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

Author

Peter Bala, Jan 06 2013

Keywords

• nonn
• easy
• cofr

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.

Examples

			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 + ...))))))).
		

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).

Crossrefs

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

Programs

Magma

m:=25; R:=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

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 *)

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

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

Extensions

More terms from Harvey P. Dale, Feb 16 2014