## The Second Series (actually, probably 10th or 11th..)

Posted by Karl Sharman on July 29, 2011 – 10:43 am

Consider this series:

4, 5, 8, 8, 9, 9, 12, 13, 13, 13, 17, 18, …, …, …

What would the 50th, 63rd and 100th observations be?

August 1st, 2011 at 4:43 am

Really dislike these type of puzzles. Reminds me of the sequence where you have to sort numbers in alphabetical order.

Anyway my guess is:

55, 73 and 110

August 2nd, 2011 at 9:43 am

This is doing my head in, my best effort so far can reproduce most of the sequence using modulo and quotient functions which account nicely for the non-linearity, but i think i’m still a million miles from the answer.

So far i’ve investigated primes, binary representations, Qty and alphabetical positions of letters in the english names for each number, astrological star charts and even an Ouija board.

Anyone got any hints to steer me on the right path? Going to try some conway-style recursion next and maybe even roman notation just for kicks.

August 5th, 2011 at 11:26 am

I note I haven’t shown the answer – darn work getting in the way again!

Jan gets it right first time. The answer is

1 + 1 = 4

2 + 2 = 5

3 + 3 = 8 ands so on – the trick to the maths is that you need to add the number value to the number of letters in the word of the number….

1 + o-n-e (3) = 4

2 + t-w-o (3) = 5

3 + t-h-r-e-e (5) = 8 and so on…

August 10th, 2011 at 12:11 am

SECRET!!! haha