A219010 -id:A219010 - cp's OEIS frontend

A219011 Denominators in a product expansion for sqrt(5).

5, 15005, 792070839820228500005

Offset: 0

Peter Bala, Nov 09 2012

Apart from the initial term same as A145275.
a(3) has 105 digits and a(4) has 523 digits.
The product expansion in question is sqrt(5) = Product_{n >= 0} (1 + 2*A219010(n)/A219011(n)) = (1 + 6/5)*(1 + 246/15005)*(1 + 56287506246/792070839820228500005)*....

Amiram Eldar, Table of n, a(n) for n = 0..4

Cf. A000032, A000045, A145275, A219010, A219013, A219015.

a[n_] := LucasL[4*5^n] - LucasL[2*5^n] + 1; Array[a, 3, 0] (* Amiram Eldar, Jul 20 2025 *)

A219011(n):=fib(5^(n+1))/fib(5^n)$
makelist(A219011(n),n,0,3);

a(n) = Fibonacci(5^(n+1))/Fibonacci(5^n).
a(n) = A219010(n)^2 - A219010(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) = 5.
a(n) = Lucas(4*5^n) - Lucas(2*5^n) + 1. - Ehren Metcalfe, Jul 29 2017

A219012 Numerators in a product expansion for sqrt(3).

Original entry on oeis.org

4, 724, 198924689265124, 311489521560905211009923410118036749078665998068304623331823899816643124

Offset: 0

Peter Bala, Nov 09 2012

a(4) has 358 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)); a(n) and b(n) are integer sequences when x is a positive integer.
The present case is when x = 2. The denominators b(n) are in A219013. See also A219010 (x = 1) and A219014 (x = 4).

			The first three terms of the product give 70 correct decimal places of sqrt(3): (1 + 2*4/11)*(1 + 2*724/523451)*(1 + 2*198924689265124/39571031999225940638473470251) = 1.73205 08075 68877 29352 74463 41505 87236 69428 05253 81038 06280 55806 97945 19330 (0...).

Amiram Eldar, Table of n, a(n) for n = 0..4

Cf. A019973, A219010, A219014, A219013.

RecurrenceTable[{a[n+1] == a[n]^5 - 5*a[n]^3 + 5*a[n], a[0] == 4}, a, {n, 0, 3}] (* Amiram Eldar, Jul 20 2025 *)

Let tau = 2 + sqrt(3). 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) = 4.
a(n) = 2*T(5^n,2), where T(n,x) denotes the n-th Chebyshev polynomial of the first kind. - Peter Bala, Mar 29 2022
Let b(n) = a(n) - 4. The sequence {b(n)} appears to be a strong divisibility sequence, that is, gcd(b(n),b(m)) = b(gcd(n,m)) for n, m >= 1. - Peter Bala, Dec 06 2022

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

Original entry on oeis.org

6, 6726, 13765255184676885126

Offset: 0

Peter Bala, Nov 09 2012

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).

			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...).

Amiram Eldar, Table of n, a(n) for n = 0..4

Cf. A156035, A219010, A219012, A219015.

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 *)

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.

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A219011 Denominators in a product expansion for sqrt(5).

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A219012 Numerators in a product expansion for sqrt(3).

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

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