Subtraction of Powers and Other Observations

Exponent = 2
n=    1            2              3                4                5                 6               7              8              9              10
n^2=1            4              9               16               25              36             49            64            81            100
S1          3             5               7                9                11              13             15            17             19
S2                 2              2                 2                2                 2                2             2             2


Exponent = 3
n=    1            2              3                4                5                  6               7              8             9              10
n^3=1            8             27             64              125             216             343
S1          7             19             37             61                91             127
S2                 12            18             24               30                36
S3                          6               6               6                  6           

Exponent = 4
n=     1            2              3                4               5                6               7              8               9              10
n^4= 1            16           81              256          625           1296          2401       4096        6561       10000
S1            15           65           175             369          671           1105         1695                    
S2                    50           110           194          302            434           590
S3                            60            84             108          132            156
S4                                    24             24            24              24


Exponent=5
n=     1             2               3              4               5               6              7               8               9              10
n^5= 1            32             243        1024         3125        7776         16807
S1           31             211         781         2101         4651        9031
S2                    180          570         1320        2550        4380
S3                             390         750        1230         1830
S4                                     360         480         600
S5                                             120        120

Based on observations and a few other deductions, we can predict a similar senario for Exponent=6 or any other exponent for that matter.

Obs1: Recurring subtraction numbers in the are always in the corresponding S level for the exponent.

Obs2: The recurrent number is always the Exponent factorial. Exp=3, S3= 3!=3x2x1 or Exp=4, S4=4!=4x3x2x1

Obs3: the first number in the penultimate S level is always Exp!/2x(Exp+1)
    Ex: for Exp=4 the first number in S3 is 24/2 x (4+1) = 60  and for Exp=5, the first number in S4 is   120/2 x (5+1)=360
                 
Obs4: for Exp=5, the last number of each power is in series,because of that, all S1 numbers end in "1" and the remaining numbers in the remaining S levels are multiples of 10

Future observations pending..........

New Date: Dec 25th, 2012 :

For future reference, the level above S1 shall be termed the Power level or P level and the Exponent shall be represented by the letter "e" instead of "Exp".

  An attempt was made to represent all S level numbers in e=3 power subtraction series in terms of  e+1, e, e-1, e-2 etc.

Obs 5: what was definite was that the largest term in the S1 level needed to create the first recurring subtraction term in the Se level was the eth term in S1 and therefore required the (e+1)th term in the P level. So in the case of e=3, the first 6 in S3 exists because of the 3rd term "37" in S1 and the 4th term "64" in the P level.
Obs 6: From the highest to lowest terms, all S1 level terms had coefficients of        1 &-1
Obs 7: From the highest to lowest terms, all S2 level terms had coefficients of     1 , -2 , 1 (in red)
Obs 8: From the highest to lowest terms, all S3 level terms had coefficients of 1 , -3 , +3, -1 (in red)

Same thing was done for e=4.

Obs 9:  Obs 6,7,& 8 for S1, S2, and S3 respectively applied in this scenario as well.
Obs 10: From highest to lowest terms, all S4 terms had coefficients 1 , -4 , +6 , -4, & +1

Obs 11: Based on a possible pattern, the coefficients were structured in the form of a pyramid
                                                        1      -1
                                                    1     -2     +1
                                                1     -3    +3    -1
                                            1     -4    +6    -4    +1
Obs12: e=5 coefficients predicted by adding the 2 absolute values above and alternating signs starting with +
                                       1     -5   +10   -10   +5     -1
Obs 13: The math was added up (in red) and was = 120 --> same as 5! = 5x4x3x2x1 = 120

Obs 14: e=6 coefficients predicted.
                               1       -6    +15  -20   +15    -6       +1 
 Obs 15: The math added up (in red) = 720 same as 6!
Obs 16: The terms in the Se level can be predicted based on the terms in the P level using the coefficients in the pyramid (I'll call this the Coefficient pyramid) eg: the recurring term in S6 for e=6 can be determined by placing the corresponding coefficients with the P level terms from highest to lowest starting with the 7th ('e+1'th) term, then 6th, 5th, 4th, 3rd, 2nd, and 1st terms, then adding them up. [ or you can just do e!]
Future observations pending....................

Jammed Doors, Warped Cabinet, and Skanky Legs

So the main complaint was that both cabinet doors were jammed. Unfortunately, I did not have pictures prior to solving the problem, so to the left is an exaggerated schematic of the front of the cabinet. Upon inspection, the left door was jamming at corner A and the right door was jammed at corner B.
At first glance, the cabinet surface seemed flat and the center doorjam strip in yellow was slightly slanted to the left kinda like a backslash ( \ ). All four legs and metal caster wheels seemed level with the floor.
Upon closer inspection, what was noticed was that the entire cabinet was a little shaky so i checked and tightened all the corners braces that would have kept the entire structure square. But that still did not solve the problem. Shaving the jammed areas A and B also did not solve the inital problem.
But at long last a eureka moment occurred when I squeezed both sides of the cabined and noticed that both doors suddenly became unjammed and were able to open and close freely. Not only that, I noticed that the central door jam became vertical, corners A and B were no longer touching the top and bottom of the cabinet and that both doors were aligned with each other horizontally and vertically, and that the front right castor wheel was no longer touching the floor.

This picture shows what the right castor wheel looks like. It was well seated into the wooden front right leg. At first I thought that maybe that leg needed to be lengthened or placed on a block because if I lifted the front right leg alone, the entire cabinet structure would become square as if I had squeezed both sides to correct the original problem. Then I glanced at the other 3 legs and noticed that the front left castor wheel [C] was not seated completely into the wooden leg, offset about 1/4 in.

Could this have been the original problem all this while? A single unseated castor wheel causing warped cabinet instability resulting in jammed doors? So after exerting pressure on the cabinet to seat the castor completely into the front left leg, both doors were able to open and close freely without jamming.

 

PROBLEM SOLVED