cp's OEIS Frontend

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.

Showing 1-3 of 3 results.

A219010 Numerators in a product expansion for sqrt(5).

Original entry on oeis.org

3, 123, 28143753123, 17656721319717734662791328845675730903632844218828123
Offset: 0

Views

Author

Peter Bala, Nov 09 2012

Keywords

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} (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.

Examples

			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...).
		

Crossrefs

Programs

  • 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 *)
  • 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 */

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

A219013 Denominators in a product expansion for sqrt(3).

Original entry on oeis.org

11, 523451, 39571031999225940638473470251
Offset: 0

Views

Author

Peter Bala, Nov 09 2012

Keywords

Comments

The product expansion in question is sqrt(3) = Product_{n >= 0} (1 + 2*A219012(n)/A219013(n)) = (1 + 2*4/11)*(1 + 2*724/523451)*(1 + 2*198924689265124/39571031999225940638473470251)*....

Crossrefs

Programs

  • Mathematica
    RecurrenceTable[{a[n+1] == 5/2*(a[n]^4 - a[n]^2)*Sqrt[4*a[n] + 5] + a[n]^5 + 15/2*a[n]^4 - 25/2*a[n]^2 + 5, a[0] == 11}, a, {n, 0, 3}] (* Amiram Eldar, Jul 20 2025 *)

Formula

Let alpha = 1/2*(sqrt(2) + sqrt(6)) and put f(n) = 1/sqrt(6)*{alpha^n - (-1/alpha)^n}. Then a(n) = f(5^(n+1))/f(5^n).
a(n) = A219012(n)^2 - A219012(n) - 1.
Recurrence equation: a(n+1) = 5/2*(a(n)^4 - a(n)^2)*sqrt(4*a(n) + 5) + a(n)^5 + 15/2*a(n)^4 - 25/2*a(n)^2 + 5 with initial condition a(0) = 11.

A219014 Numerators in a product expansion for sqrt(2).

Original entry on oeis.org

6, 6726, 13765255184676885126
Offset: 0

Views

Author

Peter Bala, Nov 09 2012

Keywords

Comments

a(3) has 96 digits and a(4) has 479 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 a 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 a positive integer.
The present case is when x = 4. The denominators b(n) are in A219015. See also A219010 (x = 1) and A219012 (x = 2).

Examples

			The first two terms of the product give 18 correct decimal places of sqrt(2): (1 + 2*6/29)*(1 + 2*6726/45232349) = 1.41421 35623 73095 048(5...).
		

Crossrefs

Programs

  • Mathematica
    RecurrenceTable[{a[n+1] == a[n]^5 - 5*a[n]^3 + 5*a[n], a[0] == 6}, a, {n, 0, 3}] (* Amiram Eldar, Jul 20 2025 *)

Formula

Let tau = 3 + 2*sqrt(2). Then a(n) = tau^(5^n) + 1/tau^(5^n).
Recurrence equation: a(n+1) = a(n)^5 - 5*a(n)^3 + 5*a(n) with initial condition a(0) = 6.
Showing 1-3 of 3 results.