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16:53
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Q: Sequence of natural numbers such that $a_i = \sum_{r=1}^{i+4} d_r$

user19405892 Find a sequence of natural numbers $a_i$ such that $\displaystyle a_i = \sum_{r=1}^{i+4} d_r$ where $d_r \neq d_s$ for $r \neq s$ and $d_r$ divides $a_i$ for all $r \in \{ 1, 2, \dots, i+4\}.$ Attempt: Suppose $a_n = d_1+d_2+\cdots+d_{n+4}$ for some $n$. Then we see $a_{n+1} = d_1+d_2+\cdot...

number-theory
user19405892
user19405892
@MatthewConroy It is as defined in the summation.
@MatthewConroy That is correct.
awllower
awllower
In your construction how are you sure that $d_1$ divides $a_{n+1}?$
user19405892
user19405892
@awllower Since by assumption $d_1$ divides $d_1+d_2+\cdots+d_{n+4}$.
awllower
awllower
Alright, so it remains to find one such $a_1?$ Also isn't $a_{n+1}$ supposed to be $\sum_{r=1}^{n+5} d_r?$
user19405892
user19405892
@awllower Yeah, I realize my construction is incorrect. What would be a correct one?
awllower
awllower
16:53
How about $d_{n+5}=d_1+\cdots+d_{n+4}?$
user19405892
user19405892
@awllower How does that work? We have $a_{n+1} = 2(d_1+\cdots+d_n)+d_{n+1}+d_{n+2}+d_{n+3}+d_{n+4}$ and so would need $d_1+\cdots+d_n$ to divide $a_{n+1}$.
hi
awllower
awllower
hi
user19405892
user19405892
how does that work?
awllower
awllower
I mean $a_{n+1}=2(d_1+\cdots+d_{n+4})$
user19405892
user19405892
oh ok, so $d_{n+5} = d_1+d_2+\cdots+d_{n+4}$
awllower
awllower
16:54
that is, $d_{n+5}=d_1+\cdots+d_{n+4}.$
yes, I mis-typed.
user19405892
user19405892
yeah that works
we just need to find one that works
awllower
awllower
how about $28=1+2+4+7+14?$
user19405892
user19405892
yeah that works
awllower
awllower
glad to know that. :D
user19405892
user19405892
Then using our construction, $a_2 = 2 \cdot 28$
awllower
awllower
16:58
basically $a_{n+1}=2a_n.$
user19405892
user19405892
what about $a_3$?
$a_3 = 3\cdot 28$
awllower
awllower
It should be $a_3=2a_2=4\cdot28.$
user19405892
user19405892
I don't see why we need $2$ times
awllower
awllower
I am writing an answer; I shall add an explanation.
user19405892
user19405892
$a_1 = 28,a_2 = 2 \cdot 28, a_3 = 3 \cdot 28,\ldots$
Why doesn't it work like that?
awllower
awllower
17:05
Briefly, because $$a_{n+1}=d_1+\cdots+d_{n+4}+d_{n+5}=a_n+a_n=2a_n.$$
You may see the answer and point out any inappropriate points. :)
it is late here and I shall sleep now. Good night. :)

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