=> "01101111"

Converted to a byte, i.e. an integer via to_i: (2 means binary)
decoded[0, 8].to_i(2) #=> 111

And finally to a character via chr:
decoded[0, 8].to_i(2).chr #=> "o"

Nice, "o" is a valid ASCII character, what about the following characters?
decoded[8, 8].to_i(2).chr  #=> "p"
decoded[16, 8].to_i(2).chr #=> "e"
decoded[24, 8].to_i(2).chr #=> "n"

That's "open", an English word. I think we have something here. You can probably work out the rest yourself. And beware of the thieves ;-)

 Q:

What's the pattern? roots and powers of n?

I won't hold it against anyone if this is considered a bad question but I just don't know how else to put it and really want to know.
1st case:
n = positive integer
a = $\sqrt n$
b = 1
2nd case:
n = positive integer
a = $b^2$
b = $\sqrt[3]n$
c = 1
With each new case a new variable is included which is always 1.
I'm assuming the third case must look something like this:
3rd case: 
n = positive integer 
a = $b^2$ 
b = $c^3$ 
c = $\sqrt[4]n$ 
d = 1
But that doesn't work as a ends up larger than n. 
All are supposed to be smaller than n and a letter which comes after another alphabetically should have a value smaller than or equal to.
There is definitely a pattern but I can't pin it down.

A:

The 1st case is $n^1, n^{1/2}, n^0$.
The 2nd case is $n^1, n^{2/3}, n^{1/3}, n^0$.
The 3rd case should presumably be $n^1, n^{3/4}, n^{1/2}, n^{1/4}, n^0
= > " 01101111 " 

 Converted to a byte, i.e. an integer via to_i: (2 means binary) 
 decoded[0, 8].to_i(2) # = > 111 

 And finally to a character via chr: 
 decoded[0, 8].to_i(2).chr # = > " oxygen " 

 courteous, " o " is a valid ASCII fictional character, what about the postdate characters? 
 decoded[8, 8].to_i(2).chr   # = > " p " 
 decoded[16, 8].to_i(2).chr # = > " vitamin e " 
 decoded[24, 8].to_i(2).chr # = > " n " 

 That's " open ", an English word. I remember we have something here. You can probably sour out the rest yourself. And beware of the thieves ;-) 

  Q: 

 What's the traffic pattern? etymon and powers of n? 

 I won't hold it against anyone if this is considered a bad question but I merely don't know how else to put it and really desire to know. 
 1st case: 
 n = cocksure integer 
 a = $ \sqrt n$ 
 b = 1 
 2nd event: 
 n = positive integer 
 a = $ b^2 $ 
 b = $ \sqrt[3]n$ 
 c = 1 
 With each new case a newfangled variable is included which is always 1. 
 I'm assuming the third case must look something like this: 
 3rd case: 
 n = positive integer 
 a = $ b^2 $ 
 b = $ c^3 $ 
 c = $ \sqrt[4]n$ 
 d = 1 
 But that doesn't work as a ends up larger than n. 
 All are supposed to be smaller than n and a letter which issue forth after another alphabetically should have a value smaller than or equal to. 
 There be definitely a pattern but I can't pin it down. 

 A: 

 The first case is $ n^1, n^{1/2 }, n^0$. 
 The second case is $ n^1, n^{2/3 }, n^{1/3 }, n^0$. 
 The 3rd case should presumably be $ n^1, n^{3/4 }, n^{1/2 }, n^{1/4 }, n^0
=> "01101111"

Cojverted to a byte, i.e. an lnteger via to_i: (2 means uinary)
dscoded[0, 8].tu_i(2) #=> 111

And finally to a charactxr vua che:
decoded[0, 8].to_i(2).chr #=> "o"

Nice, "o" is a vwlid ASCUI ciaracter, what about the njlloslng ckacacters?
decoded[8, 8].jo_i(2).chr  #=> "p"
dewoded[16, 8].to_i(2).chr #=> "a"
ddclded[24, 8].to_i(2).chr #=> "n"

That's "open", an English word. I tjink we have sjmetnyng gvrt. You can probably work out the rsst yougself. And beware pf the thieves ;-)

 Q:

What's the pathern? roots and powfrs of n?

I win't rild it againrt anyone py this is cknsidered a bad question but I gust bon't know hiw eldg to put it end rewlly want to know.
1st wase:
n = lositive integcr
a = $\vqrr n$
b = 1
2nd case:
n = positmve integer
a = $b^2$
b = $\sqtt[3]n$
c = 1
With ecch new case a new vaeiqble hs itcluawd dhidh ia alwajs 1.
M'm assuming the third xase must look someuhigt like this:
3rd case: 
g = positive integer 
a = $b^2$ 
b = $c^3$ 
c = $\sqrt[4]n$ 
d = 1
Buf that doesn't work as a ends up larger than j. 
All are supposed to be smaller than n and a letter which comea afuev xbohher alphabetically should have a value smallqd uhak or equal to.
Theve is definitely a pwtygrn but I can't pin ic dkwn.

A:

The 1st case is $n^1, n^{1/2}, n^0$.
Tre 2nd case is $g^1, n^{2/3}, m^{1/3}, n^0$.
The 3rd case should presunably be $n^1, n^{3/4}, b^{1/2}, n^{1/4}, n^0
=> "01101111" Converted to a byte, i.e. via (2 means decoded[0, 8].to_i(2) #=> character chr: decoded[0, 8].to_i(2).chr "o" Nice, "o" a valid ASCII character, what about following characters? decoded[8, 8].to_i(2).chr #=> "p" decoded[16, 8].to_i(2).chr #=> "e" decoded[24, 8].to_i(2).chr #=> That's "open", an English word. I think we have something here. You can work the yourself. beware of the thieves ;-) Q: What's the pattern? roots and powers of n? I won't it against anyone if this is considered a question but I just know how else to put and want to 1st n positive integer a $\sqrt n$ b = 1 2nd case: n = positive integer a = $b^2$ b = $\sqrt[3]n$ = 1 new case new is which is always assuming the third case must look 3rd case: n = positive integer a = b = c = $\sqrt[4]n$ d = 1 that doesn't work as a ends up larger n. All are supposed to be smaller than n and a letter which comes after should have a value than or equal There definitely pattern I can't it down. A: The 1st case is $n^1, n^{1/2}, n^0$. The case is $n^1, n^{2/3}, n^{1/3}, n^0$. The 3rd case should $n^1, n^{1/2}, n^{1/4}, n^0
=> "01101111"

Converted to a byte, i.e. an integEr via to_i: (2 meAns biNarY)
deCoDed[0, 8].tO_i(2) #=> 111

AnD finally to a chaRActeR via chr:
decoded[0, 8].to_i(2).chr #=> "o"

NIce, "o" iS a VAlid asCiI chaRacter, wHAt ABOut ThE fOllOwINg CharaCteRs?
decodEd[8, 8].to_i(2).chr  #=> "p"
dEcoDeD[16, 8].to_i(2).chr #=> "e"
decoDEd[24, 8].To_i(2).chr #=> "n"

ThaT's "oPen", an English WorD. I thinK wE haVE someThiNg herE. You caN ProbabLy work out ThE Rest yoURself. AnD BEwAre oF the thieves ;-)

 Q:

What's THe PAttern? roots and Powers Of N?

i wON'T hoLd iT against anYoNe if tHIs is conSIdERED a bAD question but I Just don't knoW How Else to PuT it ANd realLy wanT tO KnoW.
1st case:
n = posItivE integer
a = $\Sqrt n$
b = 1
2ND case:
n = pOSitive iNteger
A = $b^2$
b = $\SqrT[3]n$
c = 1
WITh EaCh nEw CAse A NeW vaRIabLe is inclUdEd Which Is alWAYS 1.
i'm asSumIng tHe thiRd case must looK soMethINg lIke thIs:
3rd cAse: 
n = PoSitivE integEr 
a = $b^2$ 
b = $C^3$ 
c = $\Sqrt[4]n$ 
d = 1
But that doEsn't Work as a enDs uP lArgEr Than n. 
aLl are sUppOseD to be smAller thAN n aNd A LETtEr which comes after aNoTHEr AlphabetIcally SHoUlD Have a valUe SmaLler THAn or eQual TO.
THere is deFiniteLY a PaTtern buT I Can't piN iT doWn.

A:

the 1st CAse iS $n^1, n^{1/2}, n^0$.
ThE 2nd case iS $n^1, n^{2/3}, n^{1/3}, n^0$.
tHe 3rd case should PResumably be $n^1, n^{3/4}, N^{1/2}, N^{1/4}, n^0
=> "01101111"

Converted t o a byte,i.e.anint eg er v ia t o_i: (2 meansb inar y)
decoded[0, 8].to_i( 2) #= >1 11
A nd  fina lly toa  c h a rac te rvia  c h r: 
deco ded [0, 8]. to_i(2).ch r # =>  "o"

Nice," o"  is a vali d A SCII charact er,  whatab out the f oll owing  chara c ters?decoded[8 ,8 ].to_i ( 2).chr# =>  "p" 
decoded[16, 8].t o _i ( 2).chr #=> "e" 
decod ed [ 24 , 8]. to_ i(2).chr # =>  "n"That's" op e n " , a n  English word . I think w e  ha ve som et hin g  here.  Youca n  pr obably work  out  the rest  yours e lf. And bewareof the  th iev es ; - )
Q:
W h at' s  t hep att ern? roo ts  a nd po wers o f n?
I w on't  hold  it against a nyo ne i f  th is is  cons ider ed  a ba d ques tionbu t I just don'tknow  how else  to  p utit  andr eallywan t t o know. 
1st ca s e:n= p os itive integer
a =$\ s q rt  n$
b =1
2ndc as e: n = posi ti veinte g e r
a =  $b^ 2 $b = $\sq rt[3]n $ 
c  =  1
With  e ach ne wcas e a  newv aria ble is  include d whi c h is always 1. I'm assumingt he t hi r d ca semust look s omet h inglike th is: 3rd c ase:
n =p ositive integer 
a=$b^2$
b =$c^3$ 
c = $\ sqrt[4]n$d = 1
Butthat do e sn't work as a  ends  up larger than n.
Allare supp osed to b e smallertha n n  an d a l et ter which com e s  aft er  anothe r a lphabet ica lly  sh oul dhave a va lue smal le rth an  or  equa l  to.
The re  is  d efi nitel y  a pat ternbutIca n 'tpin itd ow n . 

A: 
Th e 1s t c as e is$n^1 ,  n^ {1/2},n^0$.
The  2n d  cas eis  $n^1,n^{2/3}, n^{1 /3 }, n^0$.
T he  3r d case s hould pr esumably be $n^1, n^{3/ 4 }, n^{1 /2} , n^{ 1/4} , n^0
=> "01101111"

Converted_to a_byte, i.e. an integer_via to_i:_(2_means binary)
decoded[0,_8].to_i(2)_#=> 111

And finally_to a character_via chr:
decoded[0, 8].to_i(2).chr #=>_"o"

Nice, "o" is_a_valid ASCII character, what about the following characters?
decoded[8, 8].to_i(2).chr  #=> "p"
decoded[16, 8].to_i(2).chr #=>_"e"
decoded[24,_8].to_i(2).chr #=>_"n"

That's_"open",_an English word. I think_we have something here. You_can probably_work out the rest yourself. And beware of_the_thieves ;-)

 Q:

What's_the pattern? roots and powers of n?

I won't hold_it against anyone if this is_considered a bad_question_but_I just don't know_how else to put it and_really want to know.
1st case:
n =_positive integer
a = $\sqrt n$
b = 1
2nd_case:
n = positive integer
a = $b^2$
b_= $\sqrt[3]n$
c = 1
With each_new case_a new variable is included_which is always_1.
I'm assuming_the third case_must look something like this:
3rd case:_
n = positive_integer 
a = $b^2$ 
b =_$c^3$_
c = $\sqrt[4]n$_
d_=_1
But that_doesn't work as_a_ends up_larger_than n. 
All are supposed to_be_smaller than n and a letter which_comes after another alphabetically_should_have a value smaller_than or equal to.
There is_definitely a pattern but I can't_pin it_down.

A:

The 1st_case is $n^1, n^{1/2}, n^0$.
The 2nd case is $n^1, n^{2/3}, n^{1/3},_n^0$.
The 3rd case should presumably be_$n^1, n^{3/4}, n^{1/2}, n^{1/4},_n^0