A174502 -id:A174502 - cp's OEIS frontend

A174501 Continued fraction expansion for exp( Sum_{n>=1} 1/(n*A003499(n)) ), where A003499(n) = (3+sqrt(8))^n + (3-sqrt(8))^n.

1, 4, 1, 32, 1, 196, 1, 1152, 1, 6724, 1, 39200, 1, 228484, 1, 1331712, 1, 7761796, 1, 45239072, 1, 263672644, 1, 1536796800, 1, 8957108164, 1, 52205852192, 1, 304278004996, 1, 1773462177792, 1, 10336495061764, 1, 60245508192800, 1, 351136554095044, 1

Offset: 1

Paul D. Hanna, Mar 20 2010

			Let L = Sum_{n>=1} 1/(n*A003499(n)) or, more explicitly,
L = 1/6 + 1/(2*34) + 1/(3*198) + 1/(4*1154) + 1/(5*6726) +...
so that L = 0.1833074113563494600094468694966574405706183998044...
then exp(L) = 1.2011836088120841844713993433258934531421726294252...
equals the continued fraction given by this sequence:
exp(L) = [1;4,1,32,1,196,1,1152,1,6724,1,39200,1,...]; i.e.,
exp(L) = 1 + 1/(4 + 1/(1 + 1/(32 + 1/(1 + 1/(196 + 1/(1 +...)))))).
Compare these partial quotients to A003499(n), n=1,2,3,...:
[6,34,198,1154,6726,39202,228486,1331714,7761798,45239074,...].

Colin Barker, Table of n, a(n) for n = 1..1000
Peter Bala, Some simple continued fraction expansions for an infinite product, Part 1
Index entries for linear recurrences with constant coefficients, signature (0,7,0,-7,0,1).

Cf. A003499, A174500, A174502.

LinearRecurrence[{0,7,0,-7,0,1},{1,4,1,32,1,196},50] (* Harvey P. Dale, Jul 14 2021 *)

{a(n)=local(L=sum(m=1,2*n+1000,1./(m*round((3+sqrt(8))^m+(3-sqrt(8))^m))));contfrac(exp(L))[n]}

Vec(-x*(x^4+4*x^3-6*x^2+4*x+1)/((x-1)*(x+1)*(x^2-2*x-1)*(x^2+2*x-1)) + O(x^50)) \\ Colin Barker, May 11 2016

a(2n-1) = 1, a(2n) = A003499(n) - 2, for n>=1 [conjecture].
The above conjectures are correct. See the Bala link for details. - Peter Bala, Jan 08 2013
a(n) = 7*a(n-2)-7*a(n-4)+a(n-6). G.f.: -x*(x^4+4*x^3-6*x^2+4*x+1) / ((x-1)*(x+1)*(x^2-2*x-1)*(x^2+2*x-1)). - Colin Barker, Jan 20 2013
a(n) = (((-1-sqrt(2))^n+(1-sqrt(2))^n+(sqrt(2)-1)^n+(1+sqrt(2))^n-4))/2 for n even. - Colin Barker, May 11 2016

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

Original entry on oeis.org

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)

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]

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)

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A174501 Continued fraction expansion for exp( Sum_{n>=1} 1/(n*A003499(n)) ), where A003499(n) = (3+sqrt(8))^n + (3-sqrt(8))^n.

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

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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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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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