A174500 -id:A174500 - cp's OEIS frontend

A086903 a(n) = 8*a(n-1) - a(n-2), starting with a(0) = 2 and a(1) = 8.

2, 8, 62, 488, 3842, 30248, 238142, 1874888, 14760962, 116212808, 914941502, 7203319208, 56711612162, 446489578088, 3515205012542, 27675150522248, 217885999165442, 1715412842801288, 13505416743244862, 106327921103157608

Offset: 0

Nikolay V. Kosinov (kosinov(AT)unitron.com.ua), Sep 21 2003

a(n+1)/a(n) converges to (4+sqrt(15)) = 7.872983... a(0)/a(1)=2/8; a(1)/a(2)=8/62; a(2)/a(3)=62/488; a(3)/a(4)=488/3842; ... etc. Lim a(n)/a(n+1) as n approaches infinity = 0.127016... = 1/(4+sqrt(15)) = (4-sqrt(15)).
Twice A001091. - John W. Layman, Sep 25 2003
Except for the first term, positive values of x (or y) satisfying x^2 - 8xy + y^2 + 60 = 0. - Colin Barker, Feb 13 2014

			a(4) = 3842 = 8*a(3) - a(2) = 8*488 - 62 = (4+sqrt(15))^4 + (4-sqrt(15))^4 = 3841.9997397 + 0.0002603 = 3842.

Vincenzo Librandi, Table of n, a(n) for n = 0..200
P. Bala, Some simple continued fraction expansions for an infinite product, Part 1
Tanya Khovanova, Recursive Sequences
Index entries for recurrences a(n) = k*a(n - 1) +/- a(n - 2)
Index entries for linear recurrences with constant coefficients, signature (8,-1).

Cf. A086594, A058316, A006245, A009271, A005248, A174502.

I:=[2,8]; [n le 2 select I[n] else 8*Self(n-1)-Self(n-2): n in [1..30]]; // Vincenzo Librandi, Feb 15 2014

a[0] = 2; a[1] = 8; a[n_] := 8a[n - 1] - a[n - 2]; Table[ a[n], {n, 0, 19}] (* Robert G. Wilson v, Jan 30 2004 *)
CoefficientList[Series[(2 - 8 x)/(1 - 8 x + x^2), {x, 0, 40}], x] (* Vincenzo Librandi, Feb 15 2014 *)
LinearRecurrence[{8,-1},{2,8},30] (* Harvey P. Dale, Jan 18 2015 *)

[lucas_number2(n,8,1) for n in range(27)] # Zerinvary Lajos, Jun 25 2008

a(n) = (4+sqrt(15))^n + (4-sqrt(15))^n.
G.f.: (2-8*x)/(1-8*x+x^2). [Philippe Deléham, Nov 02 2008]
From Peter Bala, Jan 06 2013: (Start)
Let F(x) = product {n = 0..inf} (1 + x^(4*n+1))/(1 + x^(4*n+3)). Let alpha = 4 - sqrt(15). This sequence gives the simple continued fraction expansion of 1 + F(alpha) = 2.12474 84992 41370 33639 ... = 2 + 1/(8 + 1/(62 + 1/(488 + ...))).  Cf. A174502 and A005248.
Also F(-alpha) = 0.87474 74663 84045 35032 ... has the continued fraction representation 1 - 1/(8 - 1/(62 - 1/(488 - ...))) and the simple continued fraction expansion 1/(1 + 1/((8-2) + 1/(1 + 1/((62-2) + 1/(1 + 1/((488-2) + 1/(1 + ...))))))).
F(alpha)*F(-alpha) has the simple continued fraction expansion 1/(1 + 1/((8^2-4) + 1/(1 + 1/((62^2-4) + 1/(1 + 1/((488^2-4) + 1/(1 + ...))))))).
(End)

A174502 Continued fraction expansion for exp( Sum_{n>=1} 1/(n*A086903(n)) ), where A086903(n) = (4+sqrt(15))^n + (4-sqrt(15))^n.

Original entry on oeis.org

1, 6, 1, 60, 1, 486, 1, 3840, 1, 30246, 1, 238140, 1, 1874886, 1, 14760960, 1, 116212806, 1, 914941500, 1, 7203319206, 1, 56711612160, 1, 446489578086, 1, 3515205012540, 1, 27675150522246, 1, 217885999165440, 1, 1715412842801286, 1

Offset: 0

Paul D. Hanna, Mar 21 2010

			Let L = Sum_{n>=1} 1/(n*A086903(n)) or, more explicitly,
L = 1/8 + 1/(2*62) + 1/(3*488) + 1/(4*3842) + 1/(5*30248) +...
so that L = 0.1338200441271648228100625767561479963630539052200...
then exp(L) = 1.1431870779045667085973926071888878662387686835715...
equals the continued fraction given by this sequence:
exp(L) = [1;6,1,60,1,486,1,3840,1,30246,1,238140,1,...]; i.e.,
exp(L) = 1 + 1/(6 + 1/(1 + 1/(60 + 1/(1 + 1/(486 + 1/(1 +...)))))).
Compare these partial quotients to A086903(n), n=1,2,3,...:
[8,62,488,3842,30248,238142,1874888,14760962,116212808,914941502,...].

Peter Bala, Some simple continued fraction expansions for an infinite product, Part 1
Index entries for linear recurrences with constant coefficients, signature (0,9,0,-9,0,1).

Cf. A086903, A174500, A174501, A174503, A174508.

LinearRecurrence[{0,9,0,-9,0,1},{1,6,1,60,1,486},50] (* Harvey P. Dale, Jun 09 2013 *)

{a(n)=local(L=sum(m=1,2*n+1000,1./(m*round((4+sqrt(15))^m+(4-sqrt(15))^m))));contfrac(exp(L))[n]}

a(2n-2) = 1, a(2n-1) = A086903(n) - 2, for n>=1 [conjecture].
The above conjectures are correct. See the Bala link for details. - Peter Bala, Jan 08 2013
a(n) = 9*a(n-2)-9*a(n-4)+a(n-6). G.f.: -(x^4+6*x^3-8*x^2+6*x+1) / ((x-1)*(x+1)*(x^4-8*x^2+1)). [Colin Barker, Jan 20 2013]

A174503 Continued fraction expansion for exp( Sum_{n>=1} 1/(n*A087799(n)) ), where A087799(n) = (5+sqrt(24))^n + (5-sqrt(24))^n.

Original entry on oeis.org

1, 8, 1, 96, 1, 968, 1, 9600, 1, 95048, 1, 940896, 1, 9313928, 1, 92198400, 1, 912670088, 1, 9034502496, 1, 89432354888, 1, 885289046400, 1, 8763458109128, 1, 86749292044896, 1, 858729462339848, 1, 8500545331353600, 1

Offset: 0

Paul D. Hanna, Mar 21 2010

			Let L = Sum_{n>=1} 1/(n*A087799(n)) or, more explicitly,
L = 1/10 + 1/(2*98) + 1/(3*970) + 1/(4*9602) + 1/(5*95050) +...
so that L = 0.1054740177896236251618898675297390156061405857647...
then exp(L) = 1.1112372317482311056432125938345153306039099019639...
equals the continued fraction given by this sequence:
exp(L) = [1;8,1,96,1,968,1,9600,1,95048,1,940896,1,...]; i.e.,
exp(L) = 1 + 1/(8 + 1/(1 + 1/(96 + 1/(1 + 1/(968 + 1/(1 +...)))))).
Compare these partial quotients to A087799(n), n=1,2,3,...:
[10,98,970,9602,95050,940898,9313930,92198402,912670090,9034502498,...].

Peter Bala, Some simple continued fraction expansions for an infinite product, Part 1
Index entries for linear recurrences with constant coefficients, signature (0,11,0,-11,0,1).

Cf. A087799, A174500, A174502, A174504, A174509.

{a(n)=local(L=sum(m=1,2*n+1000,1./(m*round((5+sqrt(24))^m+(5-sqrt(24))^m))));contfrac(exp(L))[n]}

a(2n-2) = 1, a(2n-1) = A087799(n) - 2, for n>=1 [conjecture].
The above conjectures are correct. See the Bala link for details. - Peter Bala, Jan 08 2013
a(n) = 11*a(n-2)-11*a(n-4)+a(n-6). G.f.: -(x^4+8*x^3-10*x^2+8*x+1) / ((x-1)*(x+1)*(x^4-10*x^2+1)). [Colin Barker, Jan 20 2013]

A221073 Simple continued fraction expansion of an infinite product.

Original entry on oeis.org

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

A174508 Continued fraction expansion for exp( Sum_{n>=1} 1/(n*A086594(n)) ), where A086594(n) = (4+sqrt(17))^n + (4-sqrt(17))^n.

Original entry on oeis.org

1, 7, 65, 1, 535, 4353, 1, 35367, 287297, 1, 2333751, 18957313, 1, 153992263, 1250895425, 1, 10161155671, 82540140801, 1, 670482282087, 5446398397505, 1, 44241669462135, 359379754094593, 1, 2919279702218887, 23713617371845697, 1

Offset: 0

Paul D. Hanna, Mar 21 2010

			Let L = Sum_{n>=1} 1/(n*A086594(n)) or, more explicitly,
L = 1/8 + 1/(2*66) + 1/(3*536) + 1/(4*4354) + 1/(5*35368) +...
so that L = 0.1332613701545977545822925541573311424901819508933...
then exp(L) = 1.1425485874089841897117810754210805471767735522069...
equals the continued fraction given by this sequence:
exp(L) = [1;7,65,1,535,4353,1,35367,287297,1,2333751,...]; i.e.,
exp(L) = 1 + 1/(7 + 1/(65 + 1/(1 + 1/(535 + 1/(4353 + 1/(1 +...)))))).
Compare these partial quotients to A086594(n), n=1,2,3,...:
[8,66,536,4354,35368,287298,2333752,18957314,153992264,...].

Peter Bala, Some simple continued fraction expansions for an infinite product, Part 1.
Peter Bala, Some simple continued fraction expansions for an infinite product, Part 2.
Index entries for linear recurrences with constant coefficients, signature (0,0,67,0,0,-67,0,0,1).

Cf. A086594, A174502, A174507, A174509.

{a(n)=local(L=sum(m=1,2*n+1000,1./(m*round((4+sqrt(17))^m+(4-sqrt(17))^m))));contfrac(exp(L))[n]}

a(3n-3) = 1, a(3n-2) = A086594(2n-1) - 1, a(3n-1) = A086594(2n) - 1, for n>=1 [conjecture].
a(n) = 67*a(n-3)-67*a(n-6)+a(n-9). G.f.: -(x^2-x+1)*(x^6-8*x^5-8*x^4-2*x^3+72*x^2+8*x+1) / ((x-1)*(x^2+x+1)*(x^6-66*x^3+1)). [Colin Barker, Jan 20 2013]
From Peter Bala, Jan 25 2013: (Start)
The above conjectures are correct. The real number exp( Sum {n>=1} 1/(n*A086594(n)) ) is equal to the infinite product F(x) := product {n >= 0} (1 + x^(4*n+3))/(1 - x^(4*n+1)) evaluated at x = sqrt(17) - 4. Ramanujan has given a continued fraction expansion for the product F(x). Using this we can find the simple continued fraction expansion of the numbers F(1/2*(sqrt(N^2 + 4) - N)), N a positive integer. The present case is when N = 8. See the Bala link for details.
The theory also provides the simple continued fraction expansion of the numbers F({sqrt(17) - 4}^(2*k+1)), k = 1, 2, 3, ...: if [1; c(1), c(2), 1, c(3), c(4), 1, ...] denotes the present sequence then the simple continued fraction expansion of F({sqrt(17) - 4}^(2*k+1)) is given by [1; c(2*k+1), c(2*(2*k+1)), 1, c(3*(2*k+1)), c(4*(2*k+1)), 1, ...].
(End)

A174509 Continued fraction expansion for exp( Sum_{n>=1} 1/(n*A086927(n)) ), where A086927(n) = (5+sqrt(26))^n + (5-sqrt(26))^n.

Original entry on oeis.org

1, 9, 101, 1, 1029, 10401, 1, 105049, 1060901, 1, 10714069, 108201601, 1, 1092730089, 11035502501, 1, 111447755109, 1125513053601, 1, 11366578291129, 114791295964901, 1, 1159279537940149, 11707586675366401, 1

Offset: 0

Paul D. Hanna, Mar 21 2010

			Let L = Sum_{n>=1} 1/(n*A086927(n)) or, more explicitly,
L = 1/10 + 1/(2*102) + 1/(3*1030) + 1/(4*10402) + 1/(5*105050) +...
so that L = 0.1052516947742519131304505213983109248819463097531...
then exp(L) = 1.1109902055968924364755807035083159869000358017128...
equals the continued fraction given by this sequence:
exp(L) = [1;9,101,1,1029,10401,1,105049,1060901,1,...]; i.e.,
exp(L) = 1 + 1/(9 + 1/(101 + 1/(1 + 1/(1029 + 1/(10401 +1/(1+...)))))).
Compare these partial quotients to A086927(n), n=1,2,3,...:
[10,102,1030,10402,105050,1060902,10714070,108201602,...].

Peter Bala, Some simple continued fraction expansions for an infinite product, Part 1.
Peter Bala, Some simple continued fraction expansions for an infinite product, Part 2.
Index entries for linear recurrences with constant coefficients, signature (0,0,103,0,0,-103,0,0,1).

Cf. A086927, A174503, A174508, A174510.

LinearRecurrence[{0,0,103,0,0,-103,0,0,1},{1,9,101,1,1029,10401,1,105049,1060901},30] (* Harvey P. Dale, Dec 24 2014 *)

{a(n)=local(L=sum(m=1,2*n+1000,1./(m*round((5+sqrt(26))^m+(5-sqrt(26))^m))));contfrac(exp(L))[n]}

a(3n-3) = 1, a(3n-2) = A086927(2n-1) - 1, a(3n-1) = A086927(2n) - 1, for n>=1 [conjecture].
a(n) = 103*a(n-3)-103*a(n-6)+a(n-9). G.f.: -(x^2-x+1)*(x^6-10*x^5-10*x^4-2*x^3+110*x^2+10*x+1) / ((x-1)*(x^2+x+1)*(x^6-102*x^3+1)). [Colin Barker, Jan 20 2013]
From Peter Bala, Jan 25 2013: (Start)
The above conjectures are correct. The real number exp( Sum {n>=1} 1/(n*A086927(n)) ) is equal to the infinite product F(x) := product {n >= 0} (1 + x^(4*n+3))/(1 - x^(4*n+1)) evaluated at x = sqrt(26) - 5. Ramanujan has given a continued fraction expansion for the product F(x). Using this we can find the simple continued fraction expansion of the numbers F(1/2*(sqrt(N^2 + 4) - N)), N a positive integer. The present case is when N = 10. See the Bala link for details.
The theory also provides the simple continued fraction expansion of the numbers F({sqrt(26) - 5}^(2*k+1)), k = 1, 2, 3, ...: if [1; c(1), c(2), 1, c(3), c(4), 1, ...] denotes the present sequence then the simple continued fraction expansion of F({sqrt(26) - 5}^(2*k+1)) is given by [1; c(2*k+1), c(2*(2*k+1)), 1, c(3*(2*k+1)), c(4*(2*k+1)), 1, ...].
(End)

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

A221364 The simple continued fraction expansion of F(x) := Product_{n >= 0} (1 - x^(4n+3))/(1 - x^(4n+1)) when x = 1/2*(3 - sqrt(5)).

Original entry on oeis.org

1, 1, 1, 5, 1, 16, 1, 45, 1, 121, 1, 320, 1, 841, 1, 2205, 1, 5776, 1, 15125, 1, 39601, 1, 103680, 1, 271441, 1, 710645, 1, 1860496, 1, 4870845, 1, 12752041, 1, 33385280, 1, 87403801, 1, 228826125, 1, 599074576, 1, 1568397605, 1, 4106118241, 1, 10749957120, 1, 28143753121

Offset: 0

Peter Bala, Jan 15 2013

The function F(x) := Product_{n >= 0} (1 - x^(4*n+3))/(1 - x^(4*n+1)) is analytic for |x| < 1. When x is a quadratic irrational of the form x = 1/2*(N - sqrt(N^2 - 4)), N an integer greater than 2, the real number F(x) has a predictable simple continued fraction expansion. The first examples of these expansions, for N = 2, 4, 6 and 8, are due to Hanna. See A174500 through A175503. The present sequence is the case N = 3. See also A221365 (N = 5), A221366 (N = 7), A221369 (N = 9).

If we denote the present sequence by [1, c(1), 1, c(2), 1, c(3), ...] then for k = 1, 2, ..., the simple continued fraction expansion of F((1/2*(3 - sqrt(5)))^k) is given by the sequence [1; c(k), 1, c(2*k), 1, c(3*k), 1, ...]. Examples are given below.

			F(1/2*(3 - sqrt(5))) = 1.53879 34992 88095 08323 ... = 1 + 1/(1 + 1/(1 + 1/(5 + 1/(1 + 1/(16 + 1/(1 + 1/(45 + ...))))))).
F((1/2*(3 - sqrt(5)))^2) = 1.16725 98258 10214 95210 ... = 1 + 1/(5 + 1/(1 + 1/(45 + 1/(1 + 1/(320 + 1/(1 + 1/(2205 + ...))))))).
F((1/2*(3 - sqrt(5)))^3) = 1.05883 42773 67371 19975 ... = 1 + 1/(16 + 1/(1 + 1/(320 + 1/(1 + 1/(5776 + 1/(1 + 1/(103680 + ...))))))).

Peter Bala, Some simple continued fraction expansions for an infinite product, Part 1
Index entries for linear recurrences with constant coefficients, signature (0,4,0,-4,0,1).

Cf. A001906, A002878, A004146, A049684, A081070, A081071, A174500 (N = 4), A221365 (N = 5), A221366 (N = 7), A221369 (N = 9).

a(2*n-1) = (1/2*(3 + sqrt(5)))^n + (1/2*(3 - sqrt(5)))^n - 2 = A004146(n); a(2*n) = 1.
a(4*n+1) = A081071(n) = A002878(n)^2;
a(4*n-1) = A081070(n) = 5*A049684(n) = 5*(A001906(n))^2.
a(n) = 4*a(n-2)-4*a(n-4)+a(n-6). G.f.: -(x^4+x^3-3*x^2+x+1) / ((x-1)*(x+1)*(x^2-x-1)*(x^2+x-1)). - Colin Barker, Jan 20 2013

More terms from Michel Marcus, Feb 21 2025

A221365 The simple continued fraction expansion of F(x) := Product_{n >= 0} (1 - x^(4n+3))/(1 - x^(4n+1)) when x = 1/2*(5 - sqrt(21)).

Original entry on oeis.org

1, 3, 1, 21, 1, 108, 1, 525, 1, 2523, 1, 12096, 1, 57963, 1, 277725, 1, 1330668, 1, 6375621, 1, 30547443, 1, 146361600, 1, 701260563, 1, 3359941221, 1, 16098445548, 1, 77132286525, 1, 369562987083, 1, 1770682648896, 1

Offset: 0

Peter Bala, Jan 15 2013

The function F(x) := Product_{n >= 0} (1 - x^(4*n+3))/(1 - x^(4*n+1)) is analytic for |x| < 1. When x is a quadratic irrational of the form x = 1/2*(N - sqrt(N^2 - 4)), N an integer greater than 2, the real number F(x) has a predictable simple continued fraction expansion. The first examples of these expansions, for N = 2, 4, 6 and 8, are due to Hanna. See A174500 through A175503. The present sequence is the case N = 5. See also A221364 (N = 3), A221366 (N = 7) and A221367 (N = 9).

If we denote the present sequence by [1, c(1), 1, c(2), 1, c(3), ...] then for k = 1, 2, ..., the simple continued fraction expansion of F((1/2*(5 - sqrt(21)))^k) is given by the sequence [1; c(k), 1, c(2*k), 1, c(3*k), 1, ...].

			F(1/2*(5 - sqrt(21))) = 1.25274 83510 08359 27965 ... = 1 + 1/(3 + 1/(1 + 1/(21 + 1/(1 + 1/(108 + 1/(1 + 1/(525 + ...))))))).
F((1/2*(5 - sqrt(21)))^2) = 1.04545 84663 16495 30047 ... = 1 + 1/(21 + 1/(1 + 1/(525 + 1/(1 + 1/(12096 + 1/(1 + 1/(277725 + ...))))))).
F((1/2*(5 - sqrt(21)))^3) = 1.00917 43188 83793 73068 ... = 1 + 1/(108 + 1/(1 + 1/(12096 + 1/(1 + 1/(1330668 + 1/(1 + 1/(146361600 + ...))))))).

Peter Bala, Some simple continued fraction expansions for an infinite product, Part 1
Index entries for linear recurrences with constant coefficients, signature (0,6,0,-6,0,1).

Cf. A004254, A030221, A054493, A174500 (N = 4), A221364 (N = 3), A221366 (N = 7), A221369 (N = 9).

LinearRecurrence[{0,6,0,-6,0,1},{1,3,1,21,1,108},40] (* Harvey P. Dale, Jun 06 2023 *)

a(2*n-1) = (1/2*(5 + sqrt(21)))^n + (1/2*(5 - sqrt(21)))^n - 2 = 3*A054493(n); a(2*n) = 1.
a(4*n+1) = 3*(A030221(n))^2; a(4*n-1) = 21*(A004254(n))^2.
a(n) = 6*a(n-2)-6*a(n-4)+a(n-6). G.f.: -(x^4+3*x^3-5*x^2+3*x+1) / ((x-1)*(x+1)*(x^4-5*x^2+1)). - Colin Barker, Jan 20 2013

cp's OEIS Frontend

A086903 a(n) = 8*a(n-1) - a(n-2), starting with a(0) = 2 and a(1) = 8.

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A174502 Continued fraction expansion for exp( Sum_{n>=1} 1/(n*A086903(n)) ), where A086903(n) = (4+sqrt(15))^n + (4-sqrt(15))^n.

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A174503 Continued fraction expansion for exp( Sum_{n>=1} 1/(n*A087799(n)) ), where A087799(n) = (5+sqrt(24))^n + (5-sqrt(24))^n.

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A221073 Simple continued fraction expansion of an infinite product.

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A174508 Continued fraction expansion for exp( Sum_{n>=1} 1/(n*A086594(n)) ), where A086594(n) = (4+sqrt(17))^n + (4-sqrt(17))^n.

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A174509 Continued fraction expansion for exp( Sum_{n>=1} 1/(n*A086927(n)) ), where A086927(n) = (5+sqrt(26))^n + (5-sqrt(26))^n.

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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}).

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}).

A221364 The simple continued fraction expansion of F(x) := Product_{n >= 0} (1 - x^(4n+3))/(1 - x^(4n+1)) when x = 1/2*(3 - sqrt(5)).

A221365 The simple continued fraction expansion of F(x) := Product_{n >= 0} (1 - x^(4n+3))/(1 - x^(4n+1)) when x = 1/2*(5 - sqrt(21)).