A174500 Continued fraction expansion for exp( Sum_{n>=1} 1/(n*A003500(n)) ), where A003500(n) = (2+sqrt(3))^n + (2-sqrt(3))^n.
1, 2, 1, 12, 1, 50, 1, 192, 1, 722, 1, 2700, 1, 10082, 1, 37632, 1, 140450, 1, 524172, 1, 1956242, 1, 7300800, 1, 27246962, 1, 101687052, 1, 379501250, 1, 1416317952, 1, 5285770562, 1, 19726764300, 1, 73621286642, 1, 274758382272, 1
Offset: 1
Examples
Let L = Sum_{n>=1} 1/(n*A003500(n)) or, more explicitly,
L = 1/4 + 1/(2*14) + 1/(3*52) + 1/(4*194) + 1/(5*724) + 1/(6*2702) +...
so that L = 0.2937696594138291094177057532058145970820225289928...
then exp(L) = 1.3414748719687236691269115428250035920032300984596...
equals the continued fraction expansion given by this sequence:
exp(L) = [1;2,1,12,1,50,1,192,1,722,1,2700,1,10082,1,...]; i.e.,
exp(L) = 1 + 1/(2 + 1/(1 + 1/(12 + 1/(1 + 1/(50 + 1/(1 +...)))))).
Compare these partial quotients to A003500(n), n=1,2,3,...:
[4,14,52,194,724,2702,10084,37634,140452,524174,1956244,...].
Links
- Peter Bala, Some simple continued fraction expansions for an infinite product, Part 1
- Index entries for linear recurrences with constant coefficients, signature (0,5,0,-5,0,1).
Programs
-
Mathematica
a[n_?OddQ] = 1; a[n_?EvenQ] := a[n] = 4*a[n-2] - a[n-4] + 4; a[2] = 2; a[4] = 12; Table[a[n], {n, 1, 41}] (* Jean-François Alcover, May 15 2014, after the first conjecture *) -
PARI
{a(n)=local(L=sum(m=1,2*n+1000,1./(m*round((2+sqrt(3))^m+(2-sqrt(3))^m))));contfrac(exp(L))[n]}
Formula
a(2n-1) = 1, a(2n) = A003500(n) - 2, for n>=1 [conjecture].
From Peter Bala, Jan 04 2013: (Start)
The above conjectures are correct. The real number exp( Sum_{n>=1} 1/(n*A003500(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 = 2 - sqrt(3). Ramanujan has given a continued fraction expansion for F(x). Using this we can find the simple continued fraction expansion of the numbers F(1/2*(N - sqrt(N^2 - 4))), N an integer greater than 3. The present case is when N = 4. See the Bala link for details.
The theory also provides the simple continued fraction expansion of the numbers F((2 - sqrt(3))^k), k = 1, 2, 3, ...: if [1; c(1), 1, c(2), 1, c(3), 1, ...] denotes the present sequence then the simple continued fraction expansion of F((2 - sqrt(3))^k) is given by [1; c(k), 1, c(2*k), 1, c(3*k), 1, ...]. (End)
a(n) = 5*a(n-2)-5*a(n-4)+a(n-6). G.f.: -x*(x^4+2*x^3-4*x^2+2*x+1) / ((x-1)*(x+1)*(x^4-4*x^2+1)). [Colin Barker, Jan 20 2013]