A025175 Jacobi polynomial P((1, 1), n, (1/2)).

1, 4, 3, -40, -190, -168, 2023, 10096, 9486, -110440, -564322, -547248, 6266884, 32468464, 32101935, -364054048, -1903389802, -1906695144, 21484821178, 113055206800, 114325154076, -1282403513776

Offset

0,2

Links

Vincenzo Librandi, Table of n, a(n) for n = 0..300

Formula

Conjecture: (n+2)*a(n) + 6*(-n-1)*a(n-1) + 12*(2*n+1)*a(n-2) + 32*(-n+1)*a(n-3) = 0. - R. J. Mathar, Mar 03 2013

a(n) ~ 8*sin(Pi*n/3+Pi/4) / (3^(3/4)*sqrt(Pi*n)) * 4^n. - Vaclav Kotesovec, Jul 30 2013

From Vladimir Reshetnikov, Nov 01 2015: (Start)

G.f.: 1/(6*x^2) + (2*x-1)/(6*x^2*sqrt(16*x^2-4*x+1)).

a(n) = 2*A012125(n+1)/(n+2).

(End)

Mathematica

Table[ 2^(2n) JacobiP[ n, 1, 1, 1/2 ], {n, 0, 24} ]
Table[4^(n+1) (LegendreP[n+1, 1/2] - 2 LegendreP[n+2, 1/2])/3, {n, 0, 20}] (* Vladimir Reshetnikov, Nov 01 2015 *)
RecurrenceTable[{(n+2)*a[n] == 6*(n+1)*a[n-1] - 12*(2*n+1)*a[n-2] + 32*(n-1)*a[n-3], a[0] == 1, a[1] == 4, a[2] == 3}, a, {n, 0, 200}] (* G. C. Greubel, Nov 01 2015 *)

Prog

(PARI) a(n)=4^(n+1)*(pollegendre(n+1, 1/2) - 2*pollegendre(n+2, 1/2))/3 \\ Charles R Greathouse IV, Mar 18 2017

Crossrefs

Cf. A012125.

Sequence in context: A120078 A096201 A275521 * A248247 A016504 A362275

Adjacent sequences: A025172 A025173 A025174 * A025176 A025177 A025178

Keyword

sign

Author

Wouter Meeussen

Extensions

Sequence name by Wouter Meeussen, Mar 03 2013

Status

approved