A335333 -id:A335333 - cp's OEIS frontend

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

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

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

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)

A335338 P_5(2n+1), the Legendre polynomial of order 5 at 2n+1.

Original entry on oeis.org

1, 1683, 23525, 129367, 458649, 1256651, 2904733, 5950575, 11138417, 19439299, 32081301, 50579783, 76767625, 112825467, 161311949, 225193951, 307876833, 413234675, 545640517, 709996599, 911764601, 1156995883, 1452361725, 1805183567, 2223463249, 2715913251, 3291986933

Offset: 0

Seiichi Manyama, Jun 02 2020

Eric Weisstein's World of Mathematics, Legendre Polynomial.
Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

Row 5 of A335333.

Cf. A160737.

a[n_] := LegendreP[5, 2*n + 1]; Array[a, 27, 0] (* Amiram Eldar, May 03 2021 *)

PARI
```
a(n) = pollegendre(5, 2*n+1)
```

a(n) = 252*n^5+630*n^4+560*n^3+210*n^2+30*n+1

N=40; x='x+O('x^N); Vec((1+x)*(1+1676*x+11766*x^2+1676*x^3+x^4)/(1-x)^6)

a(n) = A160737(2*n+1)/4.
a(n) = 252*n^5 + 630*n^4 + 560*n^3 + 210*n^2 + 30*n + 1 = (2*n + 1) * (126*n^4 + 252*n^3 + 154*n^2 + 28*n + 1).
G.f.: (1+x)*(1+1676*x+11766*x^2+1676*x^3+x^4)/(1-x)^6.

cp's OEIS Frontend

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

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

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A335338 P_5(2n+1), the Legendre polynomial of order 5 at 2n+1.

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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.