login
a(n+2) = (2*n + 3)*a(n+1) + (n + 1)^2*a(n), a(0) = 1, a(1) = 0.
5

%I #27 Aug 15 2020 15:50:01

%S 1,0,1,5,44,476,6336,99504,1803024,37019664,849418560,21539756160,

%T 598194037440,18056575823040,588622339549440,20609136708249600,

%U 771323264354361600,30729606721005830400,1298448658633614566400

%N a(n+2) = (2*n + 3)*a(n+1) + (n + 1)^2*a(n), a(0) = 1, a(1) = 0.

%C Numerators of the convergents of the generalized continued fraction expansion 4/Pi - 1 = [0; 1/3, 4/5, 9/7,..., n^2/(2*n + 1),...] = 1/(3 + 4/(5 + 9/(7 + ...))). The first 4 convergents are 1/3, 5/19, 44/160 and 476/1744.

%H Indranil Ghosh, <a href="/A054766/b054766.txt">Table of n, a(n) for n = 0..392</a>

%H K. S. Brown, <a href="https://www.mathpages.com/home/kmath381/kmath381.htm">Integer Sequences Related To Pi</a>

%F a(n) ~ (1 - Pi/4) * (1 + sqrt(2))^(n + 1/2) * n^n / (2^(1/4) * exp(n)). - _Vaclav Kotesovec_, Feb 18 2017

%t RecurrenceTable[{a[n+2] == (2*n+3)*a[n+1] + (n+1)^2*a[n], a[0] == 1, a[1] == 0}, a, {n, 0, 25}] (* _Vaclav Kotesovec_, Feb 18 2017 *)

%t t={1,0};Do[AppendTo[t,(2(n-2)+3)*t[[-1]]+(n-1)^2*t[[-2]]],{n,2,18}];t (* _Indranil Ghosh_, Feb 25 2017 *)

%Y Cf. A012244, A054765.

%K nonn,easy,frac

%O 0,4

%A _N. J. A. Sloane_, May 26 2000

%E More terms from _James A. Sellers_, May 27 2000

%E Definition expanded by Al Hakanson (hawkuu(AT)gmail.com), Dec 01 2008

%E Keyword frac added by _Michel Marcus_, Feb 25 2017