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A054765 a(n+2) = (2n+3)*a(n+1) + (n+1)^2*a(n), a(0) = 0, a(1) = 1.

%I A054765 #30 Jul 02 2025 16:01:59
%S A054765 0,1,3,19,160,1744,23184,364176,6598656,135484416,3108695040,
%T A054765 78831037440,2189265960960,66083318415360,2154235544616960,
%U A054765 75425161203302400,2822882994841190400,112463980097804697600
%N A054765 a(n+2) = (2n+3)*a(n+1) + (n+1)^2*a(n), a(0) = 0, a(1) = 1.
%C A054765 The denominators of the convergents of [1/3, 4/5, 9/7, 16/9, ...]. To produce Pi the above continued fraction is used. It is formed by n^2/(2*n+1) which starts at n=1. Most numerators of continued fractions are 1 & thus are left out of the brackets. In the case of Pi they vary. Therefore here both numerators & denominators are given. The first 4 convergents are 1/3,5/19,44/160,476/1744. The value of this continued fraction is .273239... . 4*INV(1+.273239...) is Pi. - Al Hakanson (hawkuu(AT)gmail.com), Dec 01 2008
%C A054765 Starting with offset 1 = row sums of triangle A155729. [_Gary W. Adamson_ & _Alexander R. Povolotsky_, Jan 25 2009]
%H A054765 G. C. Greubel, <a href="/A054765/b054765.txt">Table of n, a(n) for n = 0..390</a>
%H A054765 K. S. Brown, <a href="https://www.mathpages.com/home/kmath381/kmath381.htm">Integer Sequences Related To Pi</a>
%F A054765 a(n) ~ Pi * (1+sqrt(2))^(n + 1/2) * n^n / (2^(9/4) * exp(n)). - _Vaclav Kotesovec_, Feb 18 2017
%p A054765 A054765 := proc(n)
%p A054765     option remember;
%p A054765     if n <= 1 then
%p A054765         n;
%p A054765     else
%p A054765         (2*n-1)*procname(n-1)+(n-1)^2*procname(n-2) ;
%p A054765     end if;
%p A054765 end proc: # _R. J. Mathar_, Jul 13 2013
%t A054765 RecurrenceTable[{a[n + 2] == (2*n + 3)*a[n + 1] + (n + 1)^2*a[n],
%t A054765 a[0] == 0, a[1] == 1}, a, {n,0,50}] (* _G. C. Greubel_, Feb 18 2017 *)
%Y A054765 Cf. A155729, A012244, A054766.
%K A054765 nonn,easy
%O A054765 0,3
%A A054765 _N. J. A. Sloane_, May 26 2000
%E A054765 More terms from _James Sellers_, May 27 2000