A221364 -id:A221364 - cp's OEIS frontend

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)).

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

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)).

Original entry on oeis.org

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

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

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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)).

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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)).

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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)).

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cp's OEIS Frontend

A221365 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*(5 - sqrt(21)).

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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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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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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)).

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)).

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)).