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Here is the question,

If the tenth number in a Fibonacci-type sequence of increasing positive integers is 301, what is the fourth number?

How do I go about solving this without overly complex math?

EDIT | Fibonacci-type: the 2 previous numbers sum to the next in the sequence. EX: 4,5,9

Jolly
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    define "Fibonacci-type". – Angina Seng Oct 14 '18 at 14:42
  • In general, you need two conditions to specify this recursion. Maybe the "increasing, positive" part is enough to make it unique...not sure. – lulu Oct 14 '18 at 14:45
  • Did you have a look at https://www.quora.com/If-the-tenth-number-in-a-Fibonacci-type-sequence-of-increasing-positive-integers-is-301-what-is-the-fourth-number (first Google hit if you search for the problem statement) ? – Martin R Oct 14 '18 at 14:54

1 Answers1

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Since you know the sequence satisfies the same recurrence as Fibonacci, its solution $g_n$ can be written as $$ g_n = g_1 F_n + (g_2-g_1) F_{n-1} = g_1 F_{n-2} + g_2 F_{n-1},\quad n=1,2,\dots, 10 $$ where $F_n$ is the usual Fibonacci. So $$ 301 = 55 g_1 + 34 (g_2 -g_1) = 21 g_1 + 34 g_2 $$ and you know $g_1,g_2$ are positive integers. So the only solutions are $(g_1,g_2)=(3,7)$, giving the fourth term $g_4=3F_2+7F_3=17$.

user10354138
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