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

0,2

LINKS

Table of n, a(n) for n=0..27.

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 (0,0,67,0,0,-67,0,0,1).

FORMULA

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)

EXAMPLE

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,...].

PROG

(PARI) {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]}

CROSSREFS

Cf. A086594, A174502, A174507, A174509.

Sequence in context: A197787 A024095 A274823 * A253802 A220597 A264875

Adjacent sequences:  A174505 A174506 A174507 * A174509 A174510 A174511

KEYWORD

cofr,nonn,easy

AUTHOR

Paul D. Hanna, Mar 21 2010

STATUS

approved

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Last modified July 23 16:10 EDT 2019. Contains 325258 sequences. (Running on oeis4.)