OFFSET
0,2
COMMENTS
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
LINKS
T. D. Noe, 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.
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
FORMULA
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)
MAPLE
seq(orthopoly[P](n, 5), n = 0 .. 20); # Robert Israel, Aug 18 2014
MATHEMATICA
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 *)
PROG
(PARI) a(n)=subst(pollegendre(n), x, 5)
(PARI) /* 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 */
(PARI) {a(n)=sum(k=0, n, binomial(n, k)^2*2^k*3^(n-k))} /* Paul D. Hanna */
(PARI) {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
(Magma) [Evaluate(LegendrePolynomial(n), 5): n in [0..40]]; // G. C. Greubel, May 21 2023
(SageMath) [gen_legendre_P(n, 0, 5) for n in range(41)] # G. C. Greubel, May 21 2023
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
STATUS
approved