Find the nth number in the increasing sequence formed by 0,2,4,6,8?

Question

We have an increasing sequence in which each element is consist of even digits only (0, 2, 4, 6, 8). How can we find the nth number in this sequence

Is it possible to find nth number in this sequence in O(1) time.

Sequence: 0, 2, 4, 6, 8, 20, 22, 24, 26, 28, 40, 42, 44, 46, 48, 60, 62, 64, 66, 68, 80, 82, 84, 86, 88, 200, 202 and so on.

Contrail

where n is start form zero or one? – Jilljillana

I'm voting to close this question as off-topic because it's a dump of a homework assignment that shows ZERO effort from the OP. – Scheming

Your first stop on the web when looking at this exact type problem is the On-line Encyclopedia of Integer Sequences (OEIS), which has this to say for the sequence 0, 2, 4, 6, 8, 20, 22, 24, 26, 28, 40, 42. – Gentle

Answer 1

The nth number in this sequence is n in base 5, with the digits doubled.

def base5(n):
    if n == 0: return
    for x in base5(n // 5): yield x
    yield n % 5

def seq(n):
    return int(''.join(str(2 * x) for x in base5(n)) or '0')

for i in xrange(100):
    print i, seq(i)

This runs in O(log n) time. I don't think it's possible to do it in O(1) time.

It can be simplified a bit by combining the doubling of the digits with the generation of the base 5 digits of n:

def seq(n):
    return 10 * seq(n // 5) + (n % 5) * 2 if n else 0

Dhammapada

@Paul Hankin : Sir may you please help me with how did you solved this sequence ? – Meaning

@Ram it's not much help to you, but I saw straight away the relationship with numbers in base 5, and from there, writing the code is relatively straightforward. Lasse in the comments above suggested checking the sequence in OEIS -- that'd be a good way to find the pattern if you didn't see it. – Dhammapada

Answer 2

int Code()
{
    k=0;
   for(i=0;i<=10000;i++)
   {
       count=0;
       n=i;
       while(n!=0)
       {
           c=n%10;
           n=n/10;
           if(c%2!=0)
           {
               count=1;
           }
       }
       if(count==0)
       { a[k]=i;
       k++;}
    }
 }

Bandbox