This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A219010 #32 Jul 20 2025 08:59:42
%S A219010 3,123,28143753123,
%T A219010 17656721319717734662791328845675730903632844218828123
%N A219010 Numerators in a product expansion for sqrt(5).
%C A219010 a(4) has 262 digits.
%C A219010 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} (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.
%C A219010 The present case is when x = 1. The denominators b(n) are in A219011. See also A219012 (x = 2) and A219014 (x = 4).
%C A219010 For another product expansion for sqrt(5) see A002814.
%H A219010 Amiram Eldar, <a href="/A219010/b219010.txt">Table of n, a(n) for n = 0..4</a>
%F A219010 Let tau = (3 + sqrt(5))/2. Then a(n) = tau^(5^n) + 1/tau^(5^n).
%F A219010 a(n) = (1/2)*(1 + 5*(Fibonacci(5^n)*Fibonacci(5^n + 1) + (Fibonacci(5^n - 1))^2)).
%F A219010 Recurrence equation: a(n+1) = a(n)^5 - 5*a(n)^3 + 5*a(n) with initial condition a(0) = 3.
%F A219010 a(n) = Lucas(2*5^n). - _Ehren Metcalfe_, Jul 29 2017
%e A219010 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...).
%t A219010 Table[(1 + 5*(Fibonacci[5^n] Fibonacci[5^n + 1] + Fibonacci[5^n - 1]^2))/2, {n, 0, 3}] (* or *)
%t A219010 Table[LucasL[2*5^n], {n, 0, 3}] (* _Michael De Vlieger_, Jul 30 2017 *)
%o A219010 (Maxima) A219010(n):=1/2*(1 + 5*(fib(5^n)*fib(5^n+1)+(fib(5^n - 1))^2))$
%o A219010 makelist(A219010(n),n,0,3); /* _Martin Ettl_, Nov 12 2012 */
%Y A219010 Cf. A000032, A000045, A002814, A104457, A219011, A219012, A219014.
%K A219010 nonn,easy
%O A219010 0,1
%A A219010 _Peter Bala_, Nov 09 2012