A174500 -id:A174500 - cp's OEIS frontend

A221366 The simple continued fraction expansion of F(x) := Product_{n >= 0} (1 - x^(4n+3))/(1 - x^(4n+1)) when x = (1/2)(7 - 3sqrt(5)).

1, 5, 1, 45, 1, 320, 1, 2205, 1, 15125, 1, 103680, 1, 710645, 1, 4870845, 1, 33385280, 1, 228826125, 1, 1568397605, 1, 10749957120, 1, 73681302245, 1, 505019158605, 1, 3461452808000, 1, 23725150497405, 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 = 7. See also A221364 (N = 3), A221365 (N = 5) 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*(7 - sqrt(45)))^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*(7 - sqrt(45))) = 1.16725 98258 10214 95210 ... = 1 + 1/(5 + 1/(1 + 1/(45 + 1/(1 + 1/(320 + 1/(1 + 1/(2205 + ...))))))).
F((1/2*(7 - sqrt(45)))^2) = 1.02173 93445 69104 86504 ... = 1 + 1/(45 + 1/(1 + 1/(2205 + 1/(1 + 1/(103680 + 1/(1 + 1/(4870845 + ...))))))).
F((1/2*(7 - sqrt(45)))^3) = 1.00311 52648 91110 10148 ... = 1 + 1/(320 + 1/(1 + 1/(103680 + 1/(1 + 1/(33385280 + 1/(1 + 1/(10749957120 + ...))))))).

Peter Bala, Some simple continued fraction expansions for an infinite product, Part 1
Index entries for linear recurrences with constant coefficients, signature (0,8,0,-8,0,1).

Cf. A004187, A033890, A049682, A081070.

Cf. A174500 (N = 4), A221364 (N = 3), A221365 (N = 5), A221369 (N = 9).

LinearRecurrence[{0,8,0,-8,0,1},{1,5,1,45,1,320},40] (* or *) Riffle[ LinearRecurrence[{8,-8,1},{5,45,320},20],1,{1,-1,2}] (* Harvey P. Dale, Jan 04 2018 *)

a(2*n-1) = (1/2*(7 + sqrt(45)))^n + (1/2*(7 - sqrt(45)))^n - 2 = A081070(n); a(2*n) = 1.
a(4*n-1) = 45*A049682(n) = 45*(A004187(n))^2;
a(4*n+1) = 5*(A033890(n))^2.
a(n) = 8*a(n-2)-8*a(n-4)+a(n-6). G.f.: -(x^4+5*x^3-7*x^2+5*x+1) / ((x-1)*(x+1)*(x^2-3*x+1)*(x^2+3*x+1)). - Colin Barker, Jan 20 2013

A174505 Continued fraction expansion for exp( Sum_{n>=1} 1/(n*Lucas(n)) ), where Lucas(n) = A000032(n) = ((1+sqrt(5))/2)^n + ((1-sqrt(5))/2)^n.

Original entry on oeis.org

3, 1, 3, 6, 1, 10, 17, 1, 28, 46, 1, 75, 122, 1, 198, 321, 1, 520, 842, 1, 1363, 2206, 1, 3570, 5777, 1, 9348, 15126, 1, 24475, 39602, 1, 64078, 103681, 1, 167760, 271442, 1, 439203, 710646, 1, 1149850, 1860497, 1, 3010348, 4870846, 1, 7881195, 12752042, 1

Offset: 0

Paul D. Hanna, Mar 21 2010

			Let L = Sum_{n>=1} 1/(n*A000032(n)) or, more explicitly,
L = 1 + 1/(2*3) + 1/(3*4) + 1/(4*7) + 1/(5*11) + 1/(6*18) +...
so that L = 1.3240810281350207977825663314844927483236088628781...
then exp(L) = 3.7587296006215531704236522952745520722012715044952...
equals the continued fraction given by this sequence:
exp(L) = [3;1,3,6,1,10,17,1,28,46,1,75,122,1,198,321,1,...]; i.e.,
exp(L) = 3 + 1/(1 + 1/(3 + 1/(6 + 1/(1 + 1/(10 + 1/(17 +...)))))).
Compare these partial quotients to A000032(n), n=1,2,3,...:
[1,3,4,7,11,18,29,47,76,123,199,322,521,843,1364,2207,3571,5778,...].

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 (0,0,4,0,0,-4,0,0,1).

Cf. A000032 (Lucas numbers), A174500, A174504, A174506.

LinearRecurrence[{0,0,4,0,0,-4,0,0,1},{3,1,3,6,1,10,17,1,28,46},50] (* Harvey P. Dale, Feb 02 2025 *)

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

a(3n-2) = 1, a(3n-1) = A000032(2n+1) - 1, a(3n) = A000032(2n+2) - 1, for n>=1 [conjecture].
a(n) = 4*a(n-3)-4*a(n-6)+a(n-9) for n>9. G.f.: (x^9 -x^7 -5*x^6 +2*x^5 +3*x^4 +6*x^3 -3*x^2 -x -3) / ((x -1)*(x^2 +x +1)*(x^6 -3*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*Lucas(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 = 1/2*(sqrt(5) - 1). 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 = 1. See the Bala link for details.
The theory also provides the simple continued fraction expansion of the numbers F({1/2*(sqrt(5) - 1)}^(2*k+1)), k = 1, 2, 3, ...: if [3, 1, c(3), c(4), 1, c(5), c(6), 1, ...] denotes the present sequence then the simple continued fraction expansion of F({1/2*(sqrt(5) - 1)}^(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)

A174506 Continued fraction expansion for exp( Sum_{n>=1} 1/(n*A014448(n)) ), where A014448(n) = (2+sqrt(5))^n + (2-sqrt(5))^n.

Original entry on oeis.org

1, 3, 17, 1, 75, 321, 1, 1363, 5777, 1, 24475, 103681, 1, 439203, 1860497, 1, 7881195, 33385281, 1, 141422323, 599074577, 1, 2537720635, 10749957121, 1, 45537549123, 192900153617, 1, 817138163595, 3461452808001, 1

Offset: 0

Paul D. Hanna, Mar 21 2010

			Let L = Sum_{n>=1} 1/(n*A014448(n)) or, more explicitly,
L = 1/4 + 1/(2*18) + 1/(3*76) + 1/(4*322) + 1/(5*1364) +...
so that L = 0.2831229765066671850017990708479258794794782639219...
then exp(L) = 1.3272683746094012523448609429829013914921330866098...
equals the continued fraction given by this sequence:
exp(L) = [1;3,17,1,75,321,1,1363,5777,1,24475,103681,1,...]; i.e.,
exp(L) = 1 + 1/(3 + 1/(17 + 1/(1 + 1/(75 + 1/(321 + 1/(1 +...)))))).
Compare these partial quotients to A014448(n), n=1,2,3,...:
[4,18,76,322,1364,5778,24476,103682,439204,1860498,7881196,33385282,...].

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 (0,0,19,0,0,-19,0,0,1).

Cf. A014448, A174500, A174505, A174507.

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

a(3n-3) = 1, a(3n-2) = A014448(2n-1) - 1, a(3n-1) = A014448(2n) - 1, for n>=1 [conjecture].
a(n) = 19*a(n-3)-19*a(n-6)+a(n-9). G.f.: -(x^2 -x +1)*(x^6 -4*x^5 -4*x^4 -2*x^3 +20*x^2 +4*x +1) / ((x -1)*(x^2 -3*x +1)*(x^2 +x +1)*(x^4 +3*x^3 +8*x^2 +3*x +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*A014448(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(5) - 2. 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 = 4. See the Bala link for details.
The theory also provides the simple continued fraction expansion of the numbers F({sqrt(5) - 2}^(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(5) - 2}^(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)

A174507 Continued fraction expansion for exp( Sum_{n>=1} 1/(n*A085447(n)) ), where A085447(n) = (3+sqrt(10))^n + (3-sqrt(10))^n.

Original entry on oeis.org

1, 5, 37, 1, 233, 1441, 1, 8885, 54757, 1, 337433, 2079361, 1, 12813605, 78960997, 1, 486579593, 2998438561, 1, 18477210965, 113861704357, 1, 701647437113, 4323746327041, 1, 26644125399365, 164188498723237, 1, 1011775117738793

Offset: 0

Paul D. Hanna, Mar 21 2010

			Let L = Sum_{n>=1} 1/(n*A085447(n)) or, more explicitly,
L = 1/6 + 1/(2*38) + 1/(3*234) + 1/(4*1442) + 1/(5*8886) +...
so that L = 0.1814484777922995750614847484088330644558009487798...
then exp(L) = 1.1989527624251050123398509513177598419795554140316...
equals the continued fraction given by this sequence:
exp(L) = [1;5,37,1,233,1441,1,8885,54757,1,337433,...]; i.e.,
exp(L) = 1 + 1/(5 + 1/(37 + 1/(1 + 1/(233 + 1/(1441 + 1/(1 +...)))))).
Compare these partial quotients to A085447(n), n=1,2,3,...:
[6,38,234,1442,8886,54758,337434,2079362,12813606,78960998,...].

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 (0,0,39,0,0,-39,0,0,1).

Cf. A085447, A174501, A174506, A174508.

LinearRecurrence[{0,0,39,0,0,-39,0,0,1},{1,5,37,1,233,1441,1,8885,54757},30] (* Harvey P. Dale, Aug 07 2016 *)

{a(n)=local(L=sum(m=1,2*n+1000,1./(m*round((3+sqrt(10))^m+(3-sqrt(10))^m))));contfrac(exp(L))[n]}

a(3n-3) = 1, a(3n-2) = A085447(2n-1) - 1, a(3n-1) = A085447(2n) - 1, for n>=1 [conjecture].
a(n) = 39*a(n-3)-39*a(n-6)+a(n-9). G.f.: -(x^2-x+1) * (x^6 -6*x^5 -6*x^4 -2*x^3 +42*x^2 +6*x +1) / ((x-1)*(x^2+x+1)*(x^6-38*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*A085447(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(10) - 3. 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 = 6. See the Bala link for details.
The theory also provides the simple continued fraction expansion of the numbers F({sqrt(10) - 3}^(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(10) - 3}^(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)

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]

A221367 The simple continued fraction expansion of F(x) := Product_{n >= 0} (1 - x^(4n+3))/(1 - x^(4n+1)) when x = (1/2)*(9 - sqrt(77)).

Original entry on oeis.org

1, 7, 1, 77, 1, 700, 1, 6237, 1, 55447, 1, 492800, 1, 4379767, 1, 38925117, 1, 345946300, 1, 3074591597, 1, 27325378087, 1, 242853811200, 1, 2158358922727, 1, 19182376493357, 1, 170483029517500, 1, 1515164889164157, 1, 13466000972959927, 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 = 9. See also A221364 (N = 3), A221365 (N = 5) and A221366 (N = 7).

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*(9 - sqrt(77)))^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*(9 - sqrt(77))) = 1.12519 81018 34502 81936 ... = 1 + 1/(7 + 1/(1 + 1/(77 + 1/(1 + 1/(700 + 1/(1 + 1/(6237 + ...))))))).
F((1/2*(9 - sqrt(77)))^2) = 1.01282 05391 65421 74656 ... = 1 + 1/(77 + 1/(1 + 1/(6237 + 1/(1 + 1/(492800 + 1/(1 + 1/(38925117 + ...))))))).
F((1/2*(9 - sqrt(77)))^3) = 1.00142 65335 27667 24640 ... = 1 + 1/(700 + 1/(1 + 1/(492800 + 1/(1 + 1/(345946300 + 1/(1 + 1/(242853811200 + ...))))))).

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

Cf. A018193, A057081, A174500 (N = 4), A221364 (N = 3), A221365 (N = 5), A221366 (N = 7).

a(2*n-1) = (1/2*(9 + sqrt(77)))^n + (1/2*(9 - sqrt(77)))^n - 2; a(2*n) = 1.
a(4*n-1) = 77*(A018913(n))^2; a(4*n+1) = 7*(A057081(n))^2.
a(n) = 10*a(n-2)-10*a(n-4)+a(n-6). G.f.: -(x^4+7*x^3-9*x^2+7*x+1) / ((x-1)*(x+1)*(x^4-9*x^2+1)). - Colin Barker, Jan 20 2013

A174510 Continued fraction expansion for exp( Sum_{n>=1} 1/(n*A080040(n)) ), where A080040(n) = (1+sqrt(3))^n + (1-sqrt(3))^n.

Original entry on oeis.org

1, 1, 3, 1, 9, 13, 1, 37, 51, 1, 141, 193, 1, 529, 723, 1, 1977, 2701, 1, 7381, 10083, 1, 27549, 37633, 1, 102817, 140451, 1, 383721, 524173, 1, 1432069, 1956243, 1, 5344557, 7300801, 1, 19946161, 27246963, 1, 74440089, 101687053, 1, 277814197

Offset: 0

Paul D. Hanna, Mar 21 2010

			Let L = Sum_{n>=1} 1/(n*A080040(n)) or, more explicitly,
L = 1/2 + 1/(2*8) + 1/(3*20) + 1/(4*56) + 1/(5*152) + 1/(6*416) +...
so that L = 0.5855329921665857283309456463364081071245363598803...
then exp(L) = 1.7959479567807442397990076546690432122217738278933...
equals the continued fraction given by this sequence:
exp(L) = [1;1,3,1,9,13,1,37,51,1,141,193,1,529,723,1,...]; i.e.,
exp(L) = 1 + 1/(1 + 1/(3 + 1/(1 + 1/(9 + 1/(13 + 1/(1 +...)))))).
Compare these partial quotients to A080040(n)/2^[n/2], n=1,2,3,...:
[2,4,10,14,38,52,142,194,530,724,1978,2702,7382,10084,27550,...],
where A080040 begins:
[2,8,20,56,152,416,1136,3104,8480,23168,63296,172928,472448,...].

Index entries for linear recurrences with constant coefficients, signature (0,0,5,0,0,-5,0,0,1).

Cf. A080040, A174500, A174504, A174509.

{a(n)=local(L=sum(m=1,2*n+1000,1./(m*round((1+sqrt(3))^m+(1-sqrt(3))^m))));contfrac(exp(L))[n]}

a(3n-3) = 1, a(3n-2) = A080040(2n-1)/2^(n-1) - 1, a(3n-1) = A080040(2n)/2^n - 1, for n>=1 [conjecture].

a(n) = 5*a(n-3)-5*a(n-6)+a(n-9). G.f.: -(x^2-x+1)*(x^6-2*x^5-2*x^4-2*x^3+4*x^2+2*x+1) / ((x-1)*(x^2+x+1)*(x^6-4*x^3+1)). [Colin Barker, Jan 20 2013]

cp's OEIS Frontend

A221366 The simple continued fraction expansion of F(x) := Product_{n >= 0} (1 - x^(4*n+3))/(1 - x^(4*n+1)) when x = (1/2)*(7 - 3*sqrt(5)).

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A174505 Continued fraction expansion for exp( Sum_{n>=1} 1/(n*Lucas(n)) ), where Lucas(n) = A000032(n) = ((1+sqrt(5))/2)^n + ((1-sqrt(5))/2)^n.

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A174506 Continued fraction expansion for exp( Sum_{n>=1} 1/(n*A014448(n)) ), where A014448(n) = (2+sqrt(5))^n + (2-sqrt(5))^n.

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A174507 Continued fraction expansion for exp( Sum_{n>=1} 1/(n*A085447(n)) ), where A085447(n) = (3+sqrt(10))^n + (3-sqrt(10))^n.

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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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A221367 The simple continued fraction expansion of F(x) := Product_{n >= 0} (1 - x^(4*n+3))/(1 - x^(4*n+1)) when x = (1/2)*(9 - sqrt(77)).

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A221366 The simple continued fraction expansion of F(x) := Product_{n >= 0} (1 - x^(4n+3))/(1 - x^(4n+1)) when x = (1/2)(7 - 3sqrt(5)).

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.

A221367 The simple continued fraction expansion of F(x) := Product_{n >= 0} (1 - x^(4n+3))/(1 - x^(4n+1)) when x = (1/2)*(9 - sqrt(77)).