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URL: https://oeis.org/A084768

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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.
14
1, 7, 73, 847, 10321, 129367, 1651609, 21360031, 278905249, 3668760487, 48543499753, 645382441711, 8614382884849, 115367108888311, 1549456900170553, 20861640747345727, 281483386791966529, 3805228005705102151, 51527535767904810889, 698796718936034430607
OFFSET
0,2
COMMENTS
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
LINKS
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.
FORMULA
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
MATHEMATICA
Table[LegendreP[n, 7], {n, 0, 20}] (* Vaclav Kotesovec, Jul 31 2013 *)
PROG
(PARI) for(n=0, 30, print1(subst(pollegendre(n), x, 7)", "))
(PARI) {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), ", "))
(PARI) /* 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), ", "))
(Magma) [Evaluate(LegendrePolynomial(n), 7): n in [0..40]]; // G. C. Greubel, May 17 2023
(SageMath) [gen_legendre_P(n, 0, 7) for n in range(41)] # G. C. Greubel, May 17 2023
CROSSREFS
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).
Sequence in context: A071060 A092444 A099141 * A357165 A357226 A384826
KEYWORD
nonn,easy
AUTHOR
Paul D. Hanna, Jun 03 2003
STATUS
approved