cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A221193 Simple continued fraction expansion of product {k >= 0} (1 - 2*(N - sqrt(N^2-1))^(4*k+3))/(1 - 2*(N - sqrt(N^2-1))^(4*k+1)) at N = 3.

Original entry on oeis.org

1, 1, 1, 32, 1, 97, 1, 1152, 1, 3361, 1, 39200, 1, 114241, 1, 1331712, 1, 3880897, 1, 45239072, 1, 131836321, 1, 1536796800, 1, 4478554081, 1, 52205852192, 1, 152139002497, 1, 1773462177792, 1, 5168247530881, 1, 60245508192800, 1, 175568277047521, 1
Offset: 0

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Author

Peter Bala, Jan 08 2013

Keywords

Comments

Simple continued fraction expansion of product {k >= 0} (1 - 2*(N - sqrt(N^2-1))^(4*k+3))/(1 - 2*(N - sqrt(N^2-1))^(4*k+1)) at N = 3. For other cases see A221075 (N = 2), A221194 (N = 4) and A221195 (N = 5).
Denoting the present sequence by [1, c(1), 1, c(2), 1, c(3), 1, ...] then for n >= 0 the sequence [1, c(2*n+1), 1, c(2*(2*n+1)), 1, c(3*(2*n+1)), 1, ...] gives the simple continued fraction expansion of product {k >= 0} (1 - 2*((3 - 2*sqrt(2))^(2*n+1))^(4*k+3))/(1 - 2*((3 - 2*sqrt(2))^(2*n+1))^(4*k+1)).

Examples

			Product {k >= 0} (1 - 2*(3 - 2*sqrt(2))^(4*k+3))/(1 - 2*(3 - 2*sqrt(2))^(4*k+1)) = 1.50746 49374 34879 05211 ... = 1 + 1/(1 + 1/(1 + 1/(32 + 1/(1 + 1/(97 + ...))))).

Links

Crossrefs

Cf. A221075 (N = 2), A221194 (N = 4), A221195 (N = 5).

Formula

a(4*n-1) = (3 + 2*sqrt(2))^(2*n) + (3 - 2*sqrt(2))^(2*n) - 2;
a(4*n+1) = 1/2*((3 + 2*sqrt(2))^(2*n+1) + (3 - 2*sqrt(2))^(2*n+1)) - 2; a(2*n) = 1.
G.f.: -(x^8+x^7+31*x^5-34*x^4+31*x^3+x+1) / ((x-1)*(x+1)*(x^2-2*x-1)*(x^2+2*x-1)*(x^4+6*x^2+1)). [Colin Barker, Jan 14 2013]

Extensions

More terms from Colin Barker, Jan 14 2013

A221194 Simple continued fraction expansion of product {k >= 0} (1 - 2*(N - sqrt(N^2-1))^(4*k+3))/(1 - 2*(N - sqrt(N^2-1))^(4*k+1)) at N = 4.

Original entry on oeis.org

1, 2, 1, 60, 1, 242, 1, 3840, 1, 15122, 1, 238140, 1, 937442, 1, 14760960, 1, 58106402, 1, 914941500, 1, 3601659602, 1, 56711612160, 1, 223244789042, 1, 3515205012540, 1, 13837575261122, 1, 217885999165440, 1, 857706421400642, 1, 13505416743244860, 1
Offset: 0

Views

Author

Peter Bala, Jan 08 2013

Keywords

Comments

Simple continued fraction expansion of product {k >= 0} (1 - 2*(N - sqrt(N^2-1))^(4*k+3))/(1 - 2*(N - sqrt(N^2-1))^(4*k+1)) at N = 4. For other cases see A221075 (N = 2), A221193 (N = 3) and A221195 (N = 5).
Denoting the present sequence by [1, c(1), 1, c(2), 1, c(3), 1, ...] then for n >= 0 the sequence [1, c(2*n+1), 1, c(2*(2*n+1)), 1, c(3*(2*n+1)), 1, ...] gives the simple continued fraction expansion of product {k >= 0} (1 - 2*((4 - sqrt(15))^(2*n+1))^(4*k+3))/(1 - 2*((4 - sqrt(15))^(2*n+1))^(4*k+1)).

Examples

			product {k >= 0} (1 - 2*(4 - sqrt(15))^(4*k+3))/(1 - 2*(4 - sqrt(15))^(4*k+1)) = 1.33513 52548 90793 94897 ... = 1 + 1/(2 + 1/(1 + 1/(60 + 1/(1 + 1/(242 + ...))))).

Links

Crossrefs

Cf. A221075 (N = 2), A221193 (N = 3) and A221195 (N = 5).

Programs

  • Mathematica
    LinearRecurrence[{0,1,0,62,0,-62,0,-1,0,1},{1,2,1,60,1,242,1,3840,1,15122},40] (* Harvey P. Dale, Aug 03 2023 *)

Formula

a(4*n-1) = (4 + sqrt(15))^(2*n) + (4 - sqrt(15))^(2*n) - 2;
a(4*n+1) = 1/2*{(4 + sqrt(15))^(2*n+1) + (4 - sqrt(15))^(2*n+1)} - 2; a(2*n) = 1.
G.f.: -(x^4-2*x^3+12*x^2-2*x+1)*(x^4+4*x^3-4*x^2+4*x+1) / ((x-1)*(x+1)*(x^4-8*x^2+1)*(x^4+8*x^2+1)). [Colin Barker, Jan 14 2013]

Extensions

More terms from Colin Barker, Jan 14 2013
Showing 1-2 of 2 results.

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