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....................
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