What would be the Nth number? How can I find Nth number such that it is divisible by a OR b? I was thinking to find gcd (Greatest Common Divisor) for both the given numbers (a and b) and use that but I couldn't think of any approach as such.
 A: Suppose $a$ and $b$ are relatively prime and without loss of generality that $a \leq b$. Put $q = b/a$. The sequence will be 
$[a, 2a, 3a, 4a, \ldots, \lfloor q\rfloor a, b, \;\;\;(\lfloor q\rfloor+1)a, (\lfloor q\rfloor+2)a, \ldots, 2\lfloor q\rfloor a, 2b, \ldots \ldots,ab, \ldots ]$
For example, if $a=2$ and $b=11$, the sequence is:
$$[2,4,6,8,10,11,12,14,16,18,20,22,\ldots]$$
To compute arbitrary terms in this sequence, we need to keep track of (1) the number of multiples of $a$ we've seen, (2) the number of multiples of $b$ we've seen, (3) the number of multiples of $ab$ we've seen.
Consider the $n$th term in this sequence.


*

*Because $a$ and $b$ are relatively prime, we won't see a number in this sequence that's a multiple of $a$ and $b$ until after we've seen $b$ multiples of $a$, and $a$ multiples of $b$— so every $(a+b-1)$ terms. The number of multiples of both we've seen is therefore $C(n)\equiv \lfloor n/(a+b-1)\rfloor$.

*Approximately every $\lceil q\rceil $th term in this sequence is a multiple of $b$. Hence the number of multiples of $b$ we've seen is $B(n)\equiv \lfloor n/\lceil q\rceil \rfloor$.

*Every term in the sequence is a multiple of $a$, except the multiples of $b$ alone. Hence the number of multiples of $a$ we've seen is $A(n) = n-B+C$.
The computation is as follows. Given $n$, compute $A(n)$, $B(n)$, $C(n)$. If $n$ is a multiple of $\lceil b/a\rceil$, return $B(n)\cdot b$. Otherwise, return $A(n)\cdot a$.
If you wanted to expand the definitions of everything to write a formula solely in terms of $n$, $a$, and $b$, you might say:
$$D_n \equiv \begin{cases}b\lfloor n/\lceil b/a\rceil\rfloor& \text{if } \lceil b/a\rceil \text{ divides }n\\ a\left(n- \lfloor n/\lceil b/a\rceil\rfloor  + \lfloor n/(a+b-1)\rfloor\right)&\text{otherwise}\end{cases}$$

If $a$ and $b$ are not relatively prime, compute their greatest common divisor $d$. Then we can write $a = kd$ and $b=\ell d$, where $k$ and $\ell$ are relatively prime.
To compute the sequence for $a$ and $b$, instead compute the sequence for $k$ and $\ell$, but multiply every term in the sequence by $d$.
A: The process goes through this path
$(1.)$Note that the the position of multiples (of $a$ or $b$) throughout the number line of $\Bbb N$ are repeating image of  the positions of them in $(0,$ $lcm(a,b)]$ interval.
$(2.)$We know number of multiples of $a$ in $(x,y]$ is $\lfloor \frac{y}{a}\rfloor -\lfloor \frac{x}{a}\rfloor +1$
$(3.)$Then using $(2)$ in the intervals $(0,b],(b,2b],(2b,3b],...,(lcm(a,b)-b,lcm(a,b)]$ and using $(1)$, I got the following result.
The $N^{th}$ number is,
$$j[a,b]+\frac{bn+(N-ij-n)a}{2}+(-1)^{\lfloor\frac{\lfloor\frac{nb}{a}\rfloor+n}{N-ij}\rfloor}\cdot \frac{(N-ij-n)a-bn}{2}$$ where $[a,b]=$lcm of a and b ,  , $$j=\lfloor\frac{N-1}{i}\rfloor$$ where $$i=\frac{a+b-(a,b)}{(a,b)}$$  $n$ is the maximum integer satisfying $\lfloor n\cdot \frac{b}{a}\rfloor +n\le N-ij$  and $(a,b)$ being gcd of $a$ and $b$.
A: I saw this problem on SPOJ I am not sure if it the efficient way to solve this problem or not but I got my Solution accepted.
Here is the way to solve this problem.
if we generate a sequence with a,b and N.
 the first term will be min(a,b) and the last term will be N*max(a,b).
all the numbers which are either divisible by a or b will be in this range.
So we can do a binary search to find that which is Nth term of the sequence.
it will be something like this.
    l = min(a,b)
    u =  n*max(a,b)+1
    while(l != u){ 
        mid = l + (u - l) /2
        if(getDiv(a,b,mid-1) >= n)u = mid
        else l = mid + 1
    }

So the Nth term of  the sequence will be mid which is the required answer. 
where getDiv(a,b,m) are the integers between 0 and m are divisible by a or b".
floor(m/a) + floor(m/b) - floor(m/Lcm(a,b));
A: $$D_n \equiv \begin{cases}b\lfloor n/\lceil b/a\rceil\rfloor& \text{if } \lceil b/a\rceil \text{ divides }n\\ a\left(n- \lfloor n/\lceil b/a\rceil\rfloor  + \lfloor n/(a+b-1)\rfloor\right)&\text{otherwise}\end{cases}$$
This logic doesn't work for $a=5$, $b=9$ & $n=10$ right ??
is there any way to modify this to make this work or am I missing something.
Regards,
Devaraj
