Understanding Example 16 printing the powers of 2 from Big O notation - Cracking the Coding Interview

Had the author missed to compute I/O calls?

The following function prints the powers of 2 from 1 through n (inclusive). For example, if n is 4, it would print 1,2, and 4. What is its runtime?

int powersOf2(int n) {
  if (n < 1) {
    return 0;
  } else if (n == 1) {
    System.out.println(1);
    return 1; 
  } else {
    int prev = powersOf2(n / 2); 
    int curr =prev * 2; 
    System.out.println(curr); 
    return curr;
  }
}

the runtime is O(log n)

According to the Example 12 (string permutations) the length of argument of System.out.println() call makes sense:

Executing line 7 takes O(n) time since each character needs to be printed

From I/O perspective we need to print powers of 2 from 0 to K, where K is [log(N)], the amount of characters to print for 2^X is [1 + X/log(10)], so the total amount of characters to print is [K + 1 + K*(K+1)/2log(10)] and the runtime is O(log²N) but not O(log N)

PS.

The Example 15 - printing memoized Fibonacci numbers seems to have the same drawback:

void allFib(int n) {
  int[] memo = new int[n + 1]; 
  for (int i = 0; i < n; i++) {
    System.out.println(i + ": " + fib(i, memo));
  }
}

int fib(int n, int[] memo) {
  if (n <= 0) return 0;
  else if (n == 1) return 1;
  else if (memo[n] > 0) return memo[n];
  memo[n] = fib(n - 1, memo) + fib(n - 2, memo);
  return memo[n];
}

We're doing a constant amount of work N times, so this is O(n) time.

The amount of characters to print for the sequence of first N Fibonacci numbers ~ N², so the runtime should be O(N²).