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It is possible to show that the square root of two can be expressed as an infinite continued fraction.

√ 2 = 1 + 1/(2 + 1/(2 + 1/(2 + … ))) = 1.414213…

By expanding this for the first four iterations, we get:

1 + 1/2 = 3/2 = 1.5

1 + 1/(2 + 1/2) = 7/5 = 1.4

1 + 1/(2 + 1/(2 + 1/2)) = 17/12 = 1.41666…

1 + 1/(2 + 1/(2 + 1/(2 + 1/2))) = 41/29 = 1.41379…

The next three expansions are 99/70, 239/169, and 577/408, but the eighth expansion, 1393/985, is the first example where the number of digits in the numerator exceeds the number of digits in the denominator.

In the first one-thousand expansions, how many fractions contain a numerator with more digits than denominator?

^{?}View Code RSPLUS

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 |
f <- function(i) { #x <- 1/2 x <- c(1,2) ## numerator, denominator x <- gmp::as.bigz(x) # to support large sized integer values if (i > 1) { for (j in 2:i) { #x= 1/(2+x) x <- rev(c(x[2] * 2 + x[1], x[2])) } } #x <- 1 + x x <- c(x[2] + x[1], x[2]) return(x) } dec.len <- function(d) { l <- length(unlist(strsplit(as.character(d), split=""))) return(l) } is.longer <- function(x) { # x is a vector of length 2 c(numerator, denomiator) num.len <- dec.len(x[1]) denom.len <- dec.len(x[2]) if (num.len > denom.len) { return(TRUE) } else { return(FALSE) } } idx <- sapply(1:1000, function(x) is.longer(f(x))) |

> sum(idx) [1] 153

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