A084769 -id:A084769 - cp's OEIS frontend

A243007 a(n) = A084769(n)^2.

1, 81, 14641, 3272481, 806616801, 210358905201, 56912554609681, 15800522430616641, 4471485120646226881, 1284238494711502355601, 373195323236525968732401, 109489964937514282794301281, 32378265673661271315300820641, 9639042117142706280223219663281

Offset: 0

Paul D. Hanna, Aug 18 2014

nonn

In general, we have the binomial identity:
if b(n) = Sum_{k=0..n} t^k * C(2*k, k) * C(n+k, n-k),
then b(n)^2 = Sum_{k=0..n} (t^2+t)^k * C(2*k, k)^2 * C(n+k, n-k),
where the g.f. of b(n) is 1/sqrt(1 - (4*t+2)*x + x^2),
and the g.f. of b(n)^2 is 1 / AGM(1-x, sqrt((1+x)^2 - (4*t+2)^2*x)), where AGM(x,y) = AGM((x+y)/2,sqrt(x*y)) is the arithmetic-geometric mean.
Note that the g.f. of A084769 is 1/sqrt(1 - 18*x + x^2).
Limit_{n -> oo} a(n+1)/a(n) = (9 + 4*sqrt(5))^2 = 161 + 72*sqrt(5).

			G.f.: A(x) = 1 + 81*x + 14641*x^2 + 3272481*x^3 + 806616801*x^4 +...

Vincenzo Librandi, Table of n, a(n) for n = 0..100

Sequences of the form LegendreP(n, 2*m+1)^2: A000012 (m=0), A243949 (m=1), A243943 (m=2), A243944 (m=3), this sequence (m=4).

Cf. A084769.

[Evaluate(LegendrePolynomial(n),9)^2 : n in [0..30]]; // G. C. Greubel, May 17 2023

Table[SeriesCoefficient[1/Sqrt[1 -18x +x^2], {x,0,n}], {n,0,20}]^2 (* Vincenzo Librandi, Feb 14 2018 *)
LegendreP[Range[0,30], 9]^2 (* G. C. Greubel, May 17 2023 *)

{a(n) = sum(k=0, n, 20^k * binomial(2*k, k)^2 * binomial(n+k, n-k) )}
for(n=0, 20, print1(a(n), ", "))

{a(n) = sum(k=0, n, 4^k * binomial(2*k, k) * binomial(n+k, n-k) )^2}
for(n=0, 20, print1(a(n), ", "))
{a(n)=polcoeff( 1 / agm(1-x, sqrt((1+x)^2 - 18^2*x +x*O(x^n))), n)}
for(n=0, 20, print1(a(n), ", ")) \\ Paul D. Hanna, Aug 30 2014

[gen_legendre_P(n,0,9)^2 for n in range(41)] # G. C. Greubel, May 17 2023

G.f.: 1 / AGM(1-x, sqrt(1- 322*x + x^2)).  - Paul D. Hanna, Aug 30 2014
a(n) = Sum_{k=0..n} 20^k * C(2*k, k)^2 * C(n+k, n-k).
a(n)^(1/2) = Sum_{k=0..n} 4^k * C(2*k, k) * C(n+k, n-k).
a(n) ~ (2 + sqrt(5))^(4*n+2) / (8*sqrt(5)*Pi*n). - Vaclav Kotesovec, Sep 28 2019

A006442 Expansion of 1/sqrt(1 - 10*x + x^2).

Original entry on oeis.org

1, 5, 37, 305, 2641, 23525, 213445, 1961825, 18205345, 170195525, 1600472677, 15122515985, 143457011569, 1365435096485, 13033485491077, 124715953657025, 1195966908404545, 11490534389896325, 110584004488276645, 1065853221648055025

Offset: 0

N. J. A. Sloane

Number of Delannoy paths from (0,0) to (n,n) with steps U(0,1), H(1,0) and D(1,1) where H can choose from two colors. - Paul Barry, May 25 2005
Number of lattice paths from (0,0) to (n,n) using steps (0,1), (1,1), and two kinds of steps (1,0). - Joerg Arndt, Jul 01 2011
The Gauss congruences a(n*p^k) == a(n*p^(k-1)) (mod p^k) hold for prime p and positive integers n and k. - Peter Bala, Jan 09 2022

Seiichi Manyama, Table of n, a(n) for n = 0..1000 (terms 0..200 from T. D. Noe)
Hacène Belbachir and Abdelghani Mehdaoui, Recurrence relation associated with the sums of square binomial coefficients, Quaestiones Mathematicae (2021) Vol. 44, Issue 5, 615-624.
Hacène Belbachir, Abdelghani Mehdaoui and László Szalay, Diagonal Sums in the Pascal Pyramid, II: Applications, J. Int. Seq., Vol. 22 (2019), Article 19.3.5.
Ömür Deveci and Anthony G. Shannon, Some aspects of Neyman triangles and Delannoy arrays, Mathematica Montisnigri (2021) Vol. L, 36-43.
Tony D. Noe, On the Divisibility of Generalized Central Trinomial Coefficients, Journal of Integer Sequences, Vol. 9 (2006), Article 06.2.7.
T. R. S. Walsh, Number of sensed planar maps with n edges and m vertices

Column k=2 of A335333.
Sequences of the form LegendreP(n, 2*m+1): A000012 (m=0), A001850 (m=1), this sequence (m=2), A084768 (m=3), A084769 (m=4).
Cf. A098270, A243943 (a(n)^2).

[Evaluate(LegendrePolynomial(n), 5): n in [0..40]]; // G. C. Greubel, May 21 2023

seq(orthopoly[P](n,5), n = 0 .. 20); # Robert Israel, Aug 18 2014

Table[LegendreP[n, 5], {n, 0, 19}] (* Arkadiusz Wesolowski, Aug 13 2012 *)
CoefficientList[Series[1 / Sqrt[1 - 10 x + x^2], {x, 0, 20}], x] (* Vincenzo Librandi, Nov 23 2014 *)

PARI
```
a(n)=subst(pollegendre(n),x,5)
```

/* as lattice paths: same as in A092566 but use */
steps=[[1,0], [1,0], [0,1], [1,1]]; /* note the double [1,0] */
/* Joerg Arndt, Jul 01 2011 */

{a(n)=sum(k=0,n,binomial(n,k)^2*2^k*3^(n-k))} /* Paul D. Hanna */

{a(n) = sum(k=0, n, 2^k * binomial(2*k, k) * binomial(n+k, n-k) )}
for(n=0, 25, print1(a(n), ", ")) \\ Paul D. Hanna, Aug 17 2014

[gen_legendre_P(n,0,5) for n in range(41)] # G. C. Greubel, May 21 2023

Legendre polynomial evaluated at 5. - Michael Somos, Dec 04 2001
G.f.: 1/sqrt(1 - 10*x + x^2).
a(n) equals the central coefficient of (1 + 5*x + 6*x^2)^n. - Paul D. Hanna, Jun 03 2003
a(n) equals the (n+1)-th term of the binomial transform of 1/(1-2x)^(n+1). - Paul D. Hanna, Sep 29 2003
a(n) = Sum_{k=0..n} 2^k*binomial(n, k)*binomial(n+k, k). - Benoit Cloitre, Apr 13 2004
a(n) = Sum_{k=0..n} binomial(n,k)^2 * 2^k * 3^(n-k). - Paul D. Hanna, Feb 04 2012
E.g.f.: exp(5*x) * Bessel_I(0, 2*sqrt(6)*x). - Paul Barry, May 25 2005
D-finite with recurrence: n*a(n) - 5*(2n-1)*a(n-1) + (n-1)*a(n-2) = 0 [Eq (4) in the T. D. Noe article]. R. J. Mathar, Jun 26 2012
a(n) ~ (5 + 2*sqrt(6))^n/(2*sqrt(Pi*n)*sqrt(5*sqrt(6) - 12)). - Vaclav Kotesovec, Oct 05 2012
a(n) = hypergeom([-n, n+1], [1], -2). - Peter Luschny, May 23 2014
a(n) = Sum_{k=0..n} 2^k * C(2*k, k) * C(n+k, n-k). - Paul D. Hanna, Aug 17 2014
a(n) = Sum_{k=0..n} (k+1) * 3^k * (-1)^(n-k) * binomial(n,k) * binomial(n+k+1,n) / (n+k+1). - Vladimir Kruchinin, Nov 23 2014
From Peter Bala, Nov 28 2021: (Start)
a(n) = (1/3)*(1/2)^n*Sum_{k >= n} binomial(k,n)^2*(2/3)^k.
a(n) = (1/3)^(n+1)*hypergeom([n+1, n+1], [1], 2/3).
a(n) = (2^n)*hypergeom([-n, -n], [1], 3/2).
a(n) = [x^n] ((x - 1)*(3 - 2*x))^n
a(n) = (1/2)^n*A098270(n). (End)
a(n) = (-1)^n * Sum_{k=0..n} (1/10)^(n-2*k) * binomial(-1/2,k) * binomial(k,n-k). - Seiichi Manyama, Aug 28 2025
a(n) = Sum_{k=0..floor(n/2)} 6^k * 5^(n-2*k) * binomial(n,2*k) * binomial(2*k,k). - Seiichi Manyama, Aug 30 2025

A084768 a(n) = P_n(7), where P_n is n-th Legendre polynomial; also, a(n) = central coefficient of (1 + 7x + 12x^2)^n.

Original entry on oeis.org

1, 7, 73, 847, 10321, 129367, 1651609, 21360031, 278905249, 3668760487, 48543499753, 645382441711, 8614382884849, 115367108888311, 1549456900170553, 20861640747345727, 281483386791966529, 3805228005705102151, 51527535767904810889, 698796718936034430607

Offset: 0

Paul D. Hanna, Jun 03 2003

More generally, given fixed parameters b and c, we have the identities:
(1) a(n) = Sum_{k=0..n} binomial(n,k)^2 * b^k * c^(n-k);
(2) a(n) = [x^n] (1 + (b+c)*x + b*c*x^2)^n;
(3) g.f.: 1/sqrt(1 - 2*(b+c)*x + (b-c)^2*x^2);
(4) Sum_{n>=1} a(n)*x^n/n = log(G(x)) where G(x) = 1 + (b+c)*x*G(x) + b*c*x^2*G(x)^2.
Number of directed 2-D walks of length 2n starting at (0,0) and ending on the X-axis using steps NE, SE, NE, SW and avoiding NE followed by SE. - David Scambler, Jun 24 2013

Michael De Vlieger, Table of n, a(n) for n = 0..875
Hacène Belbachir, Abdelghani Mehdaoui, and László Szalay, Diagonal Sums in the Pascal Pyramid, II: Applications, J. Int. Seq., Vol. 22 (2019), Article 19.3.5.
G. Levy, Solutions of second order recurrence equations (2010) PhD Thesis, Florida State University, page 3.
Tony D. Noe, On the Divisibility of Generalized Central Trinomial Coefficients, Journal of Integer Sequences, Vol. 9 (2006), Article 06.2.7.

Column k=3 of A335333.
Sequences of the form LegendreP(n, 2*m+1): A000012 (m=0), A001850 (m=1), A006442 (m=2), this sequence (m=3), A084769 (m=4).
Cf. A084774, A243944 (a(n)^2).

[Evaluate(LegendrePolynomial(n),7): n in [0..40]]; // G. C. Greubel, May 17 2023

Table[LegendreP[n, 7], {n, 0, 20}] (* Vaclav Kotesovec, Jul 31 2013 *)

for(n=0,30,print1(subst(pollegendre(n),x,7)","))

{a(n)=sum(k=0, n, binomial(n, k)^2*3^k*4^(n-k))} \\ Paul D. Hanna, Sep 28 2012
for(n=0, 20, print1(a(n), ", "))

/* From a(n)^2 = A243944(n) (Paul D. Hanna, Aug 18 2014): */
{a(n) = sqrtint( sum(k=0, n, 12^k * binomial(2*k, k)^2 * binomial(n+k, n-k) ) )}
for(n=0, 20, print1(a(n), ", "))

[gen_legendre_P(n,0,7) for n in range(41)] # G. C. Greubel, May 17 2023

G.f.: 1/sqrt(1 - 14*x + x^2).
Also a(n) = (n+1)-th term of the binomial transform of 1/(1-3x)^(n+1).
a(n) = Sum_{k=0..n} 3^k*C(n,k)*C(n+k,k). - Benoit Cloitre, Apr 13 2004
E.g.f.: exp(7*x) * Bessel_I(0, 2*sqrt(12)*x). - Paul Barry, May 25 2005
D-finite with recurrence: n*a(n) + 7*(1-2*n)*a(n-1) + (n-1)*a(n-2) = 0. - R. J. Mathar, Sep 27 2012
a(n) = Sum_{k=0..n} C(n,k)^2 * 3^k * 4^(n-k). - Paul D. Hanna, Sep 28 2012
a(n) ~ (7+4*sqrt(3))^(n+1/2)/(2*3^(1/4)*sqrt(2*Pi*n)). - Vaclav Kotesovec, Jul 31 2013
a(n) = hypergeom([-n, n+1], [1], -3). - Peter Luschny, May 23 2014
a(n)^2 = Sum_{k=0..n} 12^k * C(2*k, k)^2 * C(n+k, n-k) = A243944(n). - Paul D. Hanna, Aug 18 2014
From Peter Bala, Apr 17 2024: (Start)
a(n) = (1/4)*(1/3)^n*Sum_{k >= n} binomial(k, n)^2*(3/4)^k.
a(n) = (1/4)^(n+1)*hypergeom([n+1, n+1], [1], 3/4).
a(n) = [x^n] ((1 + x)*(4 + 3*x))^n = [x^n] ((1 + 3*x)*(1 + 4*x))^n.
a(n) = (3^n)*hypergeom([-n, -n], [1], 4/3) = (4^n)*hypergeom([-n, -n], [1], 3/4).
The Gauss congruences hold: a(n*p^r) == a(n*p^(r-1)) (mod p^r) for all primes p and positive integers n and r.
a(n) = (-1)^n * Sum_{k = 0..n} (-4)^k*binomial(2*k, k)*binomial(n+k, n-k).
G.f: Sum_{n >= 0} (3^n)*binomial(2*n, n)*x^n/(1 - x)^(2*n+1) = 1 + 7*x + 73*x^2 + 847^x^3 + .... (End)
a(n) = (-1)^n * Sum_{k=0..n} (1/14)^(n-2*k) * binomial(-1/2,k) * binomial(k,n-k). - Seiichi Manyama, Aug 28 2025
a(n) = Sum_{k=0..floor(n/2)} 12^k * 7^(n-2*k) * binomial(n,2*k) * binomial(2*k,k). - Seiichi Manyama, Aug 30 2025

A243946 Expansion of sqrt( (1+x + sqrt(1-18x+x^2)) / (2(1-18*x+x^2)) ).

Original entry on oeis.org

1, 7, 91, 1345, 20995, 337877, 5544709, 92234527, 1549694195, 26237641045, 446925926881, 7650344197987, 131489964887341, 2267722252458475, 39224201631222475, 680160975405238145, 11820134678459908115, 205812328555924135045, 3589742656727603141425, 62707329988264214752675

Offset: 0

Paul D. Hanna, Aug 17 2014

Square each term to form a bisection of A243945.

Limit_{n->oo} a(n+1)/a(n) = 9 + 4*sqrt(5).

			G.f.: A(x) = 1 + 7*x + 91*x^2 + 1345*x^3 + 20995*x^4 + 337877*x^5 + ...,
where A(x)^2 = (1+x + sqrt(1-18*x+x^2)) / (2*(1-18*x+x^2)).

Seiichi Manyama, Table of n, a(n) for n = 0..500

Cf. A243945, A243947, A084769, A245926, A008316.

seq(add(binomial(n,k)*binomial(n+k,k)*binomial(2*n+2*k,n+k), k = 0..n)/binomial(2*n,n), n = 0..20); # Peter Bala, Mar 14 2018

a[n_] := Hypergeometric2F1[-n, n + 1/2, 1, -4];
Table[a[n], {n, 0, 19}] (* Peter Luschny, Mar 16 2018 *)
CoefficientList[Series[Sqrt[(1+x+Sqrt[1-18x+x^2])/(2(1-18x+x^2))],{x,0,20}],x] (* Harvey P. Dale, Dec 26 2019 *)
a[n_] := Sum[(5^k Gamma[2 n + 1])/(Gamma[2 k + 1]*Gamma[n - k + 1]^2), {k, 0, n}];
Flatten[Table[a[n], {n, 0, 19}]] (* Detlef Meya, May 22 2024 *)

/* From definition: */
{a(n)=polcoeff( sqrt( (1+x + sqrt(1-18*x+x^2 +x*O(x^n))) / (2*(1-18*x+x^2 +x*O(x^n))) ), n)}
for(n=0, 20, print1(a(n), ", "))

/* From a(n) = sqrt( A243945(2*n) ): */
{a(n)=sqrtint( sum(k=0, 2*n, binomial(2*k, k)^2*binomial(2*n+k, 2*n-k)) )}
for(n=0, 20, print1(a(n), ", "))

{a(n) = sum(k=0, n, 5^(n-k)*binomial(2*k, k)*binomial(2*n, 2*k))} \\ Seiichi Manyama, Aug 25 2020

from math import comb
def A243946(n): return sum(5**(n-k)*comb(m:=k<<1,k)*comb(n<<1,m) for k in range(n+1)) # Chai Wah Wu, Mar 23 2023

a(n)^2 = Sum_{k=0..2*n} C(2*k, k)^2 * C(2*n+k, 2*n-k).
a(n) ~ sqrt(2+sqrt(5)) * (9+4*sqrt(5))^n / (2*sqrt(2*Pi*n)). - Vaclav Kotesovec, Aug 18 2014. Equivalently, a(n) ~ phi^(6*n + 3/2) / (2*sqrt(2*Pi*n)), where phi = A001622 is the golden ratio. - Vaclav Kotesovec, Dec 08 2021
From Peter Bala, Mar 14 2018: (Start)
a(n) = P(2*n,sqrt(5)), where P(n,x) denotes the n-th Legendre polynomial. See A008316.
a(n) = (1/C(2*n,n))*Sum_{k = 0..n} C(n,k)*C(n+k,k)* C(2*n+2*k,n+k). In general, P(2*n,sqrt(1 + 4*x)) = (1/C(2*n,n))*Sum_{k=0..n} C(n,k)*C(n+k,k)*C(2*n+2*k,n+k)*x^k.
a(n) = Sum_{k = 0..2*n} C(2*n,k)^2 * phi^(2*n-2*k), where phi = (sqrt(5) + 1)/2.
a(n) = Sum_{k = 0..2*n} C(2*n,k)*C(2*n+k,k)*Phi^k, where Phi = (sqrt(5) - 1)/2. (End)
a(n) = hypergeom([-n, n + 1/2], [1], -4). - Peter Luschny, Mar 16 2018
D-finite with recurrence: n*(2*n-1)*(4*n-5)*a(n) -(4*n-3)*(36*n^2-54*n+11)*a(n-1) +(n-1)*(4*n-1)*(2*n-3)*a(n-2)=0. - R. J. Mathar, Jan 20 2020
a(n) = Sum_{k=0..n} 5^(n-k) * binomial(2*k,k) * binomial(2*n,2*k). - Seiichi Manyama, Aug 25 2020

A243947 Expansion of g.f. sqrt( (1+x - sqrt(1-18x+x^2)) / (10x(1-18x+x^2)) ).

Original entry on oeis.org

1, 11, 155, 2365, 37555, 610897, 10098997, 168894355, 2849270515, 48395044705, 826479148001, 14177519463191, 244109912494525, 4216385987238575, 73024851218517275, 1267712063327871245, 22052786911315216595, 384321597582115655825, 6708530714274563938225

Offset: 0

Paul D. Hanna, Aug 17 2014

Multiply the square of each term by 5 to form a bisection of A243945.

Limit_{n->oo} a(n+1)/a(n) = 9 + 4*sqrt(5).

			G.f.: A(x) = 1 + 11*x + 155*x^2 + 2365*x^3 + 37555*x^4 + 610897*x^5 + ...
where
A(x)^2 = (1+x - sqrt(1-18*x+x^2)) / (10*x*(1-18*x+x^2)).

Seiichi Manyama, Table of n, a(n) for n = 0..500

Cf. A243945, A243946, A084769, A245927, A008316.

seq(add(1/2*binomial(2*k+1,k)*binomial(n,k)*binomial(2*n+2*k+2,2*n+1)/binomial(n+k+1,n), k = 0..n), n = 0..20); # Peter Bala, Mar 15 2018

CoefficientList[Series[Sqrt[((1+x-Sqrt[1-18x+x^2]))/(10x(1-18x+x^2))],{x,0,20}],x] (* Harvey P. Dale, Jul 31 2016 *)
a[n_] := Hypergeometric2F1[-n, n + 3/2, 1, -4];
Table[a[n], {n, 0, 18}] (* Peter Luschny, Mar 16 2018 *)

/* From definition: */
{a(n)=polcoeff( sqrt( (1+x - sqrt(1-18*x+x^2 +x^2*O(x^n))) / (10*x*(1-18*x+x^2 +x*O(x^n))) ), n)}
for(n=0, 20, print1(a(n), ", "))

/* From a(n) = sqrt( A243945(2*n+1)/5 ): */
{a(n)=sqrtint( (1/5)*sum(k=0, 2*n+1, binomial(2*k, k)^2*binomial(2*n+k+1, 2*n-k+1)) )}
for(n=0, 20, print1(a(n), ", "))

from math import comb
def A243947(n): return sum(5**(n-k)*comb(m:=k<<1,k)*comb((n<<1)+1,m) for k in range(n+1)) # Chai Wah Wu, Mar 23 2023

a(n)^2 = (1/5) * Sum_{k=0..2*n+1} C(2*k, k)^2 * C(2*n+k+1, 2*n-k+1).
a(n) ~ (9+4*sqrt(5))^(n+1) / (2*5^(1/4)*sqrt(2*Pi*n) * sqrt(5+2*sqrt(5))). - Vaclav Kotesovec, Aug 18 2014. Equivalently, a(n) ~ phi^(6*n + 9/2) / (2^(3/2) * sqrt(5*Pi*n)), where phi = A001622 is the golden ratio. - Vaclav Kotesovec, Dec 08 2021
From Peter Bala, Mar 15 2018: (Start)
a(n) = (1/sqrt(5))*P(2*n+1,sqrt(5)), where P(n,x) denotes the n-th Legendre polynomial. See A008316.
a(n) = Sum_{k = 0..n} (1/2)*C(2*k+1,k)*C(n,k)*C(2*n+2*k+2,2*n+1)/C(n+k+1,n). In general, (1/sqrt(1 + 4*x))*P(2*n+1,sqrt(1+4*x)) = (1/(2*C(2*n+1,n))) * Sum_{k = 0..n} C(n,k)*C(n+k+1,k)*C(2*n+2*k+2,n+k+1)*x^k.
a(n) = (1/sqrt(5))*Sum_{k = 0..2*n+1} C(2*n+1,k)^2 * phi^(2*n-2*k+1), where phi = (sqrt(5) + 1)/2.
a(n) = (1/sqrt(5))*Sum_{k = 0..2*n+1} C(2*n+1,k)*C(2*n+1+k,k) * Phi^k, where Phi = (sqrt(5) - 1)/2. (End)
a(n) = hypergeom([-n, n + 3/2], [1], -4). - Peter Luschny, Mar 16 2018
From Peter Bala, Mar 17 2018: (Start)
a(n) = Sum_{k = 0..n} C(2*n+1,2*k)*C(2*k,k)*5^(n-k).
D-finite with recurrence: n*(4*n-3)*(2*n+1)*a(n) = (4*n-1)*(36*n^2-18*n-7)*a(n-1) - (n-1)*(2*n-1)*(4*n+1)*a(n-2). (End)

A335333 Square array T(n,k), n >= 0, k >= 0, read by antidiagonals, where column k is the expansion of 1/sqrt(1 - 2(2k+1)*x + x^2).

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 5, 13, 1, 1, 7, 37, 63, 1, 1, 9, 73, 305, 321, 1, 1, 11, 121, 847, 2641, 1683, 1, 1, 13, 181, 1809, 10321, 23525, 8989, 1, 1, 15, 253, 3311, 28401, 129367, 213445, 48639, 1, 1, 17, 337, 5473, 63601, 458649, 1651609, 1961825, 265729, 1

Offset: 0

Seiichi Manyama, Jun 02 2020

			Square array begins:
  1,    1,     1,      1,      1,       1, ...
  1,    3,     5,      7,      9,      11, ...
  1,   13,    37,     73,    121,     181, ...
  1,   63,   305,    847,   1809,    3311, ...
  1,  321,  2641,  10321,  28401,   63601, ...
  1, 1683, 23525, 129367, 458649, 1256651, ...

Seiichi Manyama, Antidiagonals n = 0..139, flattened
Eric Weisstein's World of Mathematics, Legendre Polynomial.

Columns k=0..4 give A000012, A001850, A006442, A084768, A084769.
Rows n=0..6 give A000012, A005408, A003154(n+1), A160674, A144124, A335338, A144126.
Main diagonal gives A331656.
T(n,n-1) gives A331657.
Cf. A307883, A307884.

T[n_, k_] := LegendreP[n, 2*k + 1]; Table[T[k, n - k], {n, 0, 9}, {k, 0, n}] // Flatten (* Amiram Eldar, May 03 2021 *)

PARI
```
T(n, k) = pollegendre(n, 2*k+1);
```

T(n,k) is the coefficient of x^n in the expansion of (1 + (2*k+1)*x + k*(k+1)*x^2)^n.
T(n,k) = Sum_{j=0..n} k^j * (k+1)^(n-j) * binomial(n,j)^2.
T(n,k) = Sum_{j=0..n} k^j * binomial(n,j) * binomial(n+j,j).
n * T(n,k) = (2*k+1) * (2*n-1) * T(n-1,k) - (n-1) * T(n-2,k).
T(n,k) = P_n(2*k+1), where P_n is n-th Legendre polynomial.
From Seiichi Manyama, Aug 30 2025: (Start)
T(n,k) = (-1)^n * Sum_{j=0..n} (1/(2*(2*k+1)))^(n-2*j) * binomial(-1/2,j) * binomial(j,n-j).
T(n,k) = Sum_{j=0..floor(n/2)} (k*(k+1))^j * (2*k+1)^(n-2*j) * binomial(n,2*j) * binomial(2*j,j).
E.g.f. of column k: exp((2*k+1)*x) * BesselI(0, 2*sqrt(k*(k+1))*x). (End)

A300945 Rectangular array A(n, k) = hypergeom([-k, k + n/2 - 1], [1], -4) with row n >= 0 and k >= 0, read by ascending antidiagonals.

Original entry on oeis.org

1, 1, 1, 1, 3, 25, 1, 5, 43, 425, 1, 7, 65, 661, 7025, 1, 9, 91, 965, 10515, 116625, 1, 11, 121, 1345, 15105, 171097, 1951625, 1, 13, 155, 1809, 20995, 243525, 2828101, 32903225, 1, 15, 193, 2365, 28401, 337877, 4001345, 47284251, 558265825

Offset: 0

Peter Luschny, Mar 16 2018

			[0] 1,  1,  25,  425,  7025, 116625,  1951625,  32903225, ... [A299845]
[1] 1,  3,  43,  661, 10515, 171097,  2828101,  47284251, ... [A299506]
[2] 1,  5,  65,  965, 15105, 243525,  4001345,  66622085, ...
[3] 1,  7,  91, 1345, 20995, 337877,  5544709,  92234527, ... [A243946]
[4] 1,  9, 121, 1809, 28401, 458649,  7544041, 125700129, ... [A084769]
[5] 1, 11, 155, 2365, 37555, 610897, 10098997, 168894355, ... [A243947]
[6] 1, 13, 193, 3021, 48705, 800269, 13324417, 224028877, ...

Cf. A299845, A299506, A243946, A084769, A243947.

Cf. A300946.

Arow[n_, len_] := Table[Hypergeometric2F1[-k, k + n/2 - 1, 1, -4], {k, 0, len}];
Table[Print[Arow[n, 7]], {n, 0, 6}];
T[n_, k_] := If[k==0, 1, 4^k*Sum[(5/4)^j*Binomial[k, j]*Binomial[k - 2 + ((n - k)/2), j - 2 + ((n - k)/2)] ,{j, 0, n}]]; Flatten[Table[T[n, k],{n, 0, 8}, {k, 0, n}]] (* Detlef Meya, May 28 2024 *)

T(n, k) = if k = 0 then 1, otherwise 4^k*Sum_{j=0..n} (5/4)^j * binomial(k, j) * binomial(k - 2 + ((n - k)/2), j - 2 + ((n - k)/2)). - Detlef Meya, May 28 2024

A331656 a(n) = Sum_{k=0..n} binomial(n,k) * binomial(n+k,k) * n^k.

Original entry on oeis.org

1, 3, 37, 847, 28401, 1256651, 69125869, 4548342975, 348434664769, 30463322582899, 2993348092318101, 326572612514776079, 39170287549040392369, 5123157953193993402171, 725662909285939100555101, 110662236267661479984580351, 18077209893508013563092846849

Offset: 0

Ilya Gutkovskiy, Jan 23 2020

Seiichi Manyama, Table of n, a(n) for n = 0..321
Eric Weisstein's World of Mathematics, Legendre Polynomial

Main diagonal of A335333.

Cf. A001850, A006442, A084768, A084769, A110129, A331657.

Join[{1}, Table[Sum[Binomial[n, k] Binomial[n + k, k] n^k, {k, 0, n}], {n, 1, 16}]]
Table[SeriesCoefficient[1/Sqrt[1 - 2 (2 n + 1) x + x^2], {x, 0, n}], {n, 0, 16}]
Table[LegendreP[n, 2 n + 1], {n, 0, 16}]
Table[Hypergeometric2F1[-n, n + 1, 1, -n], {n, 0, 16}]

a(n) = {sum(k=0, n, binomial(n,k) * binomial(n+k,k) * n^k)} \\ Andrew Howroyd, Jan 23 2020

a(n) = central coefficient of (1 + (2*n + 1)*x + n*(n + 1)*x^2)^n.
a(n) = [x^n] 1 / sqrt(1 - 2*(2*n + 1)*x + x^2).
a(n) = n! * [x^n] exp((2*n + 1)*x) * BesselI(0,2*sqrt(n*(n + 1))*x).
a(n) = Sum_{k=0..n} binomial(n,k)^2 * n^k * (n + 1)^(n - k).
a(n) = P_n(2*n+1), where P_n is n-th Legendre polynomial.
a(n) ~ exp(1/2) * 4^n * n^(n - 1/2) / sqrt(Pi). - Vaclav Kotesovec, Jan 28 2020
From Seiichi Manyama, Aug 30 2025: (Start)
a(n) = (-1)^n * Sum_{k=0..n} (1/(2*(2*n+1)))^(n-2*k) * binomial(-1/2,k) * binomial(k,n-k).
a(n) = Sum_{k=0..floor(n/2)} (n*(n+1))^k * (2*n+1)^(n-2*k) * binomial(n,2*k) * binomial(2*k,k). (End)

A331657 a(n) = Sum_{k=0..n} (-1)^(n - k) * binomial(n,k) * binomial(n+k,k) * n^k.

Original entry on oeis.org

1, 1, 13, 305, 10321, 458649, 25289461, 1666406209, 127779121345, 11178899075537, 1098961472475901, 119937806278590321, 14389588419704763409, 1882432013890951832425, 266678501426944160023653, 40673387011956179149166849, 6644919093900517186643470081

Offset: 0

Ilya Gutkovskiy, Jan 23 2020

Seiichi Manyama, Table of n, a(n) for n = 0..322
Eric Weisstein's World of Mathematics, Legendre Polynomial

Cf. A001850, A006442, A084768, A084769, A110129, A331656, A335333.

[&+[(-1)^(n-k)*Binomial(n,k)*Binomial(n+k,k)*n^k:k in [0..n]]:n in [0..16]]; // Marius A. Burtea, Jan 23 2020

Join[{1}, Table[Sum[(-1)^(n - k) Binomial[n, k] Binomial[n + k, k] n^k, {k, 0, n}], {n, 1, 16}]]
Table[SeriesCoefficient[1/Sqrt[1 - 2 (2 n - 1) x + x^2], {x, 0, n}], {n, 0, 16}]
Table[LegendreP[n, 2 n - 1], {n, 0, 16}]
Table[(-1)^n Hypergeometric2F1[-n, n + 1, 1, n], {n, 0, 16}]

a(n) = {sum(k=0, n, (-1)^(n - k) * binomial(n,k) * binomial(n+k,k) * n^k)} \\ Andrew Howroyd, Jan 23 2020

a(n) = central coefficient of (1 + (2*n - 1)*x + n*(n - 1)*x^2)^n.
a(n) = [x^n] 1 / sqrt(1 - 2*(2*n - 1)*x + x^2).
a(n) = n! * [x^n] exp((2*n - 1)*x) * BesselI(0,2*sqrt(n*(n - 1))*x).
a(n) = Sum_{k=0..n} binomial(n,k)^2 * n^k * (n - 1)^(n - k).
a(n) = P_n(2*n-1), where P_n is n-th Legendre polynomial.
a(n) = (-1)^n * 2F1(-n, n + 1; 1; n).
a(n) ~ 4^n * n^(n - 1/2) / (exp(1/2) * sqrt(Pi)). - Vaclav Kotesovec, Jan 26 2020
From Seiichi Manyama, Aug 30 2025: (Start)
a(n) = (-1)^n * Sum_{k=0..n} (1/(2*(2*n-1)))^(n-2*k) * binomial(-1/2,k) * binomial(k,n-k).
a(n) = Sum_{k=0..floor(n/2)} ((n-1)*n)^k * (2*n-1)^(n-2*k) * binomial(n,2*k) * binomial(2*k,k). (End)

A269732 Dimensions of the 4-polytridendriform operad TDendr_4.

Original entry on oeis.org

1, 9, 101, 1269, 17081, 240849, 3511741, 52515549, 801029681, 12414177369, 194922521301, 3094216933509, 49575333021801, 800645021406369, 13020241953611181, 213025792632813549, 3504075376813414241, 57914491106005287849, 961297812844696640581, 16017765308027639317269, 267831397282643166904601, 4492625888792276208945009, 75578709400747348254905501

Offset: 1

N. J. A. Sloane, Mar 08 2016

Gheorghe Coserea, Table of n, a(n) for n = 1..512
Samuele Giraudo, Pluriassociative algebras II: The polydendriform operad and related operads, arXiv:1603.01394 [math.CO], 2016; Adv. Appl. Math., 77, 3-85, 2016.

Cf. A001003, A005060, A001263, A126216, A269730, A269731, A084769.

I:=[1,9]; [n le 2 select I[n] else (9*(2*n-1)*Self(n-1)-(n-2)*Self(n-2))/(n+1): n in [1..30]]; // Vincenzo Librandi, Nov 29 2016

Rest[CoefficientList[Series[(1 - 9*x - Sqrt[1 - 18*x + x^2])/(40*x), {x, 0, 20}], x]] (* Vaclav Kotesovec, Apr 24 2016 *)
Table[-I*LegendreP[n, -1, 2, 9]/(2*Sqrt[5]), {n, 1, 20}] (* Vaclav Kotesovec, Apr 24 2016 *)
RecurrenceTable[{a[1] == 1, a[2] == 9, (n+1) a[n] == 9 (2 n - 1) a[n-1] - (n - 2) a[n-2]}, a, {n, 25}] (* Vincenzo Librandi, Nov 29 2016 *)

A001263(n,k) = binomial(n-1,k-1) * binomial(n, k-1)/k;
dimTDendr(n,q) = sum(k = 0, n-1, (q+1)^k * q^(n-k-1) * A001263(n,k+1));
my(q=4); vector(23, n, dimTDendr(n,q)) \\ Gheorghe Coserea, Apr 23 2016

my(q=4, x='x + O('x^24)); Vec(serreverse(x/((1+q*x)*(1+(q+1)*x)))) \\ Gheorghe Coserea, Sep 30 2017

a(n) = P_n(4), where P_n(x) is the polynomial associated with row n of triangle A126216 in order of decreasing powers of x.
Recurrence: (n+1)*a(n) = 9*(2*n-1)*a(n-1) - (n-2)*a(n-2). - Vaclav Kotesovec, Apr 24 2016
a(n) ~ sqrt(40 + 18*sqrt(5)) * (9 + 4*sqrt(5))^n / (40*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Apr 24 2016
a(n) ~ phi^(6*n + 3) / (2^(5/2) * 5^(3/4) * sqrt(Pi) * n^(3/2)), where phi = A001622 = (1+sqrt(5))/2 is the golden ratio. - Vaclav Kotesovec, Sep 23 2017
A(x) = -serreverse(A005060(x))(-x). - Gheorghe Coserea, Sep 30 2017
O.g.f.: A(x) = (1 - sqrt(1 - 18*x + x^2) - 9*x)/(40*x). - Peter Bala, Jan 25 2018
From Peter Bala, Dec 25 2020: (Start)
a(n) = (1/(2*m*(m+1))) * Integral_{x = 1..2*m+1} Legendre_P(n,x) dx at m = 4.
a(n) = (1/(2*n+1)) * (1/(2*m*(m+1))) * ( Legendre_P(n+1,2*m+1) - Legendre_P(n-1,2*m+1) ) at m = 4. (End)

More terms from Gheorghe Coserea, Apr 23 2016

cp's OEIS Frontend

A243007 a(n) = A084769(n)^2.

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A006442 Expansion of 1/sqrt(1 - 10*x + x^2).

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A084768 a(n) = P_n(7), where P_n is n-th Legendre polynomial; also, a(n) = central coefficient of (1 + 7*x + 12*x^2)^n.

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A243946 Expansion of sqrt( (1+x + sqrt(1-18*x+x^2)) / (2*(1-18*x+x^2)) ).

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A243947 Expansion of g.f. sqrt( (1+x - sqrt(1-18*x+x^2)) / (10*x*(1-18*x+x^2)) ).

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A335333 Square array T(n,k), n >= 0, k >= 0, read by antidiagonals, where column k is the expansion of 1/sqrt(1 - 2*(2*k+1)*x + x^2).

A084768 a(n) = P_n(7), where P_n is n-th Legendre polynomial; also, a(n) = central coefficient of (1 + 7x + 12x^2)^n.

A243946 Expansion of sqrt( (1+x + sqrt(1-18x+x^2)) / (2(1-18*x+x^2)) ).

A243947 Expansion of g.f. sqrt( (1+x - sqrt(1-18x+x^2)) / (10x(1-18x+x^2)) ).

A335333 Square array T(n,k), n >= 0, k >= 0, read by antidiagonals, where column k is the expansion of 1/sqrt(1 - 2(2k+1)*x + x^2).