(In reply to
re: Clarification? by Ady TZIDON)
Well I suppose it would be the binary representation of (2n+1), but only for those cases where the binary representation of n is divisible by n, otherwise the answer is just the binary representation of n.
Let bin(n) be the decimal number formed by the binary representation of n.
For example, for n = 2, bin(n) = 10. The smallest number that contains the digits "10" and isn't divisible by 2 is 101, which is the binary representation of 5, which is 2(2) + 1.
On the other hand, for n = 3, bin(n) = 11. 11 is greater than 3, isn't divisible by 3, and contains the digits "11", so that's the solution.
The values n for which bin(n) = 0 mod n are 2, 4, 8, 10, 16, 20, 21, 32, 40, 42, 64, 80, 84, 100, 128, 160, 168, 200, 256, 273...
I don't immediately see any pattern there that would lead to a closed-form solution to your problem for all n.
|
|
Posted by tomarken
on 2014-04-03 10:33:09 |
re(2): Clarification?
