Also, we are now twittering at MathFactor; each of the authors has an account of his own; mine is CGoodmanStrauss. You can tag solutions and comments with #mathfactor. See you there!

For example, in the sequence 1, 2, 4, 8, 16, … each term is exactly 2/3rds of the previous term, plus 1/3rd of the following term. What other sequences have exactly that property?

For a given value p, what sequences s₁, s₂, s₃, … s_n have the property that each s_k = p s_(k-1) + (1-p) s_(k+1) ? Does knowing s₁ also fix the remaining terms? 

Amusingly, this actually will help with last weeks puzzle!

The sum of the digits of 351 is 9 and the sum of the digits of 2 x 351 = 702 is also 9.

1) If a number has this property, can we always rearrange its digits and obtain another number with this property (513, 135, etc all have it)

2) Which powers of 2 have this property?

3) And most of all, can you give a simple characterization of the numbers with this property, in terms of just the digits themselves?

Imagine that you have immortal rabbits, Bugs Bunny’s version of Olympus perhaps.  Even if they are immortal however rabbits are famous for one thing.  They breed like, well rabbits.  Some of the rabbits are children and some adults and are divided into pairs.  Each month any child pairs become adults and any adult pairs breed to produce a new child pair.  They are immortal so no pair ever dies.  These rabbits are also a little odd.  They live on a line (don’t complain, this is no more ludicrous than that they are immortal!), but can shuffle along.  Also if you are worried about inbreeding, the male rabbit leave the family hutch and shuffle along the line past others until they find a suitably unrelated mate.  Why we would be worried about inbreeding in immortal rabbits living on a line escapes me!

Anyway we start with one pair of children.  Lets put a c.  After a month they become adults, a.  Another month passes and they now have a pair of children, but are still there themselves.  We therfore have the original pair and a pair of children: ac.  Next month the adults have another pair of children and the children become adults: aca.  Can you see how this will work?  Each month the children become adults so we replace every c with and a, each pair of adults has a new pair of children but stays as adults, so we replace every a with ac.  We can continue to get longer and longer sequences of rabbits on this line:

aca  to acaac to acaacaca to acaacacaacaac….

Now some puzzles.  Given a line with 21 adult pairs and 13 child pairs, how many pairs of adults and children would there be after one month?

Given p adults and q children how many adults and children will there be after one month?

Finally a more difficult one.  How will the ratio of adults to children behave month on month?  Will it

a) Get closer and closer to a particular number?

b) Keep on changing without pattern?

In either case can you say more?

Your task is to locate a gold coin.

You will compare the coins in “passes” with each coin not being compared more than once in a pass. In a pass, every coin can be a member of a comparison or not, but any particular coin can’t be part of more than one comparison during the pass. Your goal is to minimize the number of passes required to be sure that you locate a single gold coin.

You should be able to describe how many passes (at most) your solution will require rather than the number of passes increasing arbitrarily with the number of coins that turn out to be in the bag.

A couple more examples: if you give me 0, I’ll give you 0, and 0+0 = 0 x 0.

If you give me 2, I’ll give you 2 and 2 + 2 = 2 x 2.

If you give me -1, I’ll give you 1/2 since -1 + 1/2 = -1 x 1/2.

Can I always give a response to your number, or is there a way to stump me?