10進の回文数を生成して2進の回文数になっているかを調べます。
def pow(n :Int, e :Int) :Int = if(e == 0) 1 else { val m = pow(n, e / 2) if(e % 2 == 1) m * m * n else m * m } def palindrome(n :Int, start :Int = 1) :Iterator[Int] = n match { case 0 => Iterator(0) case 1 => Iterator.range(start, 10) case _ => { val f = pow(10, n - 1) + 1 for(d <- Iterator.range(start, 10); m <- palindrome(n - 2, 0)) yield d * f + m * 10 } } def binary(n :Int) :Iterator[Int] = if(n == 0) Iterator() else Iterator(n & 1) ++ binary(n >> 1) def number(it :Iterator[Int]) = it.foldLeft(0)((x, y) => (x << 1) | y) val N = 6 val it = Iterator.range(1, N + 1).flatMap(n => palindrome(n)) println (it.filter(n => number(binary(n)) == n).sum)
