A sequence related to the shape 2 x + 2 y + 2 z 2^{x}+2^{y}+2^{z}

Let the increasing sequence { a n } n = 1 \{a_{n}\}_{n=1}^{\infty } be the arrangement of all the elements of { 2 x + 2 y + 2 z 0 x < y < z for x , y , z Z } \{2^{x}+2^{y}+2^{z}\Big|0\leq x<y<z\ \ \text{ for }\ \ x,y,z\in \mathbb{Z}\} in ascending order.

If a 2018 = 2 p + 2 q + 2 r , a_{2018}=2^{p}+2^{q}+2^{r}, where p , q , r p,q,r are positive integers, determine the value of p + q + r . p+q+r.


The answer is 60.

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1 solution

Haosen Chen
Mar 10, 2018

Notation: here ( n k ) \left(\begin{array}{c} n \\ k \\ \end {array}\right) ( n k 0 ) (n\geq k\geq 0) is the binomial coefficient.

Let f ( x , y , z ) = 2 x + 2 y + 2 z f(x,y,z)=2^{x}+2^{y}+2^{z} .Note that f ( 21 , 22 , 23 ) f(21,22,23) is the ( 24 3 ) = 202 4 t h \left(\begin{array}{c} 24 \\ 3 \\ \end {array}\right) =2024_{th} item of seqence { a n } \{a_{n}\} ,i.e. a 2024 a_{2024} . Thus a 2018 = f ( 15 , 22 , 23 ) a_{2018}=f(15,22,23) ,the answer is 15 + 22 + 23 = 60 15+22+23=\boxed{60} .

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