The numbers 1 through 15 have been re-ordered according to a certain logic. Deduce the logic and fill in the blanks.

15, 7, 11, 3, ___, ___, ___, 1, 14, 6, ___, ___, 12, 4, 8

(In reply to

Answer by K Sengupta)

At the outset, the first term of the sequence is 15, which is

represented as 1111 in base two (binary) system. Reversing the digits,

we also obtain 1111, which when converted to its decimal representation,

becomes 15.

The third term is 11, represented as 1011 in base 2. Reversing its

digits, we obtain 1101, which when converted back to base 10 is 13.

Accordingly, we can conjecture that the given numbers are obtained

by writing the decimal numbers 15, 14, 13, ...., 2, 1 in turn, in

their respective base two representation in 4 digits, reversing the

digits, and then converting each of the said rteversed numbers to

obtain the terms of the given sequence.

Let

S(n) = nth term of the given sequence, and:

B(p) = Binary representation of the decimal number p, in four digits.

R(p) = Number constituted by reversing the digits of B(p).

D(p) = Decimal (base 10) representation of the binary number R(p).

Accordingly, we constitute the following table:

p B(p) R(p) D(p)

15 1111 1111 15

14 1110 0111 7

13 1101 1011 11

12 1100 0011 3

11 1011 1101 13

10 1010 0101 5

9 1001 1001 9

8 1000 0001 1

7 0111 1110 14

6 0110 0110 6

5 0101 1010 10

4 0100 0010 2

3 0011 1100 12

2 0010 0100 4

1 0001 1000 8

From the above table, it is verified that S(p) = D(16-p) for all the

given values of p.

Consequently, the missing 5th term, 6th term and the 7th term are

respectively 13, 5 and 9; while the missing 11th term and the 12th

term are respectively 10 and 2.