Challenge Level

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
$196$? *A
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:

$a$, $b$, $a+b$, $a+2b$, $2a+3b$, $3a+5b$, $5a+8b$, $8a+13b$,
$13a+21b$, $21a+34b$, $34a+55b$, $55a+89b$, $89a+144b$,
$\ldots$

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 large.

Here are some sequences containing $196$.

**Sequences with $196$ as the first term**

$196, b, 196+b, 196+2b, \ldots$ | for $b > 196$ |

$a, 196, \ldots$ | for $1 < a < 195$ |

$1, 195, 196, \ldots$

$2, 194, 196, \ldots$

$...$

$97, 99, 196, \ldots$

$98, 98, 196, \ldots$ 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:

$a + 2b = 196$

or

$2a + 3b = 196$

or

$3a + 5b = 196$

etc.

In general we have to find whole number values of $a$ and $b$
satisfying

$aF(n-2) + bF(n-1) = 196$.

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
reader.

For $n = 6$ we seek values of $a$ and $b$ such that $3a + 5b = 196$

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:

$21 + 5b$ | $=$ | $196$ |

$5b$ | $=$ | $175$ |

$b$ | $=$ | $35$ |

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:

- $2, 38, \ldots$
- $7, 35, \ldots$
- $12, 32, \ldots$
- $17, 29,\ldots$
- $22, 26,\ldots$

You can use the same method to find the solutions for $n = 4, 5, 7, \ldots 12$