1, 1, 2, 11, 12, 13, 112, 113, 114, 1104, 2004, 2012, 1031, 123, ...

Name the next few numbers in the sequence. What is the ultimate fate of this sequence?

(In reply to

Answer by K Sengupta)

Let T(n) = nth term of the sequence.

Then,

Units digit of T(n) = # 1's in T(n-1) and T(n-2) taken together.

Tens digit of T(n) = # 2's in T(n-1) and T(n-2) taken together.

Hundred digit of T(n) = # 3's in T(n-1) and T(n-2) taken together, and so on.

Accordingly, we must have:

T(13) = 1031 (given)

T(14) = 123 (given), so that:

T(15) = 213

T(16) = 222

T(17) = 141

T(18) = 1032

T(19) = 1113

T(20) = 214

T(21) = 1114

T(22) = 2014

T(23) = 2014

T(24) = 2022

T(25) = 1041

T(26) = 1031

T(27) = 1032

T(28) = 213

T(29) = 222

T(30) = 141

-----------

----------, and so on.

Consequently, we can now assert that:

(i) The respective missing 15th term, 16th term and the 17th term are 213, 222 and 141.

(ii) From the 18th term onwards, the given sequence will loop indefinitely.