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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. 3
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 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,2

LINKS

Table of n, a(n) for n=0..24.

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 (0,0,103,0,0,-103,0,0,1).

FORMULA

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)

EXAMPLE

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,...].

MATHEMATICA

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 *)

PROG

(PARI) {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]}

CROSSREFS

Cf. A086927, A174503, A174508, A174510.

Sequence in context: A060150 A202833 A287039 * A103461 A101563 A269732

Adjacent sequences:  A174506 A174507 A174508 * A174510 A174511 A174512

KEYWORD

cofr,nonn,easy

AUTHOR

Paul D. Hanna, Mar 21 2010

STATUS

approved

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Last modified July 22 15:12 EDT 2019. Contains 325224 sequences. (Running on oeis4.)