A221073 Simple continued fraction expansion of an infinite product.

2, 4, 1, 8, 1, 32, 1, 56, 1, 196, 1, 336, 1, 1152, 1, 1968, 1, 6724, 1, 11480, 1, 39200, 1, 66920, 1, 228484, 1, 390048, 1, 1331712, 1, 2273376, 1, 7761796, 1, 13250216, 1, 45239072, 1, 77227928, 1, 263672644, 1, 450117360, 1, 1536796800, 1, 2623476240, 1

Offset

0,1

Comments

Simple continued fraction expansion of product {n >= 0} {1 - sqrt(m)*[sqrt(m) - sqrt(m-1)]^(4*n+3)}/{1 - sqrt(m)*[sqrt(m) - sqrt(m-1)]^(4*n+1)} at m = 2. For other cases see A221074 (m = 3), A221075 (m = 4) and A221076 (m = 5).

If we denote the present sequence by [2; 4, 1, c(3), 1, c(4), 1, ...] then for k >= 1 the sequence [1; c(2*k+1), 1, c(2*(2*k+1)), 1, c(3*(2*k+1)), 1, ...] gives the simple continued fraction expansion of product {n >= 0} [1-sqrt(2)*{(sqrt(2)-1)^(2*k+1)}^(4*n+3)]/[1 - sqrt(2)*{(sqrt(2)-1)^(2*k+1)}^(4*n+1)]. An example is given below.

Links

G. C. Greubel, Table of n, a(n) for n = 0..1000

P. Bala, Some simple continued fraction expansions for an infinite product, Part 1

Index entries for linear recurrences with constant coefficients, signature (0,1,0,6,0,-6,0,-1,0,1).

Formula

a(2*n) = 1 for n >= 1. For n >= 1 we have

a(4*n - 3) = (sqrt(2) + 1)^(2*n) + (sqrt(2) - 1)^(2*n) - 2;

a(4*n - 1) = 1/sqrt(2)*{(sqrt(2) + 1)^(2*n + 1) + (sqrt(2) - 1)^(2*n + 1)} - 2.

a(4*n - 3) = 4*A001108(n); a(4*n - 1) = 4*A053141(n).

O.g.f.: 2 + x^2/(1 - x^2) + 4*x*(1 + x^2)^2/(1 - 7*x^4 + 7*x^8 - x^12) = 2 + 4*x + x^2 + 8*x^3 + x^4 + 32*x^5 + ....

O.g.f.: (x^10-2*x^8-6*x^6+12*x^4-4*x^3+x^2-4*x-2) / ((x-1)*(x+1)*(x^4-2*x^2-1)*(x^4+2*x^2-1)). - Colin Barker, Jan 10 2014

Example

Product {n >= 0} {1 - sqrt(2)*(sqrt(2) - 1)^(4*n+3)}/{1 - sqrt(2)*(sqrt(2) - 1)^(4*n+1)} = 2.20409 39255 78752 05766 ...
= 2 + 1/(4 + 1/(1 + 1/(8 + 1/(1 + 1/(32 + 1/(1 + 1/(56 + ...))))))).
We have (sqrt(2) - 1)^3 = 5*sqrt(2) - 7 so product {n >= 0} {1 - sqrt(2)*(5*sqrt(2) - 7)^(4*n+3)}/{1 - sqrt(2)*(5*sqrt(2) - 7)^(4*n+1)} = 1.11117 34981 94843 98511 ... = 1 + 1/(8 + 1/(1 + 1/(196 + 1/(1 + 1/(1968 + 1/(1 + 1/(39200 + ...))))))).

Mathematica

NProduct[( Sqrt[2]*(Sqrt[2] - 1)^(4*n + 3) - 1)/( Sqrt[2]*(Sqrt[2] - 1)^(4*n + 1) - 1), {n, 0, Infinity}, WorkingPrecision -> 200] // ContinuedFraction[#, 37] & (* Jean-François Alcover, Mar 06 2013 *)
Join[{2}, LinearRecurrence[{0, 1, 0, 6, 0, -6, 0, -1, 0, 1}, {4, 1, 8, 1, 32, 1, 56, 1, 196, 1}, 60]] (* Harvey P. Dale, Feb 16 2014 *)

Prog

(PARI) x='x+O('x^30); Vec((x^10-2*x^8-6*x^6+12*x^4-4*x^3+x^2-4*x-2)/((x-1)*(x+1)*(x^4-2*x^2-1)*(x^4+2*x^2-1))) \\ G. C. Greubel, Jul 15 2018
(Magma) m:=25; R<x>:=PowerSeriesRing(Integers(), m); Coefficients(R!((x^10-2*x^8-6*x^6+12*x^4-4*x^3+x^2-4*x-2)/((x-1)*(x+1)*(x^4-2*x^2-1)*(x^4+2*x^2-1)))); // G. C. Greubel, Jul 15 2018

Crossrefs

Cf. A001108, A053141, A174500, A221074 (m = 3), A221075 (m = 4), A221076 (m = 5).

Sequence in context: A182319 A204182 A103324 * A181266 A302192 A087060

Adjacent sequences: A221070 A221071 A221072 * A221074 A221075 A221076

Keyword

nonn,easy,cofr

Author

Peter Bala, Jan 06 2013

Extensions

More terms from Harvey P. Dale, Feb 16 2014

Status

approved