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monkukr
05-18-2008, 01:38 PM
How many arrangements are there with http://www.mcs.surrey.ac.uk/Personal/R.Knott/Fibonacci/fibpuzzles2.html#pennies2
5 pennies on bottom...I got 24 using no gaps allowed using the stated rules.
How do you calculate other than drawing out solution?

Part 2 - which puzzle is imposter??

stapel
05-18-2008, 03:08 PM
From the referenced page, the first exercise is as follows:

Pennies for your thoughts - Part 1

Here are two puzzles which are identical - but we count the solutions in two different ways. Each involves arranging pennies (coins) in rows.

The puzzle here is that only one of these two puzzles involves the Fibonacci number series! The other puzzle does not but just begins with a few of the Fibonacci numbers and then becomes something different. One of these puzzles is a fraud, a Fibonacci forgery. So which is the real Fibonacci puzzle?

Arrange pennies in rows under these two conditions:
each penny must touch the next in its row [/*:2djrwga4]
each penny except ones on the bottom row touches two pennies on the row below[/*:2djrwga4]

There is just 1 pattern with one penny, and 1 with two pennies, but 2 for three pennies and 3 with four pennies (as shown below):

__
__ __ __ _|__|_
|__||__||__| |__||__|

==========================
__ __
__ __ __ __ _|__|_ _|__|_
|__||__||__||__| |__||__||__||__|
__ __
_|__|_ __ __ _|__|_
|__||__||__| |__||__||__|
The first condition means that there are no gaps in any row and the second means that upper rows are smaller than lower ones.

The following arrangements are not proper combinations for 6 pennies:

__ __
_|__|_ _|__|_
|__||__||__||__|
__ __
|__||__|
|__|
because the first has a gap in one row and the second has a penny which is not on the bottom row and is not touching two beneath it.

If there are P(n) such arrangements for n pennies, are the P(n) numbers always Fibonacci numbers?

Pennies for your thoughts - Part 2

This puzzle is the same as the previous one and again seems to involve the Fibonacci numbers - or does it? The puzzle is exactly the same, but P(n) now counts the number of arrangements which have n pennies on the bottom row.

__
__ __ __ _|__|_
1: |__| 2: |__||__| |__||__|

__
__ __ __ _|__|_ __
3: |__||__||__| |__||__||__|
__
__ __ __ _|__|_
__ _|__|_ _|__||__|_ _|__||__|_
|__||__||__| |__||__||__| |__||__||__|
Here there is only 1 arrangement with 1 penny on the bottom row so P(1)=1 and 2 arrangements with two on the bottom row, P(2)=2 and 5 patterns with a bottom row of three coins P(3)=5.

What happened to 3? F(4)=3 is missing! You can check that P(4)=13, so P(n) is clearly not the same as the Fibonacci series since F(4)=3 and F(6)=8 are missing. This time the question is:

Are the P(n) numbers the alternate Fibonacci numbers:

i : 0 1 2 3 4 5 6 7 8
Fib(i): 1 1 2 3 5 8 13 21 34...
P(n): 1 2 5 13 ?
n : 1 2 3 4 5
Which one of these two Pennies puzzles is the forgery (it does not continue with a pattern of Fibonacci numbers after some point) and which one genuinely always has Fibonacci numbers of arrangements?