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1.2.3.5.8.13.21.34.55.89.144.233.267.500.508. 1008.1011.2019

Peror wrote: 1.2.3.5.8.13.21.34.55.89.144.233.267.500.508. 1008.1011.2019 How to prove that ?

SivmengMaths wrote: Peror wrote: 1.2.3.5.8.13.21.34.55.89.144.233.267.500.508. 1008.1011.2019 How to prove that ? Consider a Fibonacci-like sequence defined by, $A_1 =1, A_2 =2$ and $A_{n+2} = A_{n+1} + A_{n}$, for natural $n$. Now, note that by definition, $a_i \le A_i$ for all i. But, since the $18^{\text{th}}$ term of both sequences coinicide, they must have been equal all along till then. Hence the answer.

Okay, edited: Just for fun; There will be $$\prod_{i=4}^{17} (f_{k+1}-2)$$number of possible $18$-tuples of integer satisfying above condition. where $f_{i}$ is $(i)th$ fibonacci number. If you want to calculate: Given : $a_1=1$ and $a_2=2$ So by the construction $a_k=a_i+a_j$ for $1 \leq i < j < k$* - The possibility for $a_3$ is 1 i.e., $a_3=a_1+a_2=2+1=3$ - Now we'll try to construct $a_4$ in all possible ways: $a_4=a_1+a_2=3$ or $a_4=a_1+a_3=4$ or $a_4=a_2+a_3=5$ S0 possibility for $a_4$ are $3,4,5$ Now let us try for $a_5$ $a_5$ by above construction $a_5$ can have all numbers that $a_4$ have and in addition to it $a_5$ will also have $6,7,8$ - Similarly, $a_6$ : clearly $a_6$ will have all elements of $a_5$ and in addition to it, it will also have $z_6=\{a_5 -a_3\}+(\text{greatest/last \;element \;of} \;a_4)$ so possibility for $a_6$ is $$\{3,4,5,6,7,8,9,10,11,12,13\}$$Let's do one more for better understanding i.e. $a_7$ $a_7$ will have all the elements of $a_6$ and in addition to it, it will also have $z_7=\{a_6 - a_4\}+(\text{greatest/last \;element \;of} \;a_5)$ Note at every step say $ith$ we are adding $f_{i-1}$ additional consecutive numbers to the set, where $f_{i-1}$ is $(i-1)th$ fibonacci number. i.e., if we are to calculate the possibility set for $a_k$ then it will have numbers $\{a_{k-1} \}$ $\cup$ {$f_{k-1}$ consecutive numbers after last element of $a_{k-1}$} Let's see what we got: $$|a_4|=3=f_2+f_3$$$$|a_5|=|a_4|+f_4=f_2+f_3+f_4$$. . . $$|a_k|=\sum_{i=2}^{k-1}f_{i}$$ And we know that, $$\sum_{i=1}^{n}f_{i}=f_{n+2}-1$$ So, $$\sum_{i=2}^{k-1}f_{i}=f_{k+1}-2$$as $f_1=1$ Now if we seek to find to the total number of possible integers which satisfy the above condition. $a_1,a_2,a_3,a_{18}$ are fixed and have only one possibility so we will focus on $4 \leq i \leq 17$ Now by simple combination rule, we will have Total number of possible $18$-tuples $$= \prod_{i=1}^{18} |a_i|$$ Which is actually $$= \prod_{i=4}^{17} (f_{k+1}-2)$$

@Edited .

The original solution from : https://math.stackexchange.com/a/3363470/610697

1,2,3,5,8,13,18,31,49,80,85,103,188,291,479,770,1249,2019