Patterns on sum of any two consecutive counting number I am looking for patterns in this condition, fortunately I found some but I cannot prove despite of simplicity of the condition... please give me some lead on how Im going to prove any of this pattern and maybe any of these questions will lead to the proof of other patterns. Let n be a a counting number $$1, 2, 3, 4, 5, 6, 7. . $$ this will serve as the first layer. If I add every two consecutive counting numbers it will result to $$3,5,7,9,11,13,...$$ this is the second layer. If I repeat the process I came up with $$8,12, 16, 20, 24$$ 3rd layer and so on... 
The following are the patterns I observe:


*

*The difference between any two elements in kth layer is equal to $2^{k-1}$...

*The gcf of 1st and 2nd layer, 3rd and 4th layer, 5th and 6th layer and so on is 1,4,16 and so on respectively which are all power of two but with skipping...

*The sum of all elements included in a triangle of any odd numbered size will always be a multiple of the middle number on its base. For example adding those number:$$1,2,3,4,5$$$$3,5,7,9$$$$8,12,16$$$$20,28$$$$48$$ is equal to 171 which is a multiple of 3 (the middle entry on its base). 
Any idea will of great help thanks!
 A: For convenience of the presentation we start counting the "levels" from $0$,
so that
$$
\{1,2,3\dots\}
$$
is the $0$-th level.
Denote the difference at the $k$-th level as $\Delta_k$. Let us prove that every level is an arithmetic progression with common difference $\Delta_k=2^{k}\Delta_0$. For $k=0$ it is obvious. Assume it is true at the level $n$. Then it is valid for $(n+1)$-th level as well.
Proof.
By induction hypothesis the elements of the $n$-th level represent an arithmetic progression: $$a_{ni}=a_{n0}+i\Delta_n.$$
Summing the neighboring elements of the $n$-th level one obtains for the elements of the $(n+1)$-th level the expression: 
$$a_{n+1,i}=a_{ni}+a_{n,i+1}=(a_{n0}+i\Delta_n)+(a_{n0}+(i+1)\Delta_n)=(2a_{n0}+\Delta_n)+i(2\Delta_n).$$
Thus, the $(n+1)$-th level also represents an arithmetic progression with
$$
a_{n+1,0}=2a_{n0}+\Delta_n,\text{ and } \Delta_{n+1}=2\Delta_n\stackrel{I.H.}=2\cdot 2^{n}\Delta_0=2^{n+1}\Delta_0.\tag1
$$ 
It can be similarly shown by induction using $(1)$, that the initial term at the $n$-th level is:
$$
a_{n0}=2^{n-1}(2a_{00}+n\Delta_0).
$$
For $a_{00}=\Delta_0=1$ this reads:
$$
a_{n0}=2^{n-1}(n+2)\to \{1,3,8,20,48,\dots\}.
$$
