Numerical sequence of 1 2 4 - algorithm

Question

I need help in providing an algorithm for a numerical sequence which should display a series of 1 2 4 and its consecutive summations.
e.g. If my input value is 20, it should display
1 2 4 8 9 11 15 16 18
Wherein
1 = 1
2 = 1 + 1
4 = 2 + 2
8 = 4 + 4
And the summation of 1 and 2 and 4 will repeat again starting with the present number which is 8 and so on..
9 = 8 + 1
11 = 9 + 2
15 = 11 + 4
16 = 15 + 1
18 = 16 + 2
As you can see, it should not proceed to 22 (18 + 4) since our sample input value is 20. I hope you guys get my point. I'm having a problem in designing the algorithms in the for loop. What I have now which is not working is
$input = 20;
for ($i = $i; $i < $input; $i = $i+$i) {
if($i==0){
$i = 4;
$i = $i - 3;
}elseif($i % 4 == 0){
$i = $i + 1;
}
print_r("this is \$i = $i<br><br>");
}
NOTE: Only one variable and one for loop is required, it will not be accepted if we use functions or arrays. Please help me, this is one of the most difficult problems I've encountered in PHP..

Answer 1

you can use the code
$input = 20;
$current = 1;
$val = 1;
while($val < $input){
print_r("this is \$val = $val\n");
$val = $val + $current;
$current = ($current == 4 ? 1 : $current*2);
}
see the online compiler

Answer 2

Since you have mentioned Only one variable and one for loop is required
Try this,
$input = 20;
for ($i = 1; $i < $input; $i) {
if($i>$input) break;
print_r("this is \$i = $i<br><br>");
$i=$i+1;
if($i>$input) break;
print_r("this is \$i = $i<br><br>");
$i=$i+2;
if($i>$input) break;
print_r("this is \$i = $i<br><br>");
$i=$i+4;
}
Online Compiler

Answer 3

def getSeq(n):
if n == 1:
return [1]
temp = [1]
seq = [ 1, 2, 4]
count, current, prev = 0, 0, 1
while True:
current = prev + seq[count]
if current > n:
break
prev = current
temp += [current]
count = (count + 1) % 3
return temp
print getSeq(20)
I'm pretty sure that this one is going to work
the case that we have to take care of is n == 1 and return a static result [1].
in other cases the second value is repeating circularly and adding up to previous value.

Answer 4

This Python solution should be implementable in any reasonable language:
limit = 20
n = 1 << 2
while n >> 2 < limit:
print(n >> 2)
n = (((n >> 2) + (2 ** (n & 3))) << 2) + ((n & 3) + 1) % 3
Perl Equivalent (using the style of for loop you expect):
$limit = 20;
for ($n = 1 << 2; $n >> 2 < $limit; $n = ((($n >> 2) + (2 ** ($n & 3))) << 2) + (($n & 3) + 1) % 3) {
print($n >> 2, "\n");
}
OUTPUT
1
2
4
8
9
11
15
16
18
EXPLANATION
The basic solution is this:
limit = 20
n = 1
i = 0
while n < limit:
print(n)
n = n + (2 ** i)
i = (i + 1) % 3
But we need to eliminate the extra variable i. Since i only cycles through 0, 1 and 2 we can store it in two bits. So we shift n up two bits and store the value for i in the lower two bits of n, adjusting the code accordingly.
Not only one variable and one for loop, no if statements either!