A174503 -id:A174503 - cp's OEIS frontend

A174504 Continued fraction expansion for exp( Sum_{n>=1} 1/(n*A002203(n)) ), where A002203(n) = (1+sqrt(2))^n + (1-sqrt(2))^n.

1, 1, 5, 1, 13, 33, 1, 81, 197, 1, 477, 1153, 1, 2785, 6725, 1, 16237, 39201, 1, 94641, 228485, 1, 551613, 1331713, 1, 3215041, 7761797, 1, 18738637, 45239073, 1, 109216785, 263672645, 1, 636562077, 1536796801, 1, 3710155681, 8957108165, 1

Offset: 0

Paul D. Hanna, Mar 21 2010

			Let L = Sum_{n>=1} 1/(n*A002203(n)) or, more explicitly,
L = 1/2 + 1/(2*6) + 1/(3*14) + 1/(4*34) + 1/(5*82) + 1/(6*198) +...
so that L = 0.6182475539420223547415636201969551910173293917288...
then exp(L) = 1.8556732235071087707741415885016794127474675663938...
equals the continued fraction given by this sequence:
exp(L) = [1;1,5,1,13,33,1,81,197,1,477,1153,1,2785,6725,1,...]; i.e.,
exp(L) = 1 + 1/(1 + 1/(5 + 1/(1 + 1/(13 + 1/(33 + 1/(1 +...)))))).
Compare these partial quotients to A002203(n), n=1,2,3,...:
[2,6,14,34,82,198,478,1154,2786,6726,16238,39202,94642,228486,...].

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
P. Bala, Some simple continued fraction expansions for an infinite product, Part 2
Index entries for linear recurrences with constant coefficients, signature (0,0,7,0,0,-7,0,0,1).

Cf. A002203 (companion Pell numbers), A174500, A174503, A174505.

LinearRecurrence[{0,0,7,0,0,-7,0,0,1},{1,1,5,1,13,33,1,81,197},40] (* Harvey P. Dale, Sep 15 2016 *)

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

a(3n-3) = 1, a(3n-2) = A002203(2n-1) - 1, a(3n-1) = A002203(2n) - 1, for n>=1 [conjecture].
From Colin Barker, Jan 20 2013: (Start)
a(n) = 7*a(n-3) - 7*a(n-6) + a(n-9).
G.f.: -(x^2-x+1)*(x^6-2*x^5-2*x^4-2*x^3+6*x^2+2*x+1) / ((x-1)*(x^2+x+1)*(x^6-6*x^3+1)). (End)
From Peter Bala, Jan 25 2013: (Start)
The above conjectures are correct. The real number exp( Sum {n>=1} 1/(n*A002203(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(2) - 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 = 2. See the Bala link for details.
The theory also provides the simple continued fraction expansion of the numbers F({sqrt(2) - 1}^(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(2) - 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)

A087799 a(n) = 10*a(n-1) - a(n-2), starting with a(0) = 2 and a(1) = 10.

Original entry on oeis.org

2, 10, 98, 970, 9602, 95050, 940898, 9313930, 92198402, 912670090, 9034502498, 89432354890, 885289046402, 8763458109130, 86749292044898, 858729462339850, 8500545331353602, 84146723851196170, 832966693180608098, 8245520207954884810

Offset: 0

Nikolay V. Kosinov (kosinov(AT)unitron.com.ua), Oct 11 2003

a(n+1)/a(n) converges to (5+sqrt(24)) = 9.8989794... a(0)/a(1)=2/10; a(1)/a(2)=10/98; a(2)/a(3)=98/970; a(3)/a(4)=970/9602; ... etc. Lim a(n)/a(n+1) as n approaches infinity = 0.10102051... = 1/(5+sqrt(24)) = (5-sqrt(24)).
Except for the first term, positive values of x (or y) satisfying x^2 - 10xy + y^2 + 96 = 0. - Colin Barker, Feb 25 2014
A triangle whose sides are a(n) - 1, a(n) and a(n) + 1 is nearly Fleenor-Heronian since its area is the product of an integer and the square root of 2. See A003500. - Charlie Marion, Dec 18 2020

			a(4) = 9602 = 10*a(3) - a(2) = 10*970 - 98 = (5+sqrt(24))^4 + (5-sqrt(24))^4.

T. D. Noe, Table of n, a(n) for n = 0..200
Peter Bala, Some simple continued fraction expansions for an infinite product, Part 1
Hacène Belbachir, Soumeya Merwa Tebtoub, and László Németh, Ellipse Chains and Associated Sequences, J. Int. Seq., Vol. 23 (2020), Article 20.8.5.
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 (10,-1).

Cf. A005248, A006242, A036336, A086927, A135927, A174503.

I:=[2,10]; [n le 2 select I[n] else 10*Self(n-1) - Self(n-2): n in [1..30]]; // G. C. Greubel, Nov 07 2018

a[0] = 2; a[1] = 10; a[n_] := 10a[n - 1] - a[n - 2]; Table[ a[n], {n, 0, 17}] (* Robert G. Wilson v, Jan 30 2004 *)
LinearRecurrence[{10,-1}, {2,10}, 30] (* G. C. Greubel, Nov 07 2018 *)

polsym(x^2 - 10*x + 1,20) \\ Charles R Greathouse IV, Jun 11 2011

{a(n) = 2 * real( (5 + 2 * quadgen(24))^n )}; /* Michael Somos, Feb 25 2014 */

[lucas_number2(n,10,1) for n in range(27)] # Zerinvary Lajos, Jun 25 2008

a(n) = (5+sqrt(24))^n + (5-sqrt(24))^n.
G.f.: (2-10*x)/(1-10*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 = 5 - sqrt(24). This sequence gives the simple continued fraction expansion of 1 + F(alpha) = 2.09989 80642 72052 68138 ... = 2 + 1/(10 + 1/(98 + 1/(970 + ...))).
Also F(-alpha) = 0.89989 78538 78393 34715 ... has the continued fraction representation 1 - 1/(10 - 1/(98 - 1/(970 - ...))) and the simple continued fraction expansion 1/(1 + 1/((10-2) + 1/(1 + 1/((98-2) + 1/(1 + 1/((970-2) + 1/(1 + ...))))))).
F(alpha)*F(-alpha) has the simple continued fraction expansion 1/(1 + 1/((10^2-4) + 1/(1 + 1/((98^2-4) + 1/(1 + 1/((970^2-4) + 1/(1 + ...))))))). Cf. A174503 and A005248. (End)
a(-n) = a(n). - Michael Somos, Feb 25 2014
From Peter Bala, Oct 16 2019: (Start)
8*Sum_{n >= 1} 1/(a(n) - 12/a(n)) = 1.
12*Sum_{n >= 1} (-1)^(n+1)/(a(n) + 8/a(n)) = 1.
Series acceleration formulas for sums of reciprocals:
Sum_{n >= 1} 1/a(n) = 1/8 - 12*Sum_{n >= 1} 1/(a(n)*(a(n)^2 - 12))  and
Sum_{n >= 1} (-1)^(n+1)/a(n) = 1/12 + 8*Sum_{n >= 1} (-1)^(n+1)/(a(n)*(a(n)^2 + 8)).
Sum_{n >= 1} 1/a(n) = ( (theta_3(5-sqrt(24)))^2 - 1 )/4 and
Sum_{n >= 1} (-1)^(n+1)/a(n) = ( 1 - (theta_3(sqrt(24)-5))^2 )/4, where theta_3(x) = 1 + 2*Sum_{n >= 1} x^(n^2) (see A000122). Cf. A153415 and A003499. (End)
E.g.f.: 2*exp(5*x)*cosh(2*sqrt(6)*x). - Stefano Spezia, Oct 18 2019
From Peter Bala, Mar 29 2022: (Start)
a(n) = 2*T(n,5), where T(n,x) denotes the n-th Chebyshev polynomial of the first kind.
a(2^n) = A135927(n+1) and a(3^n) = A006242(n+1), both for n >= 0. (End)

More terms from Colin Barker, Feb 25 2014

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]

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)

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A174504 Continued fraction expansion for exp( Sum_{n>=1} 1/(n*A002203(n)) ), where A002203(n) = (1+sqrt(2))^n + (1-sqrt(2))^n.

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A087799 a(n) = 10*a(n-1) - a(n-2), starting with a(0) = 2 and a(1) = 10.

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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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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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