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.

Showing 1-3 of 3 results.

A221075 Simple continued fraction expansion of an infinite product.

Original entry on oeis.org

2, 12, 1, 24, 1, 192, 1, 360, 1, 2700, 1, 5040, 1, 37632, 1, 70224, 1, 524172, 1, 978120, 1, 7300800, 1, 13623480, 1, 101687052, 1, 189750624, 1, 1416317952, 1, 2642885280, 1, 19726764300, 1, 36810643320, 1
Offset: 0

Views

Author

Peter Bala, Jan 06 2013

Keywords

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 = 4. For other cases see A221073 (m = 2), A221074 (m = 3) and A221076 (m = 5).
If we denote the present sequence by [2; 12, 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} [1-2*{(2-sqrt(3))^(2*k+1)}^(4*n+3)]/[1 - 2*{(2-sqrt(3))^(2*k+1)}^(4*n+1)]. An example is given below

Examples

			Product {n >= 0} {1 - 2*(2 - sqrt(3))^(4*n+3)}/{1 - 2*(2 - sqrt(3))^(4*n+1)} = 2.07715 13807 08976 70415 ...
= 2 + 1/(12 + 1/(1 + 1/(24 + 1/(1 + 1/(192 + 1/(1 + 1/(360 + ...))))))).
Since (2 - sqrt(3))^3 = 26 - 15*sqrt(3) we have the following simple continued fraction expansion:
product {n >= 0} {1 - 2*(26 - 15*sqrt(3))^(4*n+3)}/{1 - 2*(26 - 15*sqrt(3))^(4*n+1)} = 1.04000 05921 62729 43797 ... = 1 + 1/(24 + 1/(1 + 1/(2700 + 1/(1 + 1/(70224 + 1/(1 + 1/(7300800 + ...))))))).

Links

Crossrefs

Cf. A001353, A007654, A045899, A076139, A076140, A098301, A123480, A174500, A217855, A221073 (m = 2), A221074 (m = 3), A221076 (m = 5).

Formula

a(2*n) = 1 for n >= 1. For n >= 1 we have
a(4*n - 3) = (2 + sqrt(3))^(2*n) + (2 - sqrt(3))^(2*n) - 2;
a(4*n - 1) = 1/2*{(2 + sqrt(3) )^(2*n + 1) + (2 - sqrt(3))^(2*n + 1)} - 2.
a(4*n - 3) = 12*A098301(n) = 12*A001353(n)^2 = 4*A007654(n);
a(4*n - 1) = 24*A076139(n) = 12*A217855 = 8*A076140(n) = 6*A123480(n) = 3*A045899(n).
O.g.f.: 2 + x^2/(1 - x^2) + 12*x*(1 + x^2)^2/(1 - 15*x^4 + 15*x^8 - x^12) = 2 + 12*x + x^2 + 24*x^3 + x^4 + 192*x^5 + ....
O.g.f.: (x^10-2*x^8-14*x^6+28*x^4-12*x^3+x^2-12*x-2) / ((x-1)*(x+1)*(x^4-4*x^2+1)*(x^4+4*x^2+1)). - Colin Barker, Jan 10 2014

A221074 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 = 3.

Original entry on oeis.org

2, 8, 1, 16, 1, 96, 1, 176, 1, 968, 1, 1760, 1, 9600, 1, 17440, 1, 95048, 1, 172656, 1, 940896, 1, 1709136, 1, 9313928, 1, 16918720, 1, 92198400, 1, 167478080, 1, 912670088, 1, 1657862096, 1
Offset: 0

Views

Author

Peter Bala, Jan 06 2013

Keywords

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 = 3. For other cases see A221073 (m = 2), A221075 (m = 4) and A221076 (m = 5).

Examples

			Product {n >= 0} {1 - sqrt(3)*(sqrt(3) - sqrt(2))^(4*n+3)}/{1 - sqrt(3)*(sqrt(3) - sqrt(2))^(4*n+1)} = 2.11180 16361 44098 52896 ...
= 2 + 1/(8 + 1/(1 + 1/(16 + 1/(1 + 1/(96 + 1/(1 + 1/(176 + ...))))))).
Since (sqrt(3) - sqrt(2))^3 = 9*sqrt(3) - 11*sqrt(2) we have the following simple continued fraction expansion:
product {n >= 0} {1 - sqrt(3)*(9*sqrt(3) - 11*sqrt(2))^(4*n+3)}/{1 - sqrt(3)*(9*sqrt(3) - 11*sqrt(2))^(4*n+1)} = 1 + 1/(16 + 1/(1 + 1/(968 + 1/(1 + 1/(17440 + 1/(1 + 1/(940896 + ...))))))).

Links

Crossrefs

Cf. A098297, A105438, A132596, A174500, A221073 (m = 2), A221075 (m = 4), A221076 (m = 5).

Formula

a(2*n) = 1 for n >= 1. For n >= 1 we have
a(4*n - 3) = (sqrt(3) + sqrt(2))^(2*n) + (sqrt(3) - sqrt(2))^(2*n) - 2;
a(4*n - 1) = 1/sqrt(3)*{(sqrt(3) + sqrt(2))^(2*n + 1) + (sqrt(3) - sqrt(2))^(2*n + 1)} - 2.
a(4*n - 3) = 8*A098297(n) = 4*A132596(n); a(4*n - 1) = 4*A105038(n).
O.g.f.: 2 + x^2/(1 - x^2) + 8*x*(1 + x^2)^2/(1 - 11*x^4 + 11*x^8 - x^12) = 2 + 8*x + x^2 + 16*x^3 + x^4 + 96*x^5 + ....
If we denote the present sequence by [2; 8, 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(3)*{(sqrt(3)-sqrt(2))^(2*k+1)}^(4*n+3)]/[1 - sqrt(3)*{(sqrt(3)-sqrt(2))^(2*k+1)}^(4*n+1)]. An example is given below.
O.g.f.: (x^10-2*x^8-10*x^6+20*x^4-8*x^3+x^2-8*x-2) / ((x-1)*(x+1)*(x^8-10*x^4+1)). - Colin Barker, Jan 10 2014

A221076 Continued fraction expansion of product_{n>=0} (1-sqrt(5)*[sqrt(5)-2]^{4n+3})/(1-sqrt(5)*[sqrt(5)-2]^{4n+1}).

Original entry on oeis.org

2, 16, 1, 32, 1, 320, 1, 608, 1, 5776, 1, 10944, 1, 103680, 1, 196416, 1, 1860496, 1, 3524576, 1, 33385280, 1, 63245984, 1, 599074576, 1, 1134903168, 1, 10749957120, 1, 20365011072, 1, 192900153616, 1, 365435296160, 1
Offset: 0

Views

Author

Peter Bala, Jan 06 2013

Keywords

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 = 5. For other cases see A221073 (m = 2), A221074 (m = 3) and A221075 (m = 4).
If we denote the present sequence by [2; 16, 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(5)*{(sqrt(5)-2)^(2*k+1)}^(4*n+3)]/[1 - sqrt(5)*{(sqrt(5)-2)^(2*k+1)}^(4*n+1)]. An example is given below.

Examples

			Product {n >= 0} {1 - sqrt(5)*(sqrt(5) - 2)^(4*n+3)}/{1 - sqrt(5)*(sqrt(5) - 2)^(4*n+1)} = 2.05892 54859 32105 82744 ...
= 2 + 1/(16 + 1/(1 + 1/(32 + 1/(1 + 1/(320 + 1/(1 + 1/(608 + ...))))))).
Since (sqrt(5) - 2)^3 = 17*sqrt(5) - 38 we have the following simple continued fraction expansion:
product {n >= 0} {1 - sqrt(5)*(17*sqrt(5) - 38)^(4*n+3)}/{1 - sqrt(5)*(17*sqrt(5) - 38)^(4*n+1)} = 1.03030 31892 29728 52318 ... = 1 + 1/(32 + 1/(1 + 1/(5776 + 1/(1 + 1/(196416 + 1/(1 + 1/(33385280 + ...))))))).

Links

Crossrefs

Cf. A049664, A049863, A053606, A132584, A174500, A221073 (m = 2), A221074 (m = 3), A221075 (m = 4).

Programs

  • Mathematica
    LinearRecurrence[{0,1,0,18,0,-18,0,-1,0,1},{2,16,1,32,1,320,1,608,1,5776,1},40] (* or *) Join[{2},Riffle[LinearRecurrence[{1,18,-18,-1,1},{16,32,320,608,5776},20],1]] (* Harvey P. Dale, Jun 05 2023 *)

Formula

a(2*n) = 1 for n >= 1. For n >= 1 we have:
a(4*n - 3) = (sqrt(5) + 2)^(2*n) + (sqrt(5) - 2)^(2*n) - 2;
a(4*n - 1) = 1/sqrt(5)*{(sqrt(5) + 2)^(2*n + 1) + (sqrt(5) - 2)^(2*n + 1)} - 2.
a(4*n - 3) = 16*A049863(n) = 4*A132584(n);
a(4*n - 1) = 32*A049664(n) = 4*A053606(n).
O.g.f.: 2 + x^2/(1 - x^2) + 16*x*(1 + x^2)^2/(1 - 19*x^4 + 19*x^8 - x^12) = 2 + 16*x + x^2 + 32*x^3 + x^4 + 320*x^5 + ....
O.g.f.: (x^10-2*x^8-18*x^6+36*x^4-16*x^3+x^2-16*x-2) / ((x-1)*(x+1)*(x^4-4*x^2-1)*(x^4+4*x^2-1)). - Colin Barker, Jan 10 2014
Showing 1-3 of 3 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

Content is available under The OEIS End-User License Agreement.