A221075 -id:A221075 - cp's OEIS frontend

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

Peter Bala, Jan 06 2013

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.

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

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).

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

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

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 *)

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

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

More terms from Harvey P. Dale, Feb 16 2014

A221074 Simple continued fraction expansion of product {n >= 0} {1 - sqrt(m)[sqrt(m) - sqrt(m-1)]^(4n+3)}/{1 - sqrt(m)[sqrt(m) - sqrt(m-1)]^(4n+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

Peter Bala, Jan 06 2013

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).

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

Peter Bala, Some simple continued fraction expansions for an infinite product, Part 1
Index entries for linear recurrences with constant coefficients, signature (0,1,0,10,0,-10,0,-1,0,1).

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

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

Peter Bala, Jan 06 2013

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.

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

Peter Bala, Some simple continued fraction expansions for an infinite product, Part 1
Index entries for linear recurrences with constant coefficients, signature (0,1,0,18,0,-18,0,-1,0,1).

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

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 *)

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

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

Peter Bala, Jan 08 2013

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)).

			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 + ...))))).

Peter Bala, Some simple continued fraction expansions for an infinite product, Part 1
Index entries for linear recurrences with constant coefficients, signature (0,1,0,34,0,-34,0,-1,0,1).

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

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]

More terms from Colin Barker, Jan 14 2013

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

Peter Bala, Jan 08 2013

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)).

			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 + ...))))).

Peter Bala, Some simple continued fraction expansions for an infinite product, Part 1
Index entries for linear recurrences with constant coefficients, signature (0,1,0,62,0,-62,0,-1,0,1).

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

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 *)

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]

More terms from Colin Barker, Jan 14 2013

A221195 Simple continued fraction expansion of product {k >= 0} (1 - 2(N - sqrt(N^2-1))^(4k+3))/(1 - 2(N - sqrt(N^2-1))^(4k+1)) at N = 5.

Original entry on oeis.org

1, 3, 1, 96, 1, 483, 1, 9600, 1, 47523, 1, 940896, 1, 4656963, 1, 92198400, 1, 456335043, 1, 9034502496, 1, 44716177443, 1, 885289046400, 1, 4381729054563, 1, 86749292044896, 1, 429364731169923, 1, 8500545331353600, 1

Offset: 0

Peter Bala, Jan 08 2013

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 = 5. For other cases see A221075 (N = 2), A221193 (N = 3) and A221194 (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*((5 - sqrt(24))^(2*n+1))^(4*k+3))/(1 - 2*((5 - sqrt(24))^(2*n+1))^(4*k+1)).

			Product {k >= 0} (1 - 2*(5 - sqrt(24))^(4*k+3))/(1 - 2*(5 - sqrt(24))^(4*k+1)) = 1.25063 93996 76216 17350 ... = 1 + 1/(3 + 1/(1 + 1/(96 + 1/(1 + 1/(483 + ...))))).

Peter Bala, Some simple continued fraction expansions for an infinite product, Part 1
Index entries for linear recurrences with constant coefficients, signature (0,1,0,98,0,-98,0,-1,0,1).

Cf. A221075 (N = 2), A221193 (N = 3), A221194 (N = 4).

a(4*n-1) = (5 + sqrt(24))^(2*n) + (5 - sqrt(24))^(2*n) - 2;
a(4*n+1) = 1/2*((5 + sqrt(24))^(2*n+1) + (5 - sqrt(24))^(2*n+1)) - 2; a(2*n) = 1.
G.f.: -(x^8+3*x^7+93*x^5-98*x^4+93*x^3+3*x+1) / ((x-1)*(x+1)*(x^4-10*x^2+1)*(x^4+10*x^2+1)). [Colin Barker, Jan 14 2013]

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A221073 Simple continued fraction expansion of an infinite product.

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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.

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

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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.

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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.

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A221195 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 = 5.

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A221074 Simple continued fraction expansion of product {n >= 0} {1 - sqrt(m)[sqrt(m) - sqrt(m-1)]^(4n+3)}/{1 - sqrt(m)[sqrt(m) - sqrt(m-1)]^(4n+1)} at m = 3.

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

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

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

A221195 Simple continued fraction expansion of product {k >= 0} (1 - 2(N - sqrt(N^2-1))^(4k+3))/(1 - 2(N - sqrt(N^2-1))^(4k+1)) at N = 5.