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A219010 Numerators in a product expansion for sqrt(5). 4
3, 123, 28143753123, 17656721319717734662791328845675730903632844218828123 (list; graph; refs; listen; history; text; internal format)
OFFSET
0,1
COMMENTS
a(4) has 262 digits.
Iterating the algebraic identity sqrt(1 + 4/x) = (1 + 2*(x + 2)/(x^2 + 3*x + 1)) * sqrt(1 + 4/(x*(x^2 + 5*x + 5)^2)) produces the rapidly converging product expansion sqrt(1 + 4/x) = Product_{n = 0..inf} (1 + 2*a(n)/b(n)), where a(n) and b(n) are integer sequences when x is an integer: a(n) satisfies the recurrence a(n+1) = a(n)^5 - 5*a(n)^3 + 5*a(n) with the initial condition a(0) = x + 2 and b(n) satisfies the recurrence b(n+1) = 5/2*(b(n)^4 - b(n)^2)*sqrt(4*b(n) + 5) + b(n)^5 + 15/2*b(n)^4 - 25/2*b(n)^2 + 5 with the initial condition b(0) = x^2 + 3*x + 1.
The present case is when x = 1. The denominators b(n) are in A219011. See also A219012 (x = 2) and A219014 (x = 4).
For another product expansion for sqrt(5) see A002814.
LINKS
FORMULA
Let tau = (3 + sqrt(5))/2. Then a(n) = tau^(5^n) + 1/tau^(5^n).
a(n) = (1/2)*(1 + 5*(Fibonacci(5^n)*Fibonacci(5^n + 1) + (Fibonacci(5^n - 1))^2)).
Recurrence equation: a(n+1) = a(n)^5 - 5*a(n)^3 + 5*a(n) with initial condition a(0) = 3.
a(n) = Lucas(2*5^n). - Ehren Metcalfe, Jul 29 2017
EXAMPLE
The first three terms of the product give 51 correct decimal places of sqrt(5): (1 + 2*3/5)*(1 + 2*123/15005)*(1 + 2*28143753123/792070839820228500005) = 2.23606 79774 99789 69640 91736 68731 27623 54406 18359 61152 5(4...).
MATHEMATICA
Table[(1 + 5*(Fibonacci[5^n] Fibonacci[5^n + 1] + Fibonacci[5^n - 1]^2))/2, {n, 0, 3}] (* or *)
Table[LucasL[2*5^n], {n, 0, 3}] (* Michael De Vlieger, Jul 30 2017 *)
PROG
(Maxima) A219010(n):=1/2*(1 + 5*(fib(5^n)*fib(5^n+1)+(fib(5^n - 1))^2))$
makelist(A219010(n), n, 0, 3); /* Martin Ettl, Nov 12 2012 */
CROSSREFS
Sequence in context: A258658 A222186 A124246 * A037118 A085531 A130614
KEYWORD
nonn,easy
AUTHOR
Peter Bala, Nov 09 2012
STATUS
approved

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Last modified March 28 16:12 EDT 2024. Contains 371254 sequences. (Running on oeis4.)