A006442 -id:A006442 - cp's OEIS frontend

A243943 a(n) = A006442(n)^2.

1, 25, 1369, 93025, 6974881, 553425625, 45558768025, 3848757330625, 331434586569025, 28966516730025625, 2561512789823546329, 228690489716580520225, 20579914168308199841761, 1864413002713001259355225, 169871744046114667846619929, 15554069096581207471331850625

Offset: 0

Paul D. Hanna, Aug 17 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 A006442 is 1/sqrt(1 - 10*x + x^2).
Limit_{n -> oo} a(n+1)/a(n) = (5 + 2*sqrt(6))^2 = 49 + 20*sqrt(6).

			G.f.: A(x) = 1 + 9*x + 169*x^2 + 3969*x^3 + 103041*x^4 + 2832489*x^5 +...

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

Sequences of the form LegendreP(n, 2*m+1)^2: A000012 (m=0), A243949 (m=1), this sequence (m=2), A243944 (m=3), A243007 (m=4).

Cf. A006442.

[Evaluate(LegendrePolynomial(n), 5)^2 : n in [0..40]]; // G. C. Greubel, May 17 2023

Table[Sum[6^k * Binomial[2*k, k]^2 * Binomial[n+k, n-k], {k,0,n}], {n,0,20}] (* Vaclav Kotesovec, Sep 28 2019 *)
LegendreP[Range[0,40], 5]^2 (* G. C. Greubel, May 17 2023 *)

{a(n) = sum(k=0, n, 6^k * binomial(2*k, k)^2 * binomial(n+k, n-k) )}
for(n=0, 20, print1(a(n), ", "))

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

/* Using AGM: */
{a(n)=polcoeff( 1 / agm(1-x, sqrt((1+x)^2 - 10^2*x +x*O(x^n))), n)}
for(n=0, 20, print1(a(n), ", ")) \\ Paul D. Hanna, Aug 30 2014

[gen_legendre_P(n,0,5)^2 for n in range(41)] # G. C. Greubel, May 17 2023

G.f.: 1 / AGM(1-x, sqrt(1-98*x+x^2)).  - Paul D. Hanna, Aug 30 2014
a(n) = Sum_{k=0..n} 6^k * C(2*k, k)^2 * C(n+k, n-k).
a(n)^(1/2) = Sum_{k=0..n} 2^k * C(2*k, k) * C(n+k, n-k).
a(n) ~ (5+2*sqrt(6))^(2*n+1) / (4*Pi*sqrt(6)*n). - Vaclav Kotesovec, Sep 28 2019

A330204 Composite numbers k such that P(k, 5) == 5 (mod k), where P(k, 5) = A006442(k) is the k-th Legendre polynomial evaluated at 5.

Original entry on oeis.org

4, 15, 35, 165, 255, 615, 1815, 1876, 2636, 2948, 5380, 5565, 11235, 28545, 288380, 903644, 1807995, 2486165, 2674060, 10538572, 11791595, 14145121, 28558415, 45153277, 45682751

Offset: 1

Amiram Eldar, Dec 05 2019

P(p, 5) == 5 (mod p) for all primes p. This is a special case of Schur congruences (see A330203 for references). This sequence consists of the composite numbers for which the congruence holds.

			4 is in the sequence since it is composite and P(4, 5) = 2641 == 5 (mod 4).

Cf. A006442, A008316, A330203.

Select[Range[3000], CompositeQ[#] && Divisible[LegendreP[#, 5] - 5, #] &]

isok(k) = Mod(subst(pollegendre(k), x, 5), k) == 5;
forcomposite (k=1, 10000, if (isok(k), print1(k, ", "))); \\ Michel Marcus, Dec 06 2019

a, b = 1, 5
for n in range(2, 10000):
    a, b = b, ((10*n-5)*b - (n-1)*a)//n
    if (b%n == 5%n) and (not Integer(n).is_prime()): print(n)  # Robin Visser, Aug 17 2023

a(22)-a(23) from Robin Visser, Aug 17 2023

a(24)-a(25) from Robin Visser, Sep 11 2023

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

A084769 a(n) = P_n(9), where P_n is n-th Legendre polynomial; also, a(n) = central coefficient of (1 + 9x + 20x^2)^n.

Original entry on oeis.org

1, 9, 121, 1809, 28401, 458649, 7544041, 125700129, 2114588641, 35836273449, 610897146201, 10463745263409, 179939616743121, 3104680678772409, 53721299280288201, 931852905510160449, 16198821321758152641

Offset: 0

Paul D. Hanna, Jun 03 2003

Vincenzo Librandi, Table of n, a(n) for n = 0..200
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.
Vaclav Kotesovec, Asymptotic of a sums of powers of binomial coefficients * x^k, 2012.
Tony D. Noe, On the Divisibility of Generalized Central Trinomial Coefficients, Journal of Integer Sequences, Vol. 9 (2006), Article 06.2.7.

Column k=4 of A335333.
Sequences of the form LegendreP(n, 2*m+1): A000012 (m=0), A001850 (m=1), A006442 (m=2), A084768 (m=3), this sequence (m=4).
Cf. A243007 (a(n)^2), A269732.

[Evaluate(LegendrePolynomial(n), 9) : n in [0..40]]; // G. C. Greubel, May 17 2023

Table[SeriesCoefficient[1/Sqrt[1-18*x+x^2],{x,0,n}],{n,0,20}] (* Vaclav Kotesovec, Oct 14 2012 *)
LegendreP[Range[0, 40], 9] (* G. C. Greubel, May 17 2023 *)
a[n_] := 4^n*Sum[(5/4)^k*Binomial[n, k]^2, {k, 0, n}];
Table[a[n], {n, 0, 16}]  (* Detlef Meya, May 22 2024 *)

for(n=0,30,print1(subst(pollegendre(n),x,9)","))

{a(n)=sum(k=0, n, binomial(n, k)^2*4^k*5^(n-k))} \\ Paul D. Hanna, Sep 29 2012

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

G.f.: 1/sqrt(1-18*x+x^2).
Also a(n) = (n+1)-th term of the binomial transform of 1/(1-4x)^(n+1).
From Paul Barry, May 25 2005: (Start)
E.g.f.: exp(9*x) * Bessel_I(0, 2*sqrt(20)*x).
a(n) = Sum_{k=0..n} C(n, k)*C(n+k, k)4^k. (End)
D-finite with recurrence: n*a(n) + 9*(1-2*n)*a(n-1) + (n-1)*a(n-2) = 0. - R. J. Mathar, Sep 27 2012
a(n) = Sum_{k=0..n} binomial(n,k)^2 * 4^k * 5^(n-k). - Paul D. Hanna, Sep 29 2012
a(n) ~ sqrt(200 + 90*sqrt(5))*(9 + 4*sqrt(5))^n/(20*sqrt(Pi*n)) = (2 + sqrt(5))^(2*n+1)/(5^(1/4)*2*sqrt(2*Pi*n)). - Vaclav Kotesovec, Oct 14 2012
a(n) = hypergeom([-n, n+1], [1], -4). - Peter Luschny, May 23 2014
x*exp(Sum_{n >= 1} a(n)*x^n/n) = x + 9*x^2 + 101*x^3 + 1269*x^4 + ... is an integral power series, the o.g.f. for A269732. - Peter Bala, Jan 25 2018
a(n) = (-1)^n * Sum_{k=0..n} (1/18)^(n-2*k) * binomial(-1/2,k) * binomial(k,n-k). - Seiichi Manyama, Aug 28 2025
a(n) = Sum_{k=0..floor(n/2)} 20^k * 9^(n-2*k) * binomial(n,2*k) * binomial(2*k,k). - Seiichi Manyama, Aug 30 2025

A098270 a(n) = 2^n*P_n(5), 2^n times the Legendre polynomial of order n at 5.

Original entry on oeis.org

1, 10, 148, 2440, 42256, 752800, 13660480, 251113600, 4660568320, 87140108800, 1638884021248, 30970912737280, 587599919386624, 11185644310405120, 213540626285805568, 4086692369433395200, 78378887309200261120

Offset: 0

Paul Barry, Sep 01 2004

Central coefficients of (1 + 10*x + 24*x^2)^n. 2^n*LegendreP(n,k) yields the central coefficients of (1 + 2*k*x + (k^2-1)*x^2)^n, with g.f. 1/sqrt(1 - 4*k*x + 4*x^2).

Vincenzo Librandi, Table of n, a(n) for n = 0..200
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.
Eric Weisstein's World of Mathematics, Legendre Polynomial.

Sequences of the form 2^n*LegendreP(n, 2*m+1): A000079 (m=0), A084773 (m=1), this sequence (m=2).

Cf. A006442, A059473, A069835.

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

Table[SeriesCoefficient[1/Sqrt[1-20*x+4*x^2],{x,0,n}],{n,0,20}] (* Vaclav Kotesovec, Oct 14 2012 *)
Table[2^n*LegendreP[n, 5], {n,0,40}] (* G. C. Greubel, May 21 2023 *)

a(n)=pollegendre(n,5)<Charles R Greathouse IV, Oct 25 2011

def A098270(n): return 2^n*gen_legendre_P(n, 0, 5)
[A098270(n) for n in (0..16)] # Peter Luschny, Oct 14 2012

G.f.: 1/sqrt(1-20*x+4*x^2).
a(n) = Sum_{k=0..floor(n/2)} (-1)^k*binomial(n,k)*binomial(2*(n-k), n)*5^(n-2*k).
D-finite with recurrence: n*a(n) +10*(1-2*n)*a(n-1) +4*(n-1)*a(n-2) = 0. - R. J. Mathar, Sep 26 2012
a(n) ~ sqrt(72+30*sqrt(6))*(10+4*sqrt(6))^n/(12*sqrt(Pi*n)). - Vaclav Kotesovec, Oct 14 2012
a(n) = A059473(n,n). - Alois P. Heinz, Oct 05 2017
From Peter Bala, Nov 28 2021: (Start)
a(n) = (1/3)*Sum_{k >= n} binomial(k,n)^2*(2/3)^k.
a(n) = (4^n)*Sum_{k = 0..n} binomial(n,k)^2*(3/2)^k.
a(n) = (1/3)*(2/3)^n*hypergeometric2F1([n+1, n+1], [1], 2/3).
a(n) = (4^n)*hypergeometric2F1([-n, -n], [1], 3/2)
a(n) = [x^n] ((2*x - 2)*(3 - 2*x))^n.
a(n) = (2^n)*A006442(n). (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)

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)

A269730 Dimensions of the 2-polytridendriform operad TDendr_2.

Original entry on oeis.org

1, 5, 31, 215, 1597, 12425, 99955, 824675, 6939769, 59334605, 513972967, 4501041935, 39784038517, 354455513105, 3179928556219, 28701561707675, 260447708523505, 2374690737067925, 21744508765633327, 199877846477679815, 1843718766426242221, 17060955558786455705, 158333204443000060291

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. A001047, A001263, A126216, A269731, A269732, A001003, A006442.

Rest[CoefficientList[Series[(1 - 5*x - Sqrt[1 - 10*x + x^2])/(12*x), {x, 0, 20}], x]] (* Vaclav Kotesovec, Apr 24 2016 *)
Table[-I*LegendreP[n, -1, 2, 5]/Sqrt[6], {n, 1, 20}] (* Vaclav Kotesovec, Apr 24 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=2); vector(23, n, dimTDendr(n,q)) \\ Gheorghe Coserea, Apr 23 2016

my(q=2, x='x + O('x^24)); Vec(serreverse(x/((1+q*x)*(1+(q+1)*x)))) \\ Gheorghe Coserea, Sep 30 2017

a(n) = P_n(2), 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) = 5*(2*n-1)*a(n-1) - (n-2)*a(n-2). - Vaclav Kotesovec, Apr 24 2016
a(n) ~ sqrt(12 + 5*sqrt(6)) * (5 + 2*sqrt(6))^n / (12*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Apr 24 2016
A(x) = -serreverse(A001047(x))(-x). - Gheorghe Coserea, Sep 30 2017
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 = 2.
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 = 2. (End)
G.f. A(x) = x*exp( Sum_{n >= 1} A006442(n)*x^n/n ). - Peter Bala, Jan 09 2022

More terms from Gheorghe Coserea, Apr 23 2016

A387368 a(n) = Sum_{k=0..n} 2^k * 3^(n-k) * binomial(n+1,k) * binomial(n+1,n-k).

Original entry on oeis.org

1, 10, 93, 860, 7985, 74550, 699685, 6597400, 62457921, 593346050, 5653702637, 54012503220, 517192500721, 4962377183470, 47698928343285, 459224987322800, 4427611044899585, 42744433267222650, 413145666547033213, 3997556929553596300, 38718094094951086641

Offset: 0

Seiichi Manyama, Aug 27 2025

Vincenzo Librandi, Table of n, a(n) for n = 0..500

Cf. A006442, A387369.

Cf. A331792.

[&+[2^k * 3^(n-k) * Binomial(n+1,k) * Binomial(n+1,n-k): k in [0..n]]: n in [0..25]]; // Vincenzo Librandi, Aug 29 2025

Table[Sum[2^k * 3^(n-k)*Binomial[n+1,k]*Binomial[n+1, n-k],{k,0,n}],{n,0,25}] (* Vincenzo Librandi, Aug 29 2025 *)

a(n) = sum(k=0, n, 2^k*3^(n-k)*binomial(n+1, k)*binomial(n+1, n-k));

a(n) = Sum_{k=0..n} 3^k * 2^(n-k) * binomial(n+1,k) * binomial(n+1,n-k).
n*(n+2)*a(n) = (n+1) * (5*(2*n+1)*a(n-1) - n*a(n-2)) for n > 1.
a(n) = Sum_{k=0..floor(n/2)} 6^k * 5^(n-2*k) * binomial(n+1,n-2*k) * binomial(2*k+1,k).
a(n) = [x^n] (1+5*x+6*x^2)^(n+1).
E.g.f.: exp(5*x) * BesselI(1, 2*sqrt(6)*x) / sqrt(6), with offset 1.

cp's OEIS Frontend

A243943 a(n) = A006442(n)^2.

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A330204 Composite numbers k such that P(k, 5) == 5 (mod k), where P(k, 5) = A006442(k) is the k-th Legendre polynomial evaluated at 5.

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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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A084769 a(n) = P_n(9), where P_n is n-th Legendre polynomial; also, a(n) = central coefficient of (1 + 9*x + 20*x^2)^n.

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A098270 a(n) = 2^n*P_n(5), 2^n times the Legendre polynomial of order n at 5.

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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).

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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 + 7x + 12x^2)^n.

A084769 a(n) = P_n(9), where P_n is n-th Legendre polynomial; also, a(n) = central coefficient of (1 + 9x + 20x^2)^n.

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).