1

Here is a sequence, $a_1, a_2, a_3, \ldots$ that satisfy the following property: $a_{n+2} = a_{n+1}+a_n$, where $a_m$ is a positive integer for any $m$, and it is known that $a_7 = 2015$. How many such sequences exist?

Martín-Blas Pérez Pinilla
  • 41,546
  • 4
  • 46
  • 89
Max
  • 51
  • 4

3 Answers3

1

Hint: the solution will be $$a_n = c_1\phi^n + c_2(-\phi^{-1})^n$$ for some $c_1$, $c_2$ depending of the (integer) initial conditions $a_1$, $a_2$. Put the condition $a_7 = 2015$ and you will find a relation between $c_1$ and $c_2$, i.e. a relation between $a_1$ and $a_2$.

Martín-Blas Pérez Pinilla
  • 41,546
  • 4
  • 46
  • 89
  • Another (equivalent) equation would be $a_n=c_1F_n+c_2L_n$, where $F$ and $L$ are the Fibonacci and Lucas sequences respectively and where my $c$s are different than your $c$s. – Akiva Weinberger Jan 28 '16 at 13:52
1

$$a_7\\ =a_6+a_5\\ =a_5+a_4+a_4+a_3\\ =a_4+a_3+2a_3+2a_2+a_2+a_1\\ =a_3+a_2+a_2+a_1+2a_2+2a_1+2a_2+a_2+a_1\\ =a_2+a_1+a_2+a_2+a_1+2a_2+2a_1+2a_2+a_2+a_1\\ =8a_2+5a_1$$

So the number of solutions to $8x+5y=2015$.

JMP
  • 21,771
0

$a_7=2015$, $a_5=2015-a_6$, $a_4=2a_6-2015$, $a_3=4030-3a_6$, $a_2=5a_6-6045$, $a_1=10075-8a_6$, and all you need is $a_1>0$ and $a_2>0$, and you should be able to work out how many values of $a_6$ make that happen.

Gerry Myerson
  • 179,216