A008316 -id:A008316 - cp's OEIS frontend

A372275 Decimal expansion of the middle positive zero of the Legendre polynomial of degree 7.

7, 4, 1, 5, 3, 1, 1, 8, 5, 5, 9, 9, 3, 9, 4, 4, 3, 9, 8, 6, 3, 8, 6, 4, 7, 7, 3, 2, 8, 0, 7, 8, 8, 4, 0, 7, 0, 7, 4, 1, 4, 7, 6, 4, 7, 1, 4, 1, 3, 9, 0, 2, 6, 0, 1, 1, 9, 9, 5, 5, 3, 5, 1, 9, 6, 7, 4, 2, 9, 8, 7, 4, 6, 7, 2, 1, 8, 0, 5, 1, 3, 7, 9, 2, 8, 2, 6

Offset: 0

Pontus von Brömssen, Apr 25 2024

			0.741531185599394439863864773280788407074147647141390260119955...

Paolo Xausa, Table of n, a(n) for n = 0..10000
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]. Table 25.4, n=7.
A.H.M. Smeets, Python program for Legendre-Gauss quadrature constants.
Eric Weisstein's World of Mathematics, Legendre Polynomial.
Eric Weisstein's World of Mathematics, Legendre-Gauss Quadrature.
Wikipedia, Legendre polynomials.
Index entries for algebraic numbers, degree 6.

Cf. A008316, A100258.
There are floor(k/2) positive zeros of the Legendre polynomial of degree k:
   k | zeros                           | corresponding weights for Legendre-Gauss quadrature
  ---+---------------------------------+----------------------------------------------------
   2 | A020760                         | A000007*10
   3 | A010513/10                      | A010716
   4 | A372267, A372268                | A382103, A382104
   5 | A372269, A372270                | A382105, A382106
   6 | A372271, A372272, A372273       | A382107, A382686, A382687
   7 | A372274, this sequence, A372276 | A382688, A382689, A382690

First[RealDigits[Root[LegendreP[7, #] &, 6], 10, 100]] (* Paolo Xausa, Feb 27 2025 *)

solve (x = 0.6, 0.8, 429*x^6 - 693*x^4 + 315*x^2 - 35) \\ A.H.M. Smeets, May 31 2025

Middle positive root of 429*x^6 - 693*x^4 + 315*x^2 - 35 = 0.

A372276 Decimal expansion of the largest positive zero of the Legendre polynomial of degree 7.

Original entry on oeis.org

9, 4, 9, 1, 0, 7, 9, 1, 2, 3, 4, 2, 7, 5, 8, 5, 2, 4, 5, 2, 6, 1, 8, 9, 6, 8, 4, 0, 4, 7, 8, 5, 1, 2, 6, 2, 4, 0, 0, 7, 7, 0, 9, 3, 7, 6, 7, 0, 6, 1, 7, 7, 8, 3, 5, 4, 8, 7, 6, 9, 1, 0, 3, 9, 1, 3, 0, 6, 3, 3, 3, 0, 3, 5, 4, 8, 4, 0, 1, 4, 0, 8, 0, 5, 7, 3, 0

Offset: 0

Pontus von Brömssen, Apr 25 2024

			0.949107912342758524526189684047851262400770937670617783548769...

Paolo Xausa, Table of n, a(n) for n = 0..10000
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]. Table 25.4, n=7.
A.H.M. Smeets, Python program for Legendre-Gauss quadrature constants.
Eric Weisstein's World of Mathematics, Legendre Polynomial.
Eric Weisstein's World of Mathematics, Legendre-Gauss Quadrature.
Wikipedia, Legendre polynomials.
Index entries for algebraic numbers, degree 6.

Cf. A008316, A100258.
There are floor(k/2) positive zeros of the Legendre polynomial of degree k:
   k | zeros                           | corresponding weights for Legendre-Gauss quadrature
  ---+---------------------------------+----------------------------------------------------
   2 | A020760                         | A000007*10
   3 | A010513/10                      | A010716
   4 | A372267, A372268                | A382103, A382104
   5 | A372269, A372270                | A382105, A382106
   6 | A372271, A372272, A372273       | A382107, A382686, A382687
   7 | A372274, A372275, this sequence | A382688, A382689, A382690

First[RealDigits[Root[LegendreP[7, #] &, 7], 10, 100]] (* Paolo Xausa, Feb 27 2025 *)

solve (x = 0.8, 1.0, 429*x^6 - 693*x^4 + 315*x^ - 35) \\ A.H.M. Smeets, May 31 2025

Largest positive root of 429*x^6 - 693*x^4 + 315*x^2 - 35 = 0.

A245926 Expansion of g.f. sqrt( (1-x + sqrt(1-14x+x^2)) / (2(1-14*x+x^2)) ).

Original entry on oeis.org

1, 5, 51, 587, 7123, 89055, 1135005, 14660805, 191253843, 2513963567, 33244446601, 441772827105, 5894323986301, 78912561223553, 1059543126891027, 14261959492731387, 192392702881384275, 2600355510685245087, 35206018016510388345, 477377227987055971905

Offset: 0

Paul D. Hanna, Aug 15 2014

Square each term to form a bisection of A245925.

Limit a(n+1)/a(n) = 7 + 4*sqrt(3).

			G.f.: A(x) = 1 + 5*x + 51*x^2 + 587*x^3 + 7123*x^4 + 89055*x^5 +...
where
A(x)^2 = (1-x + sqrt(1-14*x+x^2)) / (2*(1-14*x+x^2)).
Explicitly,
A(x)^2 = 1 + 10*x + 127*x^2 + 1684*x^3 + 22717*x^4 + 309214*x^5 + 4231675*x^6 + 58117672*x^7 + 800173945*x^8 +...+ A245923(n)*x^n +...

G. C. Greubel, Table of n, a(n) for n = 0..870

Column k=3 of A337389.

Cf. A245925, A245927, A245923, A243946, A337422.

A245926 := n -> sqrt(add(binomial(4*n-2*k, 2*n-k)*binomial(4*n-k, k)^2, k=0..2*n)); seq(A245926(n), n=0..20); # Peter Luschny, Aug 17 2014

CoefficientList[Series[Sqrt[(1 - x + Sqrt[1 - 14*x + x^2])/(2*(1 - 14*x + x^2))], {x,0,50}], x] (* G. C. Greubel, Jan 29 2017 *)
a[n_] := (-1)^n Hypergeometric2F1[-n, n + 1/2, 1, 4];
Table[a[n], {n, 0, 19}] (* Peter Luschny, Mar 16 2018 *)

/* From definition: */
{a(n)=polcoeff( sqrt( (1-x + sqrt(1-14*x+x^2 +x*O(x^n))) / (2*(1-14*x+x^2 +x*O(x^n))) ), n)}
for(n=0, 20, print1(a(n), ", "))

/* From formula for a(n)^2: */
{a(n)=sqrtint(sum(k=0, 2*n, sum(j=0, 4*n-2*k, (-1)^(j+k)*binomial(4*n-k,j+k)^2*binomial(j+k, k)^2)))}
for(n=0, 20, print1(a(n), ", "))

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

a(n)^2 = Sum_{k=0..2*n} Sum_{j=0..4*n-2*k} (-1)^(j+k) * C(4*n-k,j+k)^2 * C(j+k,k)^2.
a(n) ~ (3*sqrt(3)-5) * (7+4*sqrt(3))^(n+1) / (4*sqrt(Pi*n)). - Vaclav Kotesovec, Aug 16 2014
a(n)^2 = C(4*n,2*n)*hyper4F3([-2*n,-2*n,-2*n,-2*n+1/2],[1,-4*n,-4*n],4). - Peter Luschny, Aug 17 2014
a(n)^2 = Sum_{k=0..2*n} binomial(4*n-2*k, 2*n-k)*binomial(4*n-k, k)^2. - Peter Luschny, Aug 17 2014
a(n)^2 = Sum_{k=0..2*n} (-1)^k * C(2*k, k)^2 * C(2*n+k, 2*n-k). - Paul D. Hanna, Aug 17 2014
From Peter Bala, Mar 14 2018: (Start)
a(n) = (-1)^n*P(2*n,sqrt(-3)), where P(n,x) denotes the n-th Legendre polynomial. See A008316.
a(n) = (1/C(2*n,n))*Sum_{k = 0..n} (-1)^(n+k)*C(n,k)*C(n+k,k)* C(2*n+2*k,n+k) = Sum_{k = 0..n} (-1)^(n+k)*C(2*k,k)*C(n,k) *C(2*n+2*k,2*n)/C(n+k,n). 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 * u^(n-k), where u = (1 - sqrt(-3))/2 is a primitive sixth root of unity.
a(n) = (-1)^n*Sum_{k = 0..2*n} C(2*n,k)*C(2*n+k,k)*u^(2*k).
(End)
a(n) = (-1)^n*hypergeom([-n, n + 1/2], [1], 4). - Peter Luschny, Mar 16 2018
a(0) = 1, a(1) = 5 and n * (2*n-1) * (4*n-5) * a(n) = (4*n-3) * (28*n^2-42*n+9) * a(n-1) - (n-1) * (2*n-3) * (4*n-1) * a(n-2) for n > 1. - Seiichi Manyama, Aug 28 2020
From Peter Bala, May 03 2022: (Start)
Conjecture: a(n) = [x^n] ( (1 + x + x^2)*(1 + x)^2/(1 - x)^2 )^n. Equivalently, a(n) = Sum_{k = 0..n} Sum_{i = 0..k} Sum_{j = 0..n-k-i} C(n, k)*C(k, i)*C(2*n, j)*C(3*n-k-i-j-1, n-k-i-j).
If the conjecture is true then the Gauss congruences a(n*p^k) == a(n*p^(k-1)) (mod p^k) hold for all primes p and positive integers n and k. Calculation suggests that the supercongruences a(n*p^k) == a(n*p^(k-1)) (mod p^(2*k)) hold for primes p >= 5 and positive integers n and k. (End)

A253283 Triangle read by rows: coefficients of the partial fraction decomposition of [d^n/dx^n] (x/(1-x))^n/n!.

Original entry on oeis.org

1, 0, 1, 0, 2, 3, 0, 3, 12, 10, 0, 4, 30, 60, 35, 0, 5, 60, 210, 280, 126, 0, 6, 105, 560, 1260, 1260, 462, 0, 7, 168, 1260, 4200, 6930, 5544, 1716, 0, 8, 252, 2520, 11550, 27720, 36036, 24024, 6435, 0, 9, 360, 4620, 27720, 90090, 168168, 180180, 102960, 24310

Offset: 0

Peter Luschny, Mar 20 2015

The rows give (up to sign) the coefficients in the expansion of the integer-valued polynomial (x+1)^2*(x+2)^2*(x+3)^2*...*(x+n)^2*(x+n+1) / (n!*(n+1)!) in the basis made of the binomial(x+i,i). - F. Chapoton, Oct 31 2022

This is related to the cluster fans of type B (see Fomin and Zelevinsky reference) - F. Chapoton, Nov 17 2022.

			[1]
[0, 1]
[0, 2,   3]
[0, 3,  12,   10]
[0, 4,  30,   60,   35]
[0, 5,  60,  210,  280,  126]
[0, 6, 105,  560, 1260, 1260,  462]
[0, 7, 168, 1260, 4200, 6930, 5544, 1716]
.
R_0(x) = 1/(x-1)^0.
R_1(x) = 0/(x-1)^1 + 1/(x-1)^2.
R_2(x) = 0/(x-1)^2 + 2/(x-1)^3 + 3/(x-1)^4.
R_3(x) = 0/(x-1)^3 + 3/(x-1)^4 + 12/(x-1)^5 + 10/(x-1)^6.
Then k!*[x^k] R_n(x) is A001286(k+2) and A001754(k+3) for n = 2, 3 respectively.
.
Seen as an array A(n, k) = binomial(n + k, k)*binomial(n + 2*k - 1, n + k):
[0] 1, 1,   3,   10,    35,    126,     462, ...
[1] 0, 2,  12,   60,   280,   1260,    5544, ...
[2] 0, 3,  30,  210,  1260,   6930,   36036, ...
[3] 0, 4,  60,  560,  4200,  27720,  168168, ...
[4] 0, 5, 105, 1260, 11550,  90090,  630630, ...
[5] 0, 6, 168, 2520, 27720, 252252, 2018016, ...
[6] 0, 7, 252, 4620, 60060, 630630, 5717712, ...

Seiichi Manyama, Rows n = 0..139, flattened
F. Chapoton Enumerative Properties of Generalized Associahedra, Sém. Loth. Comb. B51b (2004).
Mark Dukes and Chris D. White, Web Matrices: Structural Properties and Generating Combinatorial Identities, arXiv:1603.01589 [math.CO], 2016.
Mark Dukes and Chris D. White, Web Matrices: Structural Properties and Generating Combinatorial Identities, Electronic Journal Of Combinatorics, 23(1) (2016), #P1.45.
S. Fomin and A. Zelevinsky Y-systems and generalized associahedra, Ann. of Math. (2) 158 (2003), no. 3.
Yunlong Shen and Lixin Shen, Orthogonal Fourier-Mellin moments for invariant pattern recognition, J. Opt. Soc. Am. A 11 (6) (1994) p 1748-1757, eq. (6).

T(n, n) = C(2*n-1, n) = A001700(n-1).
T(n, n-1) = A005430(n-1) for n >= 1.
T(n, n-2) = A051133(n-2) for n >= 2.
T(n, 2) = A027480(n-1) for n >= 2.
T(2*n, n) = A208881(n) for n >= 0.
A002002 (row sums).
Cf. A000142, A001286, A001754, A001755, A001777, A182626, A266732 (row k=3), A008316, A033282, A063007, A345013, A269944.

T_row := proc(n) local egf, k, F, t;
if n=0 then RETURN(1) fi;
egf := (x/(1-x))^n/n!; t := diff(egf,[x$n]);
F := convert(t,parfrac,x);
# print(seq(k!*coeff(series(F,x,20),x,k),k=0..7));
# gives A000142, A001286, A001754, A001755, A001777, ...
seq(coeff(F,(x-1)^(-k)),k=n..2*n) end:
seq(print(T_row(n)),n=0..7);
# 2nd version by R. J. Mathar, Dec 18 2016:
A253283 := proc(n,k)
    binomial(n,k)*binomial(n+k-1,k-1) ;
end proc:

Table[Binomial[n, k] Binomial[n + k - 1, k - 1], {n, 0, 9}, {k, 0, n}] // Flatten (* Michael De Vlieger, Feb 22 2017 *)

T(n,k) = binomial(n,k)*binomial(n+k-1,k-1);
tabl(nn) = for(n=0, nn, for (k=0, n, print1(T(n,k), ", ")); print); \\ Michel Marcus, Apr 29 2018

The exponential generating functions for the rows of the square array L(n,k) = ((n+k)!/n!)*C(n+k-1,n-1) (associated to the unsigned Lah numbers) are given by R_n(x) = Sum_{k=0..n} T(n,k)/(x-1)^(n+k).
T(n,k) = C(n,k)*C(n+k-1,k-1).
Sum_{k=0..n} T(n,k) = (-1)^n*hypergeom([-n,n],[1],2) = (-1)^n*A182626(n).
Row generating function: Sum_{k>=1} T(n,k)*z^k = z*n* 2F1(1-n,n+1 ; 2; -z). - R. J. Mathar, Dec 18 2016
From Peter Bala, Feb 22 2017: (Start)
G.f.: (1/2)*( 1 + (1 - t)/sqrt(1 - 2*(2*x + 1)*t + t^2) ) = 1 + x*t + (2*x + 3*x^2)*t^2 + (3*x + 12*x^2 + 10*x^3)*t^3 + ....
n-th row polynomial R(n,x) = (1/2)*(LegendreP(n, 2*x + 1) - LegendreP(n-1, 2*x + 1)) for n >= 1.
The row polynomials are the black diamond product of the polynomials x^n and x^(n+1) (see Dukes and White 2016 for the definition of this product).
exp(Sum_{n >= 1} R(n,x)*t^n/n) = 1 + x*t + x*(1 + 2*x)*t^2 + x*(1 + 5*x + 5*x^2)*t^3 + ... is a g.f. for A033282, but with a different offset.
The polynomials P(n,x) := (-1)^n/n!*x^(2*n)*(d/dx)^n(1 + 1/x)^n begin 1, 3 + 2*x , 10 + 12*x + 3*x^2, ... and are the row polynomials for the row reverse of this triangle. (End)
Let Q(n, x) = Sum_{j=0..n} (-1)^(n - j)*A269944(n, j)*x^(2*j - 1) and P(x, y) = (LegendreP(x, 2*y + 1) - LegendreP(x-1, 2*y + 1)) / 2 (see Peter Bala above). Then n!*(n - 1)!*[y^n] P(x, y) = Q(n, x) for n >= 1. - Peter Luschny, Oct 31 2022
From Peter Bala, Apr 18 2024: (Start)
G.f.: Sum_{n >= 0} binomial(2*n-1, n)*(x*t)^n/(1 - t)^(2*n) = 1 + x*t + (2*x + 3*x^2)*t^2 + (3*x + 12*x^2 + 10*x^3)*t^3 + ....
n-th row polynomial R(n, x) = [t^n] ( (1 - t)/(1 - (1 + x)*t) )^n.
It follows that for integer x, the sequence {R(n, x) : n >= 0} satisfies the Gauss congruences: R(n*p^r, x) == R(n*p^(r-1), x) (mod p^r) for all primes p and positive integers n and r.
R(n, -2) = (-1)^n * A002003(n) for n >= 1.
R(n, 3) = A299507(n). (End)

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)

A001801 Coefficients of Legendre polynomials.

Original entry on oeis.org

3, 15, 105, 315, 6930, 18018, 90090, 218790, 2078505, 4849845, 22309287, 50702925, 1825305300, 4071834900, 18032411700, 39671305740, 347123925225, 755505013725, 3273855059475, 7064634602025, 121511715154830, 260382246760350, 1112542327066950, 2370198870707850, 20146690401016725

Offset: 0

N. J. A. Sloane

nonn

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 798.
G. Prévost, Tables de Fonctions Sphériques. Gauthier-Villars, Paris, 1933, pp. 156-157.
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).

G. C. Greubel, Table of n, a(n) for n = 0..1000
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].
Milan Janjic, Some classes of numbers and derivatives, JIS 12 (2009) 09.8.3.

Cf. A001790, A001796, A008316, A011371.
Bisection of A004733.
Diagonal 3 of triangle A100258.

A001801:= func< n | 3*Binomial(n+3,3)*Catalan(n+2)*2^(Valuation(Factorial(n+4),2)-n-4) >;
[A001801(n): n in [0..30]]; // G. C. Greubel, Apr 26 2025

A001801[n_]:= 3*2^(2*n+1)*Binomial[n+3/2, n]/2^DigitCount[n+4,2,1];
Table[A001801[n], {n,0,40}] (* G. C. Greubel, Apr 26 2025 *)

a(n)=if(n<0,0,polcoeff(pollegendre(n+4),n)*2^valuation((n\2*2+4)!,2))

def A001801(n): return 3*2^(n-3)*binomial(n+3/2,n)*2^valuation(factorial(n+4), 2)
print([A001801(n) for n in range(31)]) # G. C. Greubel, Apr 26 2025

a(n) = 3*2^(n-3)*binomial(n + 3/2, n)*2^A011371(n+4). - G. C. Greubel, Apr 26 2025

More terms from Michael Somos, Oct 25 2002

A001800 Coefficients of Legendre polynomials.

Original entry on oeis.org

1, 3, 30, 70, 315, 693, 12012, 25740, 109395, 230945, 1939938, 4056234, 16900975, 35102025, 1163381400, 2404321560, 9917826435, 20419054425, 167890003050, 344616322050, 1412926920405, 2893136075115, 47342226683700, 96742811049300, 395033145117975

Offset: 0

N. J. A. Sloane

nonn

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 798.
G. Prévost, Tables de Fonctions Sphériques. Gauthier-Villars, Paris, 1933, pp. 156-157.
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).

Alois P. Heinz, Table of n, a(n) for n = 0..500
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].
Eric Weisstein's World of Mathematics, Legendre Polynomial, eq. 28.

Cf. A001790, A001796, A001803, A008316.

Diagonal 2 of triangle A100258.

A001800:= func< n | (n+1)*(n+2)*Catalan(n+1)/2^(&+Intseq(n+2, 2)) >;
[A001800(n): n in [0..30]]; // G. C. Greubel, Apr 25 2025

wt:= proc(n) local m, r; m:=n; r:=0;
       while m>0 do r:= r+irem(m, 2, 'm') od; r
     end:
a:= n-> (n+1) *binomial(2*n+2, n+1)/2^wt(n+2):
seq(a(n), n=0..30);  # Alois P. Heinz, May 29 2013

a[n_] := (n+1)*Binomial[2*n+2, n+1]/2^DigitCount[n+2, 2, 1]; Table[a[n], {n, 0, 24}] (* Jean-François Alcover, Mar 13 2014 *)

a(n)=if(n<0,0,-polcoeff(pollegendre(n+2),n)*2^valuation((n\2*2)!,2))

def A001800(n): return (n+1)*binomial(2*n+2,n+1)//2^sum((n+2).digits(2))
print([A001800(n) for n in range(31)]) # G. C. Greubel, Apr 25 2025

a(n) = (n+1) * C(2n+2, n+1) / 2^A000120(n+2).

More terms from Michael Somos, Oct 25 2002

A349077 a(n) = 4^n * P(2*n, n), where P(n, x) is n-th Legendre polynomial.

Original entry on oeis.org

1, 4, 886, 575296, 748553926, 1638884021248, 5430931463592636, 25386301852394340352, 159203574262026117932614, 1290247693627696897075707904, 13126820230906199855332092508756, 163819123650250694146607819756929024, 2460884002303138397686849151579559249436

Offset: 0

Vaclav Kotesovec, Nov 07 2021

nonn

Vaclav Kotesovec, Table of n, a(n) for n = 0..175
Eric Weisstein's World of Mathematics, Legendre Polynomial.
Wikipedia, Legendre polynomials.

Cf. A008316, A110129.

Table[4^n*LegendreP[2*n, n], {n, 0, 15}]

a(n) = 4^n*pollegendre(2*n, n); \\ Michel Marcus, Nov 08 2021

a(n) ~ 2^(4*n - 1/2) * n^(2*n - 1/2) / sqrt(Pi).

A121448 Triangle read by rows: T(n,k) is the number of binary trees with n edges and having k vertices of outdegree 1 (n>=0, k>=0). A binary tree is a rooted tree in which each vertex has at most two children and each child of a vertex is designated as its left or right child.

Original entry on oeis.org

1, 0, 2, 1, 0, 4, 0, 6, 0, 8, 2, 0, 24, 0, 16, 0, 20, 0, 80, 0, 32, 5, 0, 120, 0, 240, 0, 64, 0, 70, 0, 560, 0, 672, 0, 128, 14, 0, 560, 0, 2240, 0, 1792, 0, 256, 0, 252, 0, 3360, 0, 8064, 0, 4608, 0, 512, 42, 0, 2520, 0, 16800, 0, 26880, 0, 11520, 0, 1024, 0, 924, 0, 18480, 0

Offset: 0

Emeric Deutsch, Jul 31 2006

T(2n,0) = binomial(2n,n)/(n+1) (the Catalan numbers; A000108); T(2n+1,0)=0. T(n,n)=2^n (A000079). Sum(k*T(n,k),k=0..n)=2*binomial(2n,n-1)=2*A001791(n). After deleting the zeros, reflection of A091894.
From Tom Copeland, Feb 07 2016: (Start)
A shifted o.g.f. is OG(x,t) = [1 - 2tx - sqrt[(1-2tx)^2-4x^2]] / (2x) = x + 2t x^2 + (1+4t^2) x^3 + ... with compositional inverse OGinv(x,t) =  x / (1 + 2tx + x^2), the shifted o.g.f. for A053117 (mod signs).
For x > 0 and choosing the positive square root, OG(x^2,t) = H(x,t) = x^2 + 2t x^4 + (1+4t^2) x^6 + ... has the compositional inverse Hinv(x,t) = sqrt[x / (1 + 2tx + x^2)] , which satisfies Hinv(H(x, t), t) = x, and which is the generating function for the Legendre polynomials (mod signs, cf. A008316) times sqrt(x).
In general, GB(x,t,b) =  [x / (1 - 2tx + x^2)]^b is a generator for the Gegenbauer polynomials times x^b for positive roots with compositional inverse about the origin GBinv(x,t,b) = OG(x^(1/b),-t) for x>0. Cf. A097610.
(End)
From Tom Copeland, Feb 09 2016: (Start)
z1 = OG(x,t) is the zero that vanishes for x=0 for the quadratic polynomial Q(z;z1(x,t),z2(x,t)) =(z-z1)(z-z2) = z^2 - (z1+z2) z + (z1*z2) = z^2 - e1 z + e2 = z^2 - [(1-2tx)/x] z + 1, where e1 and e2 are the elementary symmetric polynomials for two indeterminates.
The other zero is given by z2(x,t) = [1 - 2tx + sqrt[(1-2tx)^2-4x^2]] / (2x) = (1 - 2tx)/x - z1(x,t).
The two are zeros of the elliptic curve in Legendre normal form y^2 = z (z-z1)(z-z2). (Added Feb 13 2016. See Landweber et al., p 14. Cf. A097610.)
(End)

			T(2,2)=4 because, denoting by L (R) an edge going from a vertex to a left (right) child, we have the paths: LL, LR, RL and RR.
Triangle starts:
  1;
  0,2;
  1,0,4;
  0,6,0,8;
  2,0,24,0,16;

Colin Defant, Postorder Preimages, arXiv preprint arXiv:1604.01723 [math.CO], 2016.
FindStat - Combinatorial Statistic Finder, The number of vertices with out-degree 1 in a binary tree.
P. Landweber, D. Ravenel, and R. Stong, Periodic cohomology theories defined by elliptic curves

Cf. A000108, A000079, A001791, A091894.
Cf. A038207, A053117, A097610, A126120, A145271, A180874, A218272.
Cf. A008316.

T:=proc(n,k) if n-k mod 2 = 0 then 2^k*binomial(n+1,k)*binomial(n+1-k,(n-k)/2)/(n+1) else 0 fi end: for n from 0 to 12 do seq(T(n,k),k=0..n) od; # yields sequence in triangular form

nn=10;Drop[CoefficientList[Series[(1-2x y - ((-4x^2+(1-2x y)^2))^(1/2))/(2 x),{x,0,nn}],{x,y}],1]//Grid  (* Geoffrey Critzer, Feb 20 2013 *)

T(n,k) = 2^k*binomial(n+1,k)binomial(n+1-k,(n-k)/2)/(n+1) if n-k is even; otherwise, T(n,k) = 0. G.f. G=G(t,z) satisfies G=1+2tzG+z^2*G^2.
T(n,k) = 2^k*A097610(n,k). - Philippe Deléham, Aug 17 2006
From Tom Copeland, Feb 09 2016: (Start)
The following is from the formalism in A097610 with h1 = 2t, h2 = 1, and MT(n,h1,h2) = MT(n,2t,1) and with OG(x,t) defined above.
E.g.f.: M(x,t) = e^(2tx) AC(x) = exp[x MT(.,2t,1)] = exp[x P(.,t)], where AC(x) = I_1(2x)/x = Sum_{n>=0} x^(2n)/(n!(n+1)!) = exp(c.x) is the e.g.f. of A126120.
P(n,t) = MT(n,2t,1) = (c. + 2t)^n = Sum_{k=0..n} binomial(n,k) c(n-k) (2t)^k with c(k) = A126120(k). P(n,t+s) = (c. + 2t + 2s)^n = (P(.,t) + 2s)^n.
P(n,t) = t^n FC(n,c./t) = t^n (2 + c./t)^n, where FC(n,t) = (2 + t)^n are the face polynomials (vectors) of the hypercubes of A038207, i.e., the row polynomials of this entry can be obtained as the umbral composition of the reverse face polynomials with the aerated Catalan numbers of A000108.
The lowering and raising operators for the row polynomials P(n,t) of this entry are L = (1/2) d/dt = (1/2) D and R = 2t + dlog{AC(L)}/dL = 2t + Sum_{n>=0} b(n) L^(2n+1)/(2n+1)! = 2t + L - L^3/3! + 5 L^5/5! - ...  with b(n) = (-1)^n A180874(n+1).
Let CP(n,t) = P(n+1,t) with CP(0,t) = 0. Then the infinitesimal generator for CP(n,t) is g(x) d/dx with g(x) = 1 /[dOGinv(x,t)/dx] = x^2 / [(OGinv(x,t))^2 (1 - x^2)] = (1 + 2t x + x^2)^2 / (1 - x^2) so that [g(x)d/dx]^n/n! x evaluated at x = 0 gives the row polynomial CP(n,t), i.e., exp[x g(u)d/du] u |_(u=0) = OG(x,t) = 1 /[1 - x P(.,t)]. Cf. A145271.
g(x) = 1 + 4t x + (3+4t) x^2 + 8t x^3 + 4(1+t^2) x^4 + 8t x^5 + 4(1+t^2) x^6 + 8t x^7 + ... has the repeating coefficients of the vector V = (1, 4t, 3+4t, 8t, 4(1+t^2), 8t, 4(1+t^2), 8t, ...). Form the lower triangular matrix U with all ones on the diagonal and below. Multiply the n-th diagonal of U by V(n), giving the matrix VU with VU(n,k) = V(n-k). Then (1,0,0,0,..) [VU * DM]^n/n! (0,1,0,0,..)^T = CP(n,t) = P(n-1,t) for n>0 with DM being the matrix A218272 representing differentiation of a power series.
(End)

cp's OEIS Frontend

A372275 Decimal expansion of the middle positive zero of the Legendre polynomial of degree 7.

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A372276 Decimal expansion of the largest positive zero of the Legendre polynomial of degree 7.

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A245926 Expansion of g.f. sqrt( (1-x + sqrt(1-14*x+x^2)) / (2*(1-14*x+x^2)) ).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

PARI

PARI

PARI

Formula

A253283 Triangle read by rows: coefficients of the partial fraction decomposition of [d^n/dx^n] (x/(1-x))^n/n!.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

PARI

Formula

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

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

PARI

PARI

PARI

Python

Formula

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

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

PARI

PARI

A245926 Expansion of g.f. sqrt( (1-x + sqrt(1-14x+x^2)) / (2(1-14*x+x^2)) ).

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)) ).