cp's OEIS Frontend

Let h(x) = x*(floor(1/x) + (floor(1/x))^2) - floor(1/x). Let x = 1/2*(1 + sqrt(5)) - 1. Then a(1) = 1, a(2) = ceiling(1/x) and, for n >= 1, a(n+2) = floor(1/h^(n-1)(x))*(1 + floor(1/h^(n)(x))).

Recurrence equations: For n >= 3, a(3*n) = 2. For n >= 4 we have a(3*n+2) = 2*a(3*n+1) - 2 and a(3*n+1) = 2*(a(3*n-2) - 1)^2.

Put P(n) = Product_{k = 1..n} a(k). Then we have the Egyptian fraction series expansion sqrt(2) = Sum_{n>=1} 1/P(n) = 1 + 1/2 + 1/(2*5) + 1/(2*5*8) + 1/(2*5*8*3) + 1/(2*5*8*3*4) + .... For n >= 2, the error made in truncating this series to n terms is less than the n-th term.

Let h(x) = x*(floor(1/x) + (floor(1/x))^2) - floor(1/x) where x = exp(1) - 2, then a(1) = a(2) = 1, a(3) = ceiling(1/x) and, for n >= 1, a(n+3) = floor(1/h^(n-1)(x))*(1 + floor(1/h^(n)(x))).

Put P(n) = Product_{k = 1..n} a(k). Then we have the Egyptian fraction series expansion exp(1) = Sum_{n>=1} 1/P(n) = 1/1 + 1/1 + 1/2 + 1/(2*3) + 1/(2*3*4) + 1/(2*3*4*5) + 1/(2*3*4*5*8) + .... For n >= 3, the error made in truncating this series to n terms is less than the n-th term.

Let h(x) = x*(floor(1/x) + (floor(1/x))^2) - floor(1/x). Let x = log(2). Then a(1) = 1 + floor(1/x) and, for n >= 1, a(n+1) = floor(1/h^(n-1)(x))*(1 + floor(1/h^(n)(x))).

Put P(n) = Product_{k = 1..n} a(k). Then we have the Egyptian fraction series expansion log(2) = Sum_{n>=1} 1/P(n) = 1/2 + 1/(2*3) + 1/(2*3*8) + 1/(2*3*8*6) + 1/(2*3*8*6*2) + .... The error made in truncating this series to n terms is less than the n-th term.

Let h(x) = x*(floor(1/x) + (floor(1/x))^2) - floor(1/x). Let x = gamma (see A001620). Then a(1) = 1 + floor(1/x) and, for n >= 1, a(n+1) = floor(1/h^(n-1)(x))*(1 + floor(1/h^(n)(x))).

Put P(n) = Product_{k = 1..n} a(k). Then we have the Egyptian fraction series expansion sqrt(2) = Sum_{n>=1} 1/P(n) = 1/2 + 1/(2*7) + 1/(2*7*18) + 1/(2*7*18*4) + 1/(2*7*18*4*2) + .... The error made in truncating this series to n terms is less than the n-th term.

Let h(x) = x*(floor(1/x) + (floor(1/x))^2) - floor(1/x). Let x = sqrt(2) - 1. Then a(1) = 1, a(2) = ceiling(1/x) and, for n >= 1, a(n+2) = floor(1/h^(n-1)(x))*(1 + floor(1/h^(n)(x))).

Put P(n) = Product_{k = 1..n} a(k). Then we have the Egyptian fraction series expansion sqrt(2) = Sum_{n>=1} 1/P(n) = 1 + 1/3 + 1/(3*6) + 1/(3*6*4) + 1/(3*6*4*2) + 1/(3*6*4*2*2) + .... For n >= 2, the error made in truncating this series to n terms is less than the n-th term.