k,n)^2*(-1)^k.
Conjecture: 224*n^2*(n-1)*a(n) - 48*(n-1)*(65*n^2 - 36*n - 13)*a(n-1) + 4*(-1839*n^3 + 11081*n^2 - 21932*n + 14280)*a(n-2) + 12*(-81*n^3 + 326*n^2 - 591*n + 562)*a(n-3) - (n-3)*(1853*n^2 - 7403*n + 7140)*a(n-4) - 12*(n-4)*(2*n-7)^2*a(n-5) = 0. - R. J. Mathar, Feb 10 2015
From Peter Bala, Aug 08 2024: (Start)
a(n) = Sum_{k = 0..n} (-1)^(n+k)*binomial(n+k, k)^2. Cf. A112029.
Conjecture (assuming an offset of 1): the supercongruence a(n*p^r) == a(n*p^(r-1)) (mod p^(3*r)) holds for all primes p >= 5 and all positive integers n and r [added Nov 29 2024: proved by Coster. See Theorem 4]. (End)
a(n) ~ 2^(4*n+2) / (5*Pi*n). - Vaclav Kotesovec, Aug 08 2024
a(n) = binomial(2*n, n)^2 * hypergeom([1, -n, -n], [-2*n, -2*n], -1). - Peter Luschny, Nov 29 2024

A110197 Number triangle of sums of squared binomial coefficients.

Original entry on oeis.org

1, 2, 1, 3, 5, 1, 4, 14, 10, 1, 5, 30, 46, 17, 1, 6, 55, 146, 117, 26, 1, 7, 91, 371, 517, 251, 37, 1, 8, 140, 812, 1742, 1476, 478, 50, 1, 9, 204, 1596, 4878, 6376, 3614, 834, 65, 1, 10, 285, 2892, 11934, 22252, 19490, 7890, 1361, 82, 1, 11, 385, 4917, 26334, 66352, 82994, 51990, 15761, 2107, 101, 1

Offset: 0

Paul Barry, Jul 15 2005

Alternatively, number square T(n,k) = Sum_{i=0..n} binomial(i+k,k)^2 read by antidiagonals.

			Rows start:
  1;
  2,   1;
  3,   5,   1;
  4,  14,  10,   1;
  5,  30,  46,  17,   1;
  6,  55, 146, 117,  26,   1;
  ...

Columns include A000027, A000330, A024166, A086020, A086023, A086025.
Row sums are A006134.
Antidiagonal sums are A110198.
T(2n,n) gives A112029.

T(n,k) = sum(i=0, n-k, binomial(i+k,k)^2);
tabl(nn) = for (n=0, nn, for (k=0, n, print1(T(n, k), ", ")); print();); \\ Michel Marcus, Dec 03 2016

T(n,k) = Sum_{i=0..n-k} binomial(i+k,k)^2.

G.f.: 1/((1-x)*sqrt(x^2*y^2-2*x^2*y-2*x*y+x^2-2*x+1)). - Vladimir Kruchinin, Mar 20 2025

A119307 Triangle read by rows: T(n, k) = Sum_{j=0..n} C(j, k)*C(j, n - k).

Original entry on oeis.org

1, 1, 1, 1, 5, 1, 1, 11, 11, 1, 1, 19, 46, 19, 1, 1, 29, 127, 127, 29, 1, 1, 41, 281, 517, 281, 41, 1, 1, 55, 541, 1579, 1579, 541, 55, 1, 1, 71, 946, 4001, 6376, 4001, 946, 71, 1, 1, 89, 1541, 8889, 20626, 20626, 8889, 1541, 89, 1, 1, 109, 2377, 17907, 56904, 82994

Offset: 0

Paul Barry, May 13 2006

			Triangle begins
  1,
  1, 1,
  1, 5, 1,
  1, 11, 11, 1,
  1, 19, 46, 19, 1,
  1, 29, 127, 127, 29, 1,
  1, 41, 281, 517, 281, 41, 1
  ...

Indranil Ghosh, Rows 0..100, flattened

Second column is A028387.
Row sums are A014300.
Central coefficients T(2*n, n) are A112029.

T := (n, k) -> if n = k then 1 else binomial(n, k)^2*hypergeom([1, -k, -n + k], [-n, -n], 1) fi: for n from 0 to 9 do seq(simplify(T(n, k)), k = 0..n) od;
# Peter Luschny, May 13 2024

Flatten[Table[Sum[Binomial[j,k] Binomial[j,n-k],{j,0,n}],{n,0,10},{k,0,n}]] (* Indranil Ghosh, Mar 03 2017 *)

tabl(nn)={for (n=0, nn, for(k=0, n, print1(sum(j=0, n, binomial(j,k)*binomial(j,n-k)),", ");); print(););};
tabl(10); \\ Indranil Ghosh, Mar 03 2017

T(n, k) = T(n, n - k).

T(n, k) = binomial(n, k)^2*hypergeom([1, -k, -n + k], [-n, -n], 1) for k=0..n-1. - Peter Luschny, May 13 2024

A295612 a(n) = Sum_{k=0..n} binomial(n+k,k)^k.

Original entry on oeis.org

1, 3, 40, 8105, 24053106, 1016507243472, 622366942086680904, 5608321882919220905812521, 752711651805019773658037206391596, 1518219710649896586598445898967340890577318, 46343146356260529633020448755386347142785083052620084

Offset: 0

Ilya Gutkovskiy, Nov 24 2017

nonn

Seiichi Manyama, Table of n, a(n) for n = 0..41
Eric Weisstein's World of Mathematics, Binomial Sums

Cf. A001700, A046899, A112028, A112029, A167008, A219562, A219563, A219564.

Table[Sum[Binomial[n + k, k]^k, {k, 0, n}], {n, 0, 10}]
Table[Sum[((n + k)!/(n! k!))^k, {k, 0, n}], {n, 0, 10}]

a(n) = sum(k=0, n, binomial(n+k,k)^k); \\ Michel Marcus, Nov 25 2017

a(n) = Sum_{k=0..n} A046899(n,k)^k.

a(n) ~ 2^(2*n^2) / (exp(1/8) * Pi^(n/2) * n^(n/2)). - Vaclav Kotesovec, Nov 25 2017

cp's OEIS Frontend

A045721 a(n) = binomial(3*n+1,n).

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A173774 The arithmetic mean of (21*k + 8)*binomial(2*k,k)^3 (k=0..n-1).

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A112028 a(n) = Sum_{k=0..n} binomial(n+k,k)^3.

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A219562 a(n) = Sum_{k=0..n} binomial(n+k,k)^4.

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A219563 Sum(binomial(n+k,k)^5, k=0..n).

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A219564 Sum(binomial(n+k,k)^6, k=0..n).

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A176335 Central coefficients T(2n,n) of number triangle A176331.

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A173774 The arithmetic mean of (21k + 8)binomial(2*k,k)^3 (k=0..n-1).