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A174506 Continued fraction expansion for exp( Sum_{n>=1} 1/(n*A014448(n)) ), where A014448(n) = (2+sqrt(5))^n + (2-sqrt(5))^n.

%I A174506 #17 Mar 25 2024 06:39:07
%S A174506 1,3,17,1,75,321,1,1363,5777,1,24475,103681,1,439203,1860497,1,
%T A174506 7881195,33385281,1,141422323,599074577,1,2537720635,10749957121,1,
%U A174506 45537549123,192900153617,1,817138163595,3461452808001,1
%N A174506 Continued fraction expansion for exp( Sum_{n>=1} 1/(n*A014448(n)) ), where A014448(n) = (2+sqrt(5))^n + (2-sqrt(5))^n.
%H A174506 Peter Bala, <a href="/A174500/a174500_2.pdf">Some simple continued fraction expansions for an infinite product, Part 1</a>.
%H A174506 Peter Bala, <a href="/A174504/a174504.pdf">Some simple continued fraction expansions for an infinite product, Part 2</a>.
%H A174506 <a href="/index/Rec#order_09">Index entries for linear recurrences with constant coefficients</a>, signature (0,0,19,0,0,-19,0,0,1).
%F A174506 a(3n-3) = 1, a(3n-2) = A014448(2n-1) - 1, a(3n-1) = A014448(2n) - 1, for n>=1 [conjecture].
%F A174506 a(n) = 19*a(n-3)-19*a(n-6)+a(n-9). G.f.: -(x^2 -x +1)*(x^6 -4*x^5 -4*x^4 -2*x^3 +20*x^2 +4*x +1) / ((x -1)*(x^2 -3*x +1)*(x^2 +x +1)*(x^4 +3*x^3 +8*x^2 +3*x +1)). [_Colin Barker_, Jan 20 2013]
%F A174506 From _Peter Bala_, Jan 25 2013: (Start)
%F A174506 The above conjectures are correct. The real number exp( Sum {n>=1} 1/(n*A014448(n)) ) is equal to the infinite product F(x) := product {n >= 0} (1 + x^(4*n+3))/(1 - x^(4*n+1)) evaluated at x = sqrt(5) - 2. Ramanujan has given a continued fraction expansion for the product F(x). Using this we can find the simple continued fraction expansion of the numbers F(1/2*(sqrt(N^2 + 4) - N)), N a positive integer. The present case is when N = 4. See the Bala link for details.
%F A174506 The theory also provides the simple continued fraction expansion of the numbers F({sqrt(5) - 2}^(2*k+1)), k = 1, 2, 3, ...: if [1; c(1), c(2), 1, c(3), c(4), 1, ...] denotes the present sequence then the simple continued fraction expansion of F({sqrt(5) - 2}^(2*k+1)) is given by [1; c(2*k+1), c(2*(2*k+1)), 1, c(3*(2*k+1)), c(4*(2*k+1)), 1, ...].
%F A174506 (End)
%e A174506 Let L = Sum_{n>=1} 1/(n*A014448(n)) or, more explicitly,
%e A174506 L = 1/4 + 1/(2*18) + 1/(3*76) + 1/(4*322) + 1/(5*1364) +...
%e A174506 so that L = 0.2831229765066671850017990708479258794794782639219...
%e A174506 then exp(L) = 1.3272683746094012523448609429829013914921330866098...
%e A174506 equals the continued fraction given by this sequence:
%e A174506 exp(L) = [1;3,17,1,75,321,1,1363,5777,1,24475,103681,1,...]; i.e.,
%e A174506 exp(L) = 1 + 1/(3 + 1/(17 + 1/(1 + 1/(75 + 1/(321 + 1/(1 +...)))))).
%e A174506 Compare these partial quotients to A014448(n), n=1,2,3,...:
%e A174506 [4,18,76,322,1364,5778,24476,103682,439204,1860498,7881196,33385282,...].
%o A174506 (PARI) {a(n)=local(L=sum(m=1,2*n+1000,1./(m*round((2+sqrt(5))^m+(2-sqrt(5))^m))));contfrac(exp(L))[n]}
%Y A174506 Cf. A014448, A174500, A174505, A174507.
%K A174506 cofr,nonn,easy
%O A174506 0,2
%A A174506 _Paul D. Hanna_, Mar 21 2010