A025175 - OEIS

%I #37 Mar 18 2017 23:52:31

%S 1,4,3,-40,-190,-168,2023,10096,9486,-110440,-564322,-547248,6266884,

%T 32468464,32101935,-364054048,-1903389802,-1906695144,21484821178,

%U 113055206800,114325154076,-1282403513776

%N Jacobi polynomial P((1, 1), n, (1/2)).

%H Vincenzo Librandi, <a href="/A025175/b025175.txt">Table of n, a(n) for n = 0..300</a>

%F 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

%F a(n) ~ 8*sin(Pi*n/3+Pi/4) / (3^(3/4)*sqrt(Pi*n)) * 4^n. - _Vaclav Kotesovec_, Jul 30 2013

%F From _Vladimir Reshetnikov_, Nov 01 2015: (Start)

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

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

%F (End)

%t Table[ 2^(2n) JacobiP[ n, 1, 1, 1/2 ], {n, 0, 24} ]

%t Table[4^(n+1) (LegendreP[n+1, 1/2] - 2 LegendreP[n+2, 1/2])/3, {n, 0, 20}] (* _Vladimir Reshetnikov_, Nov 01 2015 *)

%t 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 *)

%o (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

%Y Cf. A012125.

%K sign

%O 0,2

%A _Wouter Meeussen_

%E Sequence name by _Wouter Meeussen_, Mar 03 2013