Liam's house has a staircase with 12 steps. He can go down the steps one at a time or two at time. In how many different ways can Liam go down the 12 steps?
Here are some circle bugs to try to replicate with some elegant
programming, plus some sequences generated elegantly in LOGO.
Using logo to investigate spirals
Lots of people sent in the solution that
the sixth term of the Fibonacci sequence starting with $2$ and $38$
is $196$ and you found other sequences with $196$ as one of the
terms. Exactly how many other Fibonacci sequences contain the term
lot of solutions as Jimmy rightly pointed out!
The simplest Fibonacci sequence is:
$1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 253, \ldots$
and we denote the $n$th term of this sequence by $F(n)$.
Starting with the terms $a, b$ (for $a$ and $b$ positive whole
numbers and $a < b$) we get the general Fib sequence:
$a$, $b$, $a+b$, $a+2b$, $2a+3b$, $3a+5b$, $5a+8b$, $8a+13b$,
$13a+21b$, $21a+34b$, $34a+55b$, $55a+89b$, $89a+144b$,
The $n$th term of the general Fib sequence $f(n) = aF(n-2) +
bF(n-1)$ and note that, if the term $196$ occurs in the sequence,
it can't be beyond the twelfth term as after that the terms are too
Here are some sequences containing $196$.
Sequences with $196$ as the first term
$1, 195, 196, \ldots$
$2, 194, 196, \ldots$
$97, 99, 196, \ldots$
$98, 98, 196, \ldots$ etc
So far we see that there are infinitely many sequences with $196$
as the first term; exactly $195$ with $196$ as the second term;
exactly $98$ with $196$ as the third term.
To find all the remaining sequences containing $196$ we have to
find whole numbers $a$ and $b$ where:
$a + 2b = 196$
$2a + 3b = 196$
$3a + 5b = 196$
In general we have to find whole number values of $a$ and $b$
$aF(n-2) + bF(n-1) = 196$.
and so we need to find whole number solutions to these equations
for $n = 4, 5, 6, \ldots12$.
We shall consider one remaining case and leave the rest to the
For $n = 6$ we seek values of $a$ and $b$ such that $3a + 5b =
There are no solutions for $a = 1$ because then b would not be a
whole number. We have already seen that $a = 2$ and $b = 38$ gives
$196$ as the sixth term. For larger values of $a$ we have to take
smaller values of $b$. For $a = 3$ or $4$ or $5$ or $6$ there are
again no solutions because b has to be a whole number.
For $a = 7$ we have:
giving the sequence $7, 35, 42, 77, 119, 196, \ldots$
To find the remaining solutions for $n = 6$ we increase $a$ by
steps of $5$ and decrease $b$ by steps of $3$.
There are five solutions for $n = 6$, which are:
You can use the same method to find the solutions for $n = 4, 5,
7, \ldots 12$