﻿ Positive Integer power sum COefficient equation Lineare Gleichungssysteme lösen mit dem Gaußschen Eliminationsverfahren Positive Integer power sum COefficient equation
Please visit second page freeman2.com/jspico2e.htm
program 　 Arndt 　 DocA 　 first9 　 Update 2009-09-07
Coefficient triangle　 Pattern conjectures　 Column 5 pattern
Proof coef. column 01　 02　 03　 04　 05　 List
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proofread. Cannot promise correctness.
If you suspect any view point wrong,
please ask a math expert near by.
Freeman 2009-06-19-10-46

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```<a name="docA01">
proof, program, Conjecture, triangle
2009-09-01-20-08 start
This file
http://freeman2.com/jspico_e.htm
find out
Positive Integer power sum COefficient
equation. Brief as PICO program.

If we want to sum 1+2+3+4+5+...+n
to find numerical answer, write loop
code to add 1, 2, 3, .... up to n.
The other quicker method is to use
equation to get answer.
The equation is
∑[i=1 to i=n]{i} = n*(n+1)/2 ---eqn.AA01

<a name="docA02">
Next problem, how to calculate
∑[i=1 to i=n]{i2} ?
numerical method is easy. Next is the
sum equation. (omit '[i=1 to i=n]')
∑{i2} = n*(n+1)*(2*n+1)/6 ---eqn.AA02
Similarly
∑{i3} = n*n*(n+1)*(n+1)/4 ---eqn.AA03
∑{i4} = n*(n+1)*(6*n*n*n+9*n*n+n-1)/30 ---eqn.AA04

<a name="docA03">
2009-08-28-18-48 LiuHH access
http://www.jackpo.org/wp-content/uploads/2006/03/formulas.pdf
save as jackpo.org_formulas.pdf
page 2 (total 10 pages) has above four
equations.

The immediate question is
what equation for ∑{i5} ?
Guess work just waste time, no use.
Spend two days find out how to get
∑{i5} equation. This method can be
extended to higher ∑{iHI} equations.
But in realty, ∑{i9} break down due
to overflow/underflow/truncation error
etc.

<a name="docA04">
How to find out equation for ∑{i5}?
= n*n*n*n*n*n/6+n*n*n*n*n/2
+5*n*n*n*n/12-n*n/12  ---eqn.AA05
Explain as following.

Look at
∑[i=1 to i=n]{i} = n*(n+1)/2 ---eqn.AA01
Here i to first power i^1 get highest
polynomial n*n in second power.

<a name="docA05">
Observation:
The general equation for ∑{ipower}
has highest power npower+1 in 'n'

Set up a general equation for ∑{i5}
It is polynomial in 5+1 power (n*n*n*n*n*n)
5+1 power polynomial has 5+2 coefficients.
Let
∑{i5}= a*n*n*n*n*n*n +b*n*n*n*n*n
+c*n*n*n*n     +d*n*n*n
+e*n*n +f*n +g ---eqn.AA06
Here a,b,c,d,e,f,g are seven unknowns.
To solve for seven unknowns need seven
linearly independent equation. To get
this much information, let n=1 to n=7.

<a name="docA06">
proof, program, Conjecture, triangle
For n=1, eqn.AA06 become
a*1 +b*1 +c*1 +d*1 +e*1 +f*n +g
= numerical result = 1 (n=1, 15)
(Above is a rule: all coef. sum to one)

For n=2, eqn.AA06 become
a*26 +b*25 +c*24 +d*23 +e*22 +f*21 +g*20
= numerical result = 33 (n=2, 15+25)

<a name="docA07">
do the same process up to n=7
(n=7, 15+25+35+45+55+65+75=29008)
Finally get the following equation

[   1   1   1   1   1  1  1][a] [    1]
[ 2^6 2^5 2^4 2^3 2*2  2  1][b]=[   33] ---eqn.AA07
[ 3^6 3^5 3^4 3^3 3*3  3  1][c] [  276]
[ 4^6 4^5 4^4 4^3 4*4  4  1][d] [ 1300]
[ 5^6 5^5 5^4 5^3 5*5  5  1][e] [ 4425]
[ 6^6 6^5 6^4 6^3 6*6  6  1][f] [12201]
[ 7^6 7^5 7^4 7^3 7*7  7  1][g] [29008]

<a name="docA08">
eqn.AA07 right hand side column vector
is numerical result.
eqn.AA07 left hand side 7*7 matrix is
coefficient matrix.
a,b,c,d,e,f,g are seven unknowns.

<a name="docA09">
Here we ask Arndt Brünner for help

Arndt Brünner's program solve for
a,b,c,d,e,f,g
These number output to box1 in two styles.

First is coefficients only
[[
∑ i^(5) for i=1 to i=7
coefficients from high power to constant
1/6, 1/2, 5/12, 0, -1/12, 0, 0
]]

Second is mathematics equation
[[
Math equation is
+n*n*n*n*n*n*(1/6)+n*n*n*n*n*(1/2)+n*n*n*n*(5/12)+n*n*n*(0)+n*n*(-1/12)+n*(0)+(0)
]]

<a name="docA10">
Nine equations coefficients are listed at
docB01
What guide line can we find?
1: Highest power coefficients are 1/(k+1)
k is power in ∑ i^(k) for i=1 to i=n
2: Second high power coefficients are 1/2
3: Constant is always zero.
What else can we get from this triangle?
2009-09-01-21-12 stop

<a name="docA11">
proof, program, Conjecture, triangle
2009-09-01-21-24 start
One more observation
4: each coefficient set sum to 1.

1=1/2+1/2+0
1=1/3+1/2+1/6+0
1=1/4+1/2+1/4+0+0
1=1/5+1/2+1/3+0+(-1/30)+0
1=1/6+1/2+5/12+0+(-1/12)+0+0
1=1/7+1/2+1/2+0+(-1/6)+0+1/42+0
1=1/8+1/2+7/12+0+(-7/24)+0+1/12+0+0
1=1/9+1/2+2/3+0+(-7/15)+0+2/9+0+(-1/30)+0
1=1/10+1/2+3/4+0+(-7/10)+0+1/2+0+(-3/20)+0+0
2009-09-01-21-26 stop

<a name="docA12">
2009-09-02-19-26 start
Consider next equation for ∑ i^(3) from i=1 to i=n
[  1   1   1  1  1][a] [1^3                ]
[2^4 2^3 2*2  2  1][b]=[1^3+2^3            ] ---eqn.AA08
[3^4 3^3 3*3  3  1][c] [1^3+2^3+3^3        ]
[4^4 4^3 4*4  4  1][d] [1^3+2^3+3^3+4^3    ]
[5^4 5^3 5*5  5  1][e] [1^3+2^3+3^3+4^3+5^3]

<a name="docA13">
It has the following value
[  1   1   1   1  1][a] [   1]
[ 16   8   4   2  1][b]=[   9] ---eqn.AA09
[ 81  27   9   3  1][c] [  36]
[256  64  16   4  1][d] [ 100]
[625 125  25   5  1][e] [ 225]

<a name="docA14">
Please pay attention to eqn.AA08
Left  hand side is matrix evaluate equation
Right hand side is vector evaluate summation

See eqn.AA08 last row
[5^4 5^3 5*5  5  1][◎]=[1^3+2^3+3^3+4^3+5^3]
('◎' indicate not important)
<a name="docA15">
Matrix row  [5^4 5^3 5*5  5   1 ] know n only
Last row is [n^4 n^3 n^2 n^1 n^0] ---eqn.AA10
Compare above left side with right side below
Vector elem.[1^3+2^3+3^3+4^3+5^3] ---eqn.AA11
Vector element know k=3rd power only.
eqn.AA10: n=constant,     power=0,1,2,3,4
eqn.AA11: Sum from 1 to n power=3=const.
Reader should pay attention to this difference.

<a name="docA16">
proof, program, Conjecture, triangle
To solve for coefficient a,b,c,d,e we must
know right hand side value. Right hand side
is numerical data.
2009-09-02-19-52 stop
```
<a name="lowrank">
2009-09-01-15-25
Lower rank formulas are here
∑ i^(k) from i=1 to i=n, k=1,2,3,4. k=positive integer only
∑ i^(1) = n*(n+1)/2
∑ i^(2) = n*(n+1)*(2*n+1)/6
∑ i^(3) = n*n*(n+1)*(n+1)/4
∑ i^(4) = n*(n+1)*(6*n*n*n+9*n*n+n-1)/30
∑ i^(5)= n*n*n*n*n*n/6+n*n*n*n*n/2+5*n*n*n*n/12-n*n/12
<a name="program0"> 　Arndt program 　first9
To solve for Positive Integer power sum COefficient
need to use Arndt Brünner's work below. Thank you Arndt !
Program take care Arndt codes, user just click
Output to Box1.   for ∑ i^( ) from i=1 to i=n
Try ∑ i^(0) to ∑ i^(9), program break down at ∑ i^(9).
Box 1 Output

Goto freeman2.com/complex2.htm box3, define line 1 "n=8" Paste
math equation at line 2, click [test box3 command, output to box4]
<a name="sum_i_from_1_to_n"> Arndt program 　first9
Power series summation. 2009-09-02-12-30
upper bound n Lower bound is 1.

i^1.2 fraction power OK

Find answer here

```<a name="sum_eqn">
2009-09-02-11-40 sum equations
∑ i to n   = +n*n*(1/2)+n*(1/2)
∑ i*i to n = +n*n*n*(1/3)+n*n*(1/2)+n*(1/6)
∑ i^3 to n = +n*n*n*n*(1/4)+n*n*n*(1/2)+n*n*(1/4)
∑ i^4 to n = +n*n*n*n*n*(1/5)+n*n*n*n*(1/2)+n*n*n*(1/3)+n*n*(0)+n*(-1/30)
∑ i^5 to n = +n*n*n*n*n*n*(1/6)+n*n*n*n*n*(1/2)+n*n*n*n*(5/12)+n*n*n*(0)+n*n*(-1/12)
∑ i^6 to n = +n*n*n*n*n*n*n*(1/7)+n*n*n*n*n*n*(1/2)+n*n*n*n*n*(1/2)+n*n*n*n*(0)+n*n*n*(-1/6)+n*n*(0)+n*(1/42)
∑ i^7 to n = +n*n*n*n*n*n*n*n*(1/8)+n*n*n*n*n*n*n*(1/2)+n*n*n*n*n*n*(7/12)+n*n*n*n*n*(0)+n*n*n*n*(-7/24)+n*n*n*(0)+n*n*(1/12)
∑ i^8 to n = +n*n*n*n*n*n*n*n*n*(1/9)+n*n*n*n*n*n*n*n*(1/2)+n*n*n*n*n*n*n*(2/3)+n*n*n*n*n*n*(0)+n*n*n*n*n*(-7/15)+n*n*n*n*(0)+n*n*n*(2/9)+n*n*(0)+n*(-1/30)
∑ i^9 to n = +n*n*n*n*n*n*n*n*n*n*(1/10)+n*n*n*n*n*n*n*n*n*(1/2)+n*n*n*n*n*n*n*n*(3/4)+n*n*n*n*n*n*n*(0)+n*n*n*n*n*n*(-7/10)+n*n*n*n*n*(0)+n*n*n*n*(1/2)+n*n*n*(0)+n*n*(-3/20)+n*(0)+(0)
∑i^10 to n = +n*n*n*n*n*n*n*n*n*n*n*(1/11)+n*n*n*n*n*n*n*n*n*n*(1/2)+n*n*n*n*n*n*n*n*n*(10/12)+n*n*n*n*n*n*n*n*(0)+n*n*n*n*n*n*n*(-1)+n*n*n*n*n*n*(0)+n*n*n*n*n*(1)+n*n*n*n*(0)+n*n*n*(-1/2)+n*n*(0)+n*(5/66)+(0)
∑i^11 to n = +n*n*n*n*n*n*n*n*n*n*n*n*(1/12)+n*n*n*n*n*n*n*n*n*n*n*(1/2)+n*n*n*n*n*n*n*n*n*n*(11/12)+n*n*n*n*n*n*n*n*n*(0)+n*n*n*n*n*n*n*n*(-165/120)+n*n*n*n*n*n*n*(0)+n*n*n*n*n*n*(462/252)+n*n*n*n*n*(0)+n*n*n*n*(-165/120)+n*n*n*(0)+n*n*(50/120)
∑i^12 to n = +n*n*n*n*n*n*n*n*n*n*n*n*n*(1/13)+n*n*n*n*n*n*n*n*n*n*n*n*(1/2)+n*n*n*n*n*n*n*n*n*n*n*(12/12)+n*n*n*n*n*n*n*n*n*n*(0)+n*n*n*n*n*n*n*n*n*(-220/120)+n*n*n*n*n*n*n*n*(0)+n*n*n*n*n*n*n*(792/252)+n*n*n*n*n*n*(0)+n*n*n*n*n*(-396/120)+n*n*n*n*(0)+n*n*n*(2200/1320)+n*n*(0)+n*(-7601/30030)

2009-09-02-11-48 done sum equations ∑i to ∑i^10
2009-09-03-19-58 found ∑i^11
2009-09-03-20-11  add  ∑i^11
2009-09-04-10-17 found ∑i^12
2009-09-04-10-22  add  ∑i^12
```

```<a name="docB01">
proof, program, Conjecture, triangle
2009-09-01-18-30
∑ i^(1) coefficients from high power to constant
1/2, 1/2, 0

∑ i^(2)
1/3, 1/2, 1/6, 0

∑ i^(3)
1/4, 1/2, 1/4, 0, 0

∑ i^(4)
1/5, 1/2, 1/3, 0, -1/30, 0

∑ i^(5)
1/6, 1/2, 5/12, 0, -1/12, 0, 0

∑ i^(6)
1/7, 1/2, 1/2, 0, -1/6, 0, 1/42, 0

∑ i^(7)
1/8, 1/2, 7/12, 0, -7/24, 0, 1/12, 0, 0

∑ i^(8)
1/9, 1/2, 2/3, 0, -7/15, 0, 2/9, 0, -1/30, 0

∑ i^(9) coefficients from high power to constant
1/10, 1/2, 3/4, 0, -7/10, 0, 1/2, 0, -3/20, 0, 0

∑ i^(9) Math equation is
+n*n*n*n*n*n*n*n*n*n*(1/10)+n*n*n*n*n*n*n*n*n*(1/2)+n*n*n*n*n*n*n*n*(3/4)+n*n*n*n*n*n*n*(0)+n*n*n*n*n*n*(-7/10)+n*n*n*n*n*(0)+n*n*n*n*(1/2)+n*n*n*(0)+n*n*(-3/20)+n*(0)+(0)

Program break down at ∑ i^(9)
∑ i^(9) coefficients are repaired
2009-09-01-18-52

<a name="docB02">
2009-09-02-00-15
Conjectures for coef. triangle
1: Highest power coefficients are 1/(k+1)
k is power in ∑ i^(k) for i=1 to i=n
2: Second high power coefficients are 1/2
3: Constant is always zero.
4: each coefficient set sum to 1.
5: Third high power coefficients are #/12
'#' increase one when power increase one.
<a name="docB03">
6: Fourth high power coefficients are zero
7: Fifth  high power coefficients are negative.
8: Start from third high power coefficients,
next power has zero coef. Then negative,
then zero, then positive, then zero etc.
9: Column five 2009-09-03-23-58
These conjectures have no proof. We can not
predict the unknown (11)th coefficient set.
2009-09-02-00-25

<a name="docB04"> Pattern conjectures, proof
2009-09-01-22-21 Coefficient triangle
∑ i^(k) , (k)=(1) to (k)=(12)
coefficients from high power to constant
(1) : 1/2, 1/2,  0　＜＝this is ∑{i}=n*(n+1)/2=n*n/2+n/2+0
(2) : 1/3, 1/2, 2/12, 0　＜＝this is ∑{i*i}=n*(n+1)*(2*n+1)/6
(3) : 1/4, 1/2, 3/12, 0,    0　＜＝this is ∑{i^3}=[n*(n+1)/2]^2
(4) : 1/5, 1/2, 4/12, 0,  -4/120, 0
(5) : 1/6, 1/2, 5/12, 0, -10/120, 0,   0
(6) : 1/7, 1/2, 6/12, 0, -20/120, 0,  6/252, 0
(7) : 1/8, 1/2, 7/12, 0, -35/120, 0, 21/252, 0,    0
(8) : 1/9, 1/2, 8/12, 0, -56/120, 0, 56/252, 0,  -4/120, 0
(9) :1/10, 1/2, 9/12, 0, -84/120, 0,126/252, 0, -18/120, 0,    0
(10):1/11, 1/2,10/12, 0,-120/120, 0,252/252, 0, -60/120, 0, 100/1320, 0
(11):1/12, 1/2,11/12, 0,-165/120, 0,462/252, 0,-165/120, 0, 550/1320, 0,     0
(12):1/13, 1/2,12/12, 0,-220/120, 0,792/252, 0,-396/120, 0,2200/1320, 0,-7601/30030,  0
column 01   02   03  04     05   06    07   08     09   10     11    12      13      14
column 01,02,03,04,06,08,10,12,14 see Pattern conjectures
column 05 conjecture and calculator.

<a name="docB05">
(10):1/11, 1/2,10/12, 0,-120/120,0,  1/1,   0,-30/60, 0, 5/66,  0
2009-09-01-22-48 guess ∑ i^(10)
From coefficient sum to 1 find '5/66'

∑ i^(10) Math equation is
+n*n*n*n*n*n*n*n*n*n*n*(1/11)
+n*n*n*n*n*n*n*n*n*n*(1/2)
+n*n*n*n*n*n*n*n*n*(10/12)
+n*n*n*n*n*n*n*n*(0)
+n*n*n*n*n*n*n*(-1)
+n*n*n*n*n*n*(0)
+n*n*n*n*n*(1)
+n*n*n*n*(0)
+n*n*n*(-1/2)
+n*n*(0)
+n*(5/66)
+(0)
2009-09-01-22-51
<a name="docB06">
proof, program, Conjecture, triangle
+n*n*n*n*n*n*n*n*n*n*n*(1/11)+n*n*n*n*n*n*n*n*n*n*(1/2)+n*n*n*n*n*n*n*n*n*(10/12)+n*n*n*n*n*n*n*n*(0)+n*n*n*n*n*n*n*(-1)+n*n*n*n*n*n*(0)+n*n*n*n*n*(1)+n*n*n*n*(0)+n*n*n*(-1/2)+n*n*(0)+n*(5/66)+(0)

<a name="docB07">
2009-09-02-06-15
4: each coefficient set sum to 1.
is easy to explain.
Any summation equation must be true for
n=1 and 1^any_power=1, this condition
force all coefficient sum to 1.
2009-09-02-06-18

<a name="docB08">
2009-09-02-06-42
3: Constant is always zero.
any equation is valid for n=0
right hand side is 0
then constant must be = 0
2009-09-02-06-43

<a name="docB09">
2009-09-03-10-59
'update 2009-09-03' add the following
code
[[
ouStr1=ouStr1.replace(/, /g,'; ');//9809031050
ouStr1=ouStr1.replace(/,/g,'.');
ouStr1=ouStr1.replace(/; /g,', ');
]]

<a name="docB10">
Before add '9809031050' code,
Math equation is (too long, omit)
***** big error equation *****
∑ i^(9) from i=1 to i=10. get
8.115895624042362e+44
but correct answer is
From eqn. get: 1574304985

<a name="docB11">
proof, program, Conjecture, triangle
After add '9809031050' code,
Math equation is (too long, omit)
***** correct equation *****
∑ i^(9) from i=1 to i=10. get
1574304985.0000004
correct answer is
1574304985

<a name="docB12">
The reason that before 'update 2009-09-03'
generate big error answer
8.115895624042362e+44
is that when request ∑ i^(9)
Arndt Brünner's work output to the following
[[
coefficients from high power to constant
6164081965611661/61640819656134050, 5155004252058715/10310008504085338, 4908405681047445/6544540908389189, 2471137929879305/4,811589347427001e+24, -6252256789683410/8931795356868823, 766295064850303/3,0021419880618152e+22, 4994368678553267/9988739288667978, 1612987475556151/6,778429079461882e+21, -4555028353434601/30366782630979680, 6036212294932357/2,0070355172700415e+22, -8816619631634209/8,586325069295574e+22
]]
where comma in [...3, 7...]
and  comma in [8,586325069295574e+22]
have different meaning.
LiuHH did not pay attention to this difference.
Use both comma as separator and cause big error.

<a name="docB13"> 2009-09-03-14-13 repeat words deleted,
2009-09-03-11-12

<a name="docB14">
"Update 2009-09-04" add ∑ i^(11) equation
"Update 2009-09-04" add huge to small code.
But no help. 2009-09-03-22-01 record
"Update 2009-09-04" add ∑ i^(12) equation. 2009-09-04-11-18 record
"Update 2009-09-04" add column 5 calculator. 2009-09-04-15-43 record

[[
∑ i^(12) for i=1 to i=n
coefficients from high power to constant
6291476941929729/81789200323615760, 2024007280027114/4048014431611459, 4889929392869138/4889934481241627, 2820674523245057/80225921439448660000, -12676584503961494/6911755417179109, 698226521061421/69727488742124080, 3033039204646851/995193537484100, 3919984334779285/6162049301525286, -44938138956937790/7138933101912657, 26108525088125380/2665864212641073, -8541165113894874/429708028862003, 156590788714843520/5221488374424287, -172590431458016930/7285082543775481, 41899526063808296/5482558770388439

Math equation is
+n*n*n*n*n*n*n*n*n*n*n*n*n*(6291476941929729/81789200323615760)+n*n*n*n*n*n*n*n*n*n*n*n*(2024007280027114/4048014431611459)+n*n*n*n*n*n*n*n*n*n*n*(4889929392869138/4889934481241627)+n*n*n*n*n*n*n*n*n*n*(2820674523245057/80225921439448660000)+n*n*n*n*n*n*n*n*n*(-12676584503961494/6911755417179109)+n*n*n*n*n*n*n*n*(698226521061421/69727488742124080)+n*n*n*n*n*n*n*(3033039204646851/995193537484100)+n*n*n*n*n*n*(3919984334779285/6162049301525286)+n*n*n*n*n*(-44938138956937790/7138933101912657)+n*n*n*n*(26108525088125380/2665864212641073)+n*n*n*(-8541165113894874/429708028862003)+n*n*(156590788714843520/5221488374424287)+n*(-172590431458016930/7285082543775481)+(41899526063808296/5482558770388439)
constant term
41899526063808296/5482558770388439
=7.642330491760459
]]
2009-09-03-13-55
∑ i^(12) from i=1 to i=n
Why constant is not zero ??!!
dig !!

<a name="docB15">
2009-09-03-20-00
LiuHH target at ∑ i^(12)
Why  ∑ i^(12)  constant is not zero?

Find ∑ i^(11) first
half try and error,
half follow pattern
start fron 2009-09-03-13-55
on 2009-09-03-19-58 found
<a name="docB16">
proof, program, Conjecture, triangle
∑ i^(11) Math equation is
+n*n*n*n*n*n*n*n*n*n*n*n*(1/12)
+n*n*n*n*n*n*n*n*n*n*n*(1/2)
+n*n*n*n*n*n*n*n*n*n*(11/12)
+n*n*n*n*n*n*n*n*n*(0)
+n*n*n*n*n*n*n*n*(-165/120)
+n*n*n*n*n*n*n*(0)
+n*n*n*n*n*n*(462/252)
+n*n*n*n*n*(0)
+n*n*n*n*(-165/120)
+n*n*n*(0)
+n*n*(50/120)
+n*(0)
+(0)

∑ i^11 to n =
+n*n*n*n*n*n*n*n*n*n*n*n*(1/12)+n*n*n*n*n*n*n*n*n*n*n*(1/2)+n*n*n*n*n*n*n*n*n*n*(11/12)+n*n*n*n*n*n*n*n*n*(0)+n*n*n*n*n*n*n*n*(-165/120)+n*n*n*n*n*n*n*(0)+n*n*n*n*n*n*(462/252)+n*n*n*n*n*(0)+n*n*n*n*(-165/120)+n*n*n*(0)+n*n*(50/120)
2009-09-03-20-03

<a name="docB17">
2009-09-03-20-57
∑ i^(12) for i=1 to i=n
is out of reach !!!
Too much deviation from the pattern !!
2009-09-03-20-59

<a name="docB18">
2009-09-03-23-22
Coefficient triangle column 1,2,3,4 pattern are
easy to observe.
Observed pattern for column five is next.
copy column five to below at left column
-4/120 ,  4+ 6=10  ,  6+4=10
-10/120 , 10+10=20  , 10+5=15
-20/120 , 20+15=35  , 15+6=21
-35/120 , 35+21=56  , 21+7=28
-56/120 , 56+28=84  , 28+8=36
-84/120 , 84+36=120 , 36+9=45 !!! RULE 2009-09-03-23-27
-120/120 ,120+45=165 ,   +10   <==future '+10'
-165/120 ,

<a name="docB19">
2009-09-03-23-38
for ∑ i^(12) column five
predict_step_1 45+10=55
predict_step_2 165+55=220
predict_step_3 -220/120

Observed rule for column five is CORRECT
In LiuHH's search for ∑ i^(12) work
record show (-220/120) used
+n*n*n*n*n*n*n*n*n*(-220/120)
2009-09-03-23-40
```
program <a name="getCol5"> Do not confuse i with n
∑i^5 [for i=1 to i=3] = 1^5+2^5+3^5 = 276 (n=3, k=5)
Above is numerical calculation use 1,2,3 fifth power.
Below is same problem formula calculation for n=3, k=5
n^6*(1/6)+n^5*(1/2)+n^4*(5/12)+n*n*(-1/12) = 276
Above is power=k=5 equation. It is true for any upper
bound n. If let n=3, above equation give answer 276.
[col 1-9] button give you first nine coefficients.
2009-09-04-15-19 A utility help decide column 5
∑i^( )

```<a name="docB20">
proof, program, Conjecture, triangle
2009-09-04-10-46
Motivation for finding ∑ i^(12) equation is that
Arndt Brünner's output for ∑ i^(12) equation has
non-zero constant. It get reasonable answer.
So the conjecture that constant be zero break
down? LiuHH hope to find the answer. Dug for
∑ i^(11) equation then ∑ i^(12) equation. Final
answer is that ∑ i^(12) equation constant is zero.

<a name="docB21">
Find ∑ i^(12) equation, first follow the pattern
and guess coefficient constant. Failed. Twice
decide give up.
The success path is the following.
Build full matrix equation.

<a name="docB22">
2009-09-04-08-06 ∑ i^(12) a,b,c,d,e,f,g,h,i,j,k,l,m,n 14 variables
[   1     1     1     1     1    1    1    1    1    1    1    1    1  1][a] [    1]
[  2^13  2^12  2^11  2^10  2^9  2^8  2^7  2^6  2^5  2^4  2^3  2*2   2  1][b]=[ 4097] ---eqn.AA101
[  3^13  3^12  3^11  3^10  3^9  3^8  3^7  3^6  3^5  3^4  3^3  3*3   3  1][c] [ 535538]
[  4^13  4^12  4^11  4^10  4^9  4^8  4^7  4^6  4^5  4^4  4^3  4*4   4  1][d] [ 17312754]
[  5^13  5^12  5^11  5^10  5^9  5^8  5^7  5^6  5^5  5^4  5^3  5*5   5  1][e] [ 261453379]
[  6^13  6^12  6^11  6^10  6^9  6^8  6^7  6^6  6^5  6^4  6^3  6*6   6  1][f] [2438235715]
[  7^13  7^12  7^11  7^10  7^9  7^8  7^7  7^6  7^5  7^4  7^3  7*7   7  1][g] [16279522916]
[  8^13  8^12  8^11  8^10  8^9  8^8  8^7  8^6  8^5  8^4  8^3  8*8   8  1][h] [84998999652]
[  9^13  9^12  9^11  9^10  9^9  9^8  9^7  9^6  9^5  9^4  9^3  9*9   9  1][i] [367428536133]
[ 10^13 10^12 10^11 10^10 10^9 10^8 10^7 10^6 10^5 10^4 10^3 10*10 10  1][j] [1367428536133]
[ 11^13 11^12 11^11 11^10 11^9 11^8 11^7 11^6 11^5 11^4 11^3 11*11 11  1][k] [4505856912854]
[ 12^13 12^12 12^11 12^10 12^9 12^8 12^7 12^6 12^5 12^4 12^3 12*12 12  1][l] [13421957361110]
[ 13^13 13^12 13^11 13^10 13^9 13^8 13^7 13^6 13^5 13^4 13^3 13*13 13  1][m] [36720042483591]
[ 14^13 14^12 14^11 14^10 14^9 14^8 14^7 14^6 14^5 14^4 14^3 14*14 14  1][n] [93413954858887]
2009-09-04-08-29

<a name="docB23">
No need to solve whole equation, because
[[
a=1/13
b=1/2
c=12/12
d=0
e=-220/120
f=0
g=w=792/252
h=0
i=x=-396/120
j=0
k=y=2200/1320
l=0
m=z=-7601/30030
n=0
2009-09-04-10-15
]]
a=1/13
b=1/2
c=12/12
d=0
f=0
h=0
j=0
l=0
n=0
are described in Conjectures
e=-220/120 is described here.
only w,x,y,z four coefficients are uncertain.

<a name="docB24">
Build four equations
[[
2009-09-04-09-40
1/13   +1/2    +1    -1*(220/120)   1*w    1*x     1*y    1*z =        1
2^13/13 +2^12/2 +2^11 -2^9*(220/120) 2^7*w  2^5*x  2^3*y   2*z =     4097 (=1^12+2^12)
3^13/13 +3^12/2 +3^11 -3^9*(220/120) 3^7*w  3^5*x  3^3*y   3*z =   535538 (=1^12+2^12+3^12)
4^13/13 +4^12/2 +4^11 -4^9*(220/120) 4^7*w  4^5*x  4^3*y   4*z = 17312754 (=1^12+2^12+3^12+4^12) [=1+pow(2,12)+pow(3,12)+pow(4,12)]

2009-09-04-09-58
1*w+   1*x+  1*y+ 1*z =     1.2564102564102564
128*w+  32*x+  8*y+ 2*z =   309.5128205128208
2187*w+ 243*x+ 27*y+ 3*z =  6115.769230769249
16384*w+1024*x+ 64*y+ 4*z = 48219.02564102411
]]
<a name="docB25">
Ask Arndt Brünner for help, get solution.
[[
2009-09-04-09-59
w = 9104177589384788/2896783778440913
x = -17220659735585978/5218381738063757
y = 9187617697792860/5512570618730461
z = -768920039355504/3037846175892421
]]

<a name="docB26">
proof, program, Conjecture, triangle
w is column 7. Previous work, lower rank
equation column 7 denominator is 252. then
w = 9104177589384788/2896783778440913
become
g=w=792/252

Similarly refer to previous denominator
find
i=x=-396/120
k=y=2200/1320

<a name="docB27">
Only
m=z=-7601/30030
no previous denominator.
How to find z ? (coef. m=z unknown)
Working record is next
[[
2009-09-04-10-05
1/13+1/2+1-220/120+792/252-396/120+2200/1320+z=1

z=1-1/13-1/2-1+220/120-792/252+396/120-2200/1320
z=-0.2531135531135533

<a name="docB28">
1320=10*11*12
120=10*12
252=3*12*7
10*11*12*13*7=120120
z=-30404/120120
2009-09-04-10-12
z=-7601/30030
]]
Up to here, general equation for ∑ i^(12)
is solved.
2009-09-04-11-12
```
<a name="huge_to_sml"> proof, program, Conjecture, triangle
2009-09-03-14-16
∑ i^(k) when k>8, program do not output small integer rational number. But output huge integer rational number. Box below try to find equivalent small integer rational number (polynomial coefficients)
huge rational:
Iterations:
Box 2 Output

Box2 not really help, if denominator blur a little bit,
Program will not return to the right denominator. 2009-09-03-20-32
```<a name="docB29">
2009-09-04-21-50
Following is a small calculation for
∑ i^3 to n = +n*n*n*n*(1/4)+n*n*n*(1/2)+n*n*(1/4) ---eqn.AA102
case, Find out coefficient values.
The key point to get answer
is that n=arbitrary.
Following is working record.
2009-09-04-21-53

<a name="docB30">
2009-09-04-21-12 for power=3
∑ i^3 to n = +n*n*n*n*(1/4)+n*n*n*(1/2)+n*n*(1/4)

SUM i*i = n*n*n*n*a + n*n*n*b + n*n*c + n*d + e
=1^3 + 2^3 + 3^3 + 4^3 + .... + n^3 ---eqn.AA103
why a=1/4 ?
why b=1/2 ?
big puzzle.
<a name="docB31">
proof, program, Conjecture, triangle

KEY observation
1^3 +...+ n^3 use
n*n*n*n*a + n*n*n*b + n*n*c + n*d + e ---eqn.AA104

1^3 +...+ (n-1)^3 use
(n-1)*(n-1)*(n-1)*(n-1)*a
+ (n-1)*(n-1)*(n-1)*b
+ (n-1)*(n-1)*c + (n-1)*d + e ---eqn.AA105

<a name="docB32">
move n^3 from right hand side to left hand
side.
n*n*n*n*a + n*n*n*(b-1) + n*n*c + n*d + e ---eqn.AA106
=1^3 + 2^3 + 3^3 + 4^3 + ....
=
use KEY observation at this step
=
(n-1)*(n-1)*(n-1)*(n-1)*a ---eqn.AA107
+ (n-1)*(n-1)*(n-1)*b
+ (n-1)*(n-1)*c + (n-1)*d + e
=
n*n*n*n*a + n*n*n*(b-1) + n*n*c + n*d + e ---eqn.AA106

e cancel here , e is undetermined here

<a name="docB33">
2009-09-04-21-18 here
(n-1)*(n-1)*(n-1)*(n-1)*a
+ (n-1)*(n-1)*(n-1)*b
+ (n-1)*(n-1)*c + (n-1)*d
=                         ---eqn.AA108
n*n*n*n*a + n*n*n*(b-1) + n*n*c + n*d

<a name="docB34">
2009-09-04-21-20
[[
n=18
a=1/4
b=1/2
c=3/12
d=0
(n-1)*(n-1)*(n-1)*(n-1)*a + (n-1)*(n-1)*(n-1)*b + (n-1)*(n-1)*c + (n-1)*d -n*n*n*n*a - n*n*n*(b-1) - n*n*c - n*d //---eqn.AA109
]]
Goto freeman2.com/complex2.htm
paste above six lines to box3, change
value of n, answer is always zero.

<a name="docB35">
eqn.AA109 is zero for ANY n

(n*n-2*n+1)*(n*n-2*n+1)*a
+ (n*n-2*n+1)*(n-1)*b
+ (n*n-2*n+1)*c + (n-1)*d
-n*n*n*n*a - n*n*n*(b-1)
- n*n*c - n*d  ---eqn.AA110
=0
=
<a name="docB36">
proof, program, Conjecture, triangle
(n*n*n*n-2*n*n*n+n*n
-2*n*n*n+2*2*n*n-2*n*1
n*n-2*n+1
)*a
+ (
n*n*n-2*n*n+1*n
-n*n+2*n-1
)*b
+ (n*n-2*n+1)*c
+ (n-1)*d
-n*n*n*n*a - n*n*n*(b-1)
- n*n*c - n*d    ---eqn.AA111
<a name="docB37">
=0
=
n*n*n*n*(a-a)
+n*n*n*(-2a-2a+b-b+1)
+n*n*(6a-3b+c-c)
+n*(-4a+3b-2c+d-d)
+a-b+c-d          ---eqn.AA112

<a name="docB38">
2009-09-04-21-34 since n is arbitrary
must its coef be zero, get
n*n*n*n*(a-a)         ==> a-a=0  no help  ---eqn.AA113
+n*n*n*(-2a-2a+b-b+1) ==> -2a-2a+b-b+1=0  ---eqn.AA114
+n*n*(6a-3b+c-c)      ==> 6a-3b+c-c=0     ---eqn.AA115
+n*(-4a+3b-2c+d-d)    ==> -4a+3b-2c+d-d=0 ---eqn.AA116
+a-b+c-d =0                               ---eqn.AA117
<a name="docB39">
2009-09-04-21-36 here
eqn.AA114:  -2a-2a+1=0 ==> a=1/4          ---eqn.AA118
eqn.AA115:  6a-3b =0   ==> b=2a=2/4=1/2   ---eqn.AA119
eqn.AA116: -4a+3b-2c=0 ==> 2c=-4a+3b
==> 2c=-4/4+3/2=1/2 ==> c=1/4     ---eqn.AA120
eqn.AA117: +a-b+c-d =0 ==> d=a-b+c
==> d=1/4-1/2+1/4=0               ---eqn.AA121

<a name="docB40">
Now find e
SUM i*i = n*n*n*n*a + n*n*n*b + n*n*c + n*d + e
since n is arbitrary, and solution is unique
it must be e=0                            ---eqn.AA122

Put all coefficient together get
∑ i^3 to n = +n*n*n*n*(1/4)+n*n*n*(1/2)+n*n*(1/4)

Above reasoning is special case for power=3
not general for power=any
power=any derivation is similar.
2009-09-04-21-47

<a name="docB41">
The following is proof section.
Update 2009-09-05 add the following proof.
Proof coef. column 01 02 03 04 05 List

proof, program, Conjecture, triangle
2009-09-05-08-32
Please view coefficient triangle.
Column 01 coefficient is 1/(power+1)
Column 02 coefficient is 1/2
Column 03 coefficient is power/12
etc.
<a name="docB42">
Is this pattern a coincidence?
Is this pattern a must?
If it is a must result, how to prove?
The following is an explanation.

<a name="docB43">
First observe a special case
∑ i^3 from i=1 to n

Answer from equation, use formula ∑ i^3 = ...
= n*n*n*n*(1/4)+n*n*n*(1/2)+n*n*(1/4) ---eqn.AA201
calculate only i=n th term.
from n_power_0 to n_power_k+1 (here k=3)
Formula contain coefficients (1/4), (1/2), (1/4).
For large n, formula method is quick.

<a name="docB44">
In contrast,
Numerical calculation, use next equation
1^3 + 2^3 + 3^3 + 4^3 + ..... + n^3 ---eqn.AA202
Answer from summation calculate only 3 rd power.
from i=1 to i=n
This summation do not use coefficients.
For large n, summation method is slow.

<a name="docB45">
Calculation use only n th term and
calculation use only 3 rd power are very
different. They get same answer!

For ∑ i^3 we write eqn.AA201=eqn.AA202 as below

+n*n*n*n*(1/4)+n*n*n*(1/2)+n*n*(1/4)
=                                  ---eqn.AA203
1^3 + 2^3 + 3^3 + 4^3 + ..... + n^3

2009-09-05-08-59 here
<a name="docB46">
proof, program, Conjecture, triangle
Above equation is for third power ∑ i^3.
For general case ∑ i^k , write as next

Answer from equation, use formula ∑ i^k = ...
∑[i=1 to n] i^k = ∑[q=0,k+1]pow(n,q)*aq ---eqn.AA204

Answer from summation use loop sum
∑[i=1 to n] i^k = ∑[i=1,n]i^k  ---eqn.AA205

<a name="docB47">
Since eqn.AA204 is general formula, it cover
the case k=3. Check if this is true.
In eqn.AA204 set k=3 get
∑[i=1 to n] i^3 = ∑[q=0,3+1]pow(n,q)*aq
expand ∑[q=0,3+1] get
∑[i=1 to n] i^3 = pow(n,0)*a0 + pow(n,1)*a1
+ pow(n,2)*a2 + pow(n,3)*a3 + pow(n,4)*a4
<a name="docB48">
It is same as next more familiar form
∑[i=1 to n] i^3 = n^0*a0 + n^1*a1
+ n^2*a2 + n^3*a3 + n^4*a4  ---eqn.AA206
for "∑[i=1 to n] i^3" coefficients are
a4=1/4, a3=1/2, a2=1/4, a1=0, a0=0
Then eqn.AA206 is same as eqn.AA201.
<a name="docB49">
How to find a4=1/4 etc. ?
Please goto program input '3'
for third power summation, then click
[calculate equation] button. Output to box1.

Above checked eqn.AA204 it cover
answer from equation power=3 case

<a name="docB50">
Below check   eqn.AA205 see whether it cover
answer from summation power=3 case
∑[i=1 to n] i^k = ∑[i=1,n]i^k  ---eqn.AA205
In eqn.AA205, set k=3 for power=3 case, get
∑[i=1 to n] i^3 = ∑[i=1,n]i^3
= 1^3 + 2^3 + 3^3 + ..... + n^3
It is same as eqn.AA202

Now "eqn.AA204 and eqn.AA205 is general" is
true for k=3 case. For k=other_value, exam
procedure is the same.
2009-09-05-09-33 here

<a name="docB51">
proof, program, Conjecture, triangle
eqn.AA204 and eqn.AA205 get same answer for
power k. We can write eqn.AA204 = eqn.AA205

General equation for positive integer power
sum.
∑[i=1 to n] i^k
= ∑[q=0,k+1]pow(n,q)*aq
= ∑[i=1,n]i^k        ---eqn.AA207

<a name="docB52">
eqn.AA207 is general,
eqn.AA207 is true for sum to n
eqn.AA207 is true for sum to n+1 too
We have next equation for sum to n+1

∑[i=1 to n+1] i^k
= ∑[q=0,k+1]pow(n+1,q)*aq
= ∑[i=1,n+1]i^k        ---eqn.AA208
<a name="docB53">
To prove
Column 01 coefficient is 1/(power+1)
Column 02 coefficient is 1/2
Column 03 coefficient is power/12
etc.
key step is
to relate eqn.AA207 and eqn.AA208

Clearly eqn.AA207 NOT EQUAL eqn.AA208
eqn.AA207 sum from 1^k up to n^k
eqn.AA208 sum from 1^k up to (n+1)^k
Both equation sum to power k. (200909061142)

<a name="docB54">
Two answer from equations are hard to
relate, because expansion of
∑[q=0,k+1]pow(n+1,q)
is complicate

Two answer from summations are next
∑[i=1,n]i^k   ---eqn.AA207 right hand side
∑[i=1,n+1]i^k ---eqn.AA208 right hand side
What a simple job ?!

<a name="docB55">
Sum to n+1 is
∑[i=1,n+1]i^k = ∑[i=1,n]i^k + (n+1)^k
from equation right hand side we get
eqn.AA208 = eqn.AA207 + (n+1)^k
then goto equation left hand side, write

∑[q=0,k+1]pow(n+1,q)*aq = ... ---eqn.AA209
= ∑[q=0,k+1]pow(n,q)*aq + (n+1)^k
2009-09-05-10-12 here

<a name="docB56">
proof, program, Conjecture, triangle
Above walk long way, not start yet.
eqn.AA209 is our start equation.

Slow part in eqn.AA209 is expansion of
pow(n+1,q). Assume q=3 then pow(n+1,3)
is next
(n+1)^3 = (n+1)*(n+1)*(n+1)
=
1*n*n*n+3*n*n+3*n+1

<a name="docB57">
Define bino(s,t) as following.```

 bino(s,t) := ( s 　 t ) ＝ s! t!(s-t)!
---binomial definition
---eqn.AA210
width of above equation
```<a name="docB58">
2009-09-05-10-30 here
Then for q=3
pow(n+1,q) = (n+1)^3 = (n+1)*(n+1)*(n+1)
can be written as
pow(n+1,3) = bino(3,3)*n^3 + bino(3,2)*n^2
+ bino(3,1)*n^1 + bino(3,0)*n^0

Please goto freeman2.com/tute0010.htm
in "binomial m[  ], n[  ]" box,
fill '3' for m and '2' for n
click [get binomial] button for answer.

<a name="docB59">
Numerical value is
pow(n+1,3) = 1*n^3 + 3*n^2
+ 3*n^1 + 1*n^0

Numerical value is good for understanding.
Symbol equation is good for proof for
general case.

<a name="docB60">
Now back to eqn.AA209

∑[q=0,k+1]pow(n+1,q)*aq = ... ---eqn.AA209
= ∑[q=0,k+1]pow(n,q)*aq + (n+1)^k
Expand get
pow(n+1,k+1)*ak+1
+pow(n+1,k)*ak
+pow(n+1,k-1)*ak-1
+pow(n+1,k-2)*ak-2
+ ∑[q=0,k-3]pow(n+1,q)*aq
=                 ---eqn.AA211 (whole eqn.)
(n+1)^k
+ pow(n,k+1)*ak+1
+ pow(n,k)*ak
+ pow(n,k-1)*ak-1
+ pow(n,k-2)*ak-2
+ ∑[q=0,k-3]pow(n,q)*aq
2009-09-05-10-48 here

<a name="docB61">
proof, program, Conjecture, triangle
eqn.AA211 both side let highest four terms
stand out of summation sign.

eqn.AA211 left hand side is (n+1)^power terms
need to use binomial coef. to expand.

eqn.AA211 right hand side is (n)^power terms
no need expansion.

<a name="docB62">
Do expansion one by one
pow(n+1,k+1)*ak+1
=              ---eqn.AA212 (one term)
[
bino(k+1,k+1)*pow(n,k+1)
+bino(k+1,k)  *pow(n,k)
+bino(k+1,k-1)*pow(n,k-1)
+bino(k+1,k-2)*pow(n,k-2)
]
*ak+1
+ ∑[q=0,k-3]bino(k+1,q)*pow(n,q)*ak+1

<a name="docB63">
Next
+pow(n+1,k)*ak
=              ---eqn.AA213 (one term)
[
bino(k,k)  *pow(n,k)
+bino(k,k-1)*pow(n,k-1)
+bino(k,k-2)*pow(n,k-2)
]
*ak
+ ∑[q=0,k-3]bino(k,q)*pow(n,q)*ak

<a name="docB64">
Next
+pow(n+1,k-1)*ak-1
=              ---eqn.AA214 (one term)
[
bino(k-1,k-1)*pow(n,k-1)
+bino(k-1,k-2)*pow(n,k-2)
]
*ak-1
+ ∑[q=0,k-3]bino(k,q)*pow(n,q)*ak-1

<a name="docB65">
Next is the special term which link
∑[i=1,n]i^k and  ∑[i=1,n+1]i^k
expand
(n+1)^k
=           ---eqn.AA215 (one term)
bino(k,k)*pow(n,k)
+bino(k,k-1)*pow(n,k-1)
+bino(k,k-2)*pow(n,k-2)
+ ∑[q=0,k-3]bino(k,q)*pow(n,q)

this term (n+1)^k no ak coef.

<a name="docB66">
proof, program, Conjecture, triangle
Now put
eqn.AA212 (one term)
eqn.AA213 (one term)
eqn.AA214 (one term)
eqn.AA215 (one term)
back to
eqn.AA211 (whole eqn.)

get
<a name="docB67">
[
bino(k+1,k+1)*pow(n,k+1)
+bino(k+1,k)  *pow(n,k)
+bino(k+1,k-1)*pow(n,k-1)
+bino(k+1,k-2)*pow(n,k-2)
]
*ak+1
+ ∑[q=0,k-3]bino(k+1,q)*pow(n,q)*ak+1
+
[
bino(k,k)  *pow(n,k)
+bino(k,k-1)*pow(n,k-1)
+bino(k,k-2)*pow(n,k-2)
]
*ak
+ ∑[q=0,k-3]bino(k,q)*pow(n,q)*ak
+
[
bino(k-1,k-1)*pow(n,k)
+bino(k-1,k-2)*pow(n,k-1)
]
*ak-1
+ ∑[q=0,k-3]bino(k,q)*pow(n,q)*ak-1

+ bino(k-2,k-2)*pow(n,k-2)*ak-2
+ ∑[q=0,k-3]bino(k-2,q)*pow(n,q)*ak-2

<a name="docB68">
=                 ---eqn.AA216 (whole eqn.)
bino(k,k)*pow(n,k)
+bino(k,k-1)*pow(n,k-1)
+bino(k,k-2)*pow(n,k-2)
+ ∑[q=0,k-3]bino(k,q)*pow(n,q)

+ pow(n,k+1)*ak+1
+ pow(n,k)*ak
+ pow(n,k-1)*ak-1
+ pow(n,k-2)*ak-2
+ ∑[q=0,k-3]pow(n,q)*aq

<a name="docB69">
2009-09-05-11-08 here

The important point for eqn.AA216 (whole eqn.)
is that
eqn.AA216 satisfy n=arbitrary_value
for this reason
n^(k+1) coefficient MUST BE ZERO
n^(k)   coefficient MUST BE ZERO
n^(k-1) coefficient MUST BE ZERO
etc.

<a name="docB70">
We need group coefficients together
rewrite eqn.AA216 (whole eqn.)
move everything to one side. The other
side is '=0', get the following

<a name="docB71">
proof, program, Conjecture, triangle
[  ---eqn.AA217 (whole eqn.) START (END)
bino(k+1,k+1)*pow(n,k+1)
+bino(k+1,k)  *pow(n,k)
+bino(k+1,k-1)*pow(n,k-1)
+bino(k+1,k-2)*pow(n,k-2)
]
*ak+1
+ ∑[q=0,k-3]bino(k+1,q)*pow(n,q)*ak+1
+
[
bino(k,k)  *pow(n,k)
+bino(k,k-1)*pow(n,k-1)
+bino(k,k-2)*pow(n,k-2)
]
*ak
+ ∑[q=0,k-3]bino(k,q)*pow(n,q)*ak
<a name="docB72">
+
[
bino(k-1,k-1)*pow(n,k-1)
+bino(k-1,k-2)*pow(n,k-2)
]
*ak-1
+ ∑[q=0,k-3]bino(k,q)*pow(n,q)*ak-1

+ bino(k-2,k-2)*pow(n,k-2)*ak-2
+ ∑[q=0,k-3]bino(k-2,q)*pow(n,q)*ak-1

<a name="docB73">
- bino(k,k)*pow(n,k)
-bino(k,k-1)*pow(n,k-1)
-bino(k,k-2)*pow(n,k-2)
- ∑[q=0,k-3]bino(k,q)*pow(n,q)

- pow(n,k+1)*ak+1
- pow(n,k)*ak
- pow(n,k-1)*ak-1
- pow(n,k-2)*ak-2
- ∑[q=0,k-3]pow(n,q)*aq

= ZERO ---eqn.AA217 (whole eqn.) END (START)

<a name="docB74">
from eqn.AA217
collect coefficients for highest term
pow(n,k+1) *
[
bino(k+1,k+1)*ak+1
- ak+1
]
We know bino(k+1,k+1)=1, so highest term
pow(n,k+1) * (zero)
Nothing new.

<a name="docB75">
Next from eqn.AA217
collect coefficients for second highest term
(bold red terms)
pow(n,k) *
[
+bino(k+1,k)*ak+1
+bino(k,k)  *ak
-bino(k,k)
-            ak
]

Where bino(k,k)=1
and ak term cancel
left
pow(n,k)*[bino(k+1,k)*ak+1-1]

<a name="docB76">
proof, program, Conjecture, triangle
Equation must be true for ANY n
We require pow(n,k) coefficients to be zero
pow(n,k)*
[bino(k+1,k)*ak+1-1] = 0 ---eqn.AA218
This requirement result
bino(k+1,k)*ak+1-1 = 0
or
ak+1 = 1/bino(k+1,k) = 1/(k+1) ---eqn.AA219

<a name="docB77">
Up to here
Proved Coefficient triangle
column 01 coefficient = 1/bino(k+1,k)
column 01 coefficient = 1/(k+1)
2009-09-05-11-39 stop

<a name="docB78">
2009-09-05-13-00 start
Above proved column 01 coefficient = 1/(k+1)
Next  prove  column 02 coefficient = 1/2

From eqn.AA217
collect coefficients for third highest term
(bold blue terms)
pow(n,k-1) *
[
+bino(k+1,k-1)*ak+1
+bino(k,k-1)  *ak
+bino(k-1,k-1)*ak-1
-bino(k,k-1)
-              ak-1
]

Where bino(k-1,k-1)=1 and ak-1 drop
left
<a name="docB79">
pow(n,k-1) * ---eqn.AA220
[
+bino(k+1,k-1)*ak+1
+bino(k,k-1)  *ak
-bino(k,k-1)
]
eqn.AA220 has three terms. they have the
following values
bino(k+1,k-1)=(k+1)!/[(k-1)!*2!]=(k+1)*k/2
from eqn.AA219, ak+1=1/(k+1)
bino(k+1,k-1)*ak+1=(k+1)*k/2/(k+1)=k/2 ---eqn.AA221

<a name="docB80">
bino(k,k-1)*ak=k*ak ---eqn.AA222
-bino(k,k-1)=-k ---eqn.AA223

eqn.AA220 become
pow(n,k-1) * ---eqn.AA224
[
+k/2
+k*ak
-k
]

<a name="docB81">
proof, program, Conjecture, triangle
eqn.AA217 demand equal to zero for ANY n.
It is possible only if coefficients are
all zero. We require eqn.AA224 equal to
zero, get

+k/2 +k*ak -k = 0
then ak=1/2  ---eqn.AA225

<a name="docB82">
Up to here
Proved Coefficient triangle
column 02 coefficient = 1/2
2009-09-05-13-23 here

Above proved column 01 coefficient = 1/(k+1)
Above prove  column 02 coefficient = 1/2
<a name="docB83">
Next  prove  column 03 coefficient = k/12

From eqn.AA217
collect coefficients for fourth highest term
(bold purple terms)
pow(n,k-2) *
[
+bino(k+1,k-2)*ak+1
+bino(k,k-2)*ak
+bino(k-1,k-2)*ak-1
+bino(k-2,k-2)*ak-2
-bino(k,k-2)
-ak-2
]

<a name="docB84">
Evaluate bino(), get
pow(n,k-2) *   ---eqn.AA226
[
+(k+1)!/[(k-2)!*3!]*1/(k+1)
+k!/[(k-2)!*2!]*1/2
+(k-1)*ak-1
+ak-2
-k!/[(k-2)!*2!]
-ak-2
]

<a name="docB85">
Here +bino(k-2,k-2)*ak-2-ak-2 cancel
Demand coefficient to be zero get

+(k+1)!/[(k-2)!*3!]*1/(k+1)
+k!/[(k-2)!*2!]*1/2
+(k-1)*ak-1
-k!/[(k-2)!*2!]
=0

<a name="docB86">
proof, program, Conjecture, triangle
+(k-1)*ak-1
=
k!/[(k-2)!*2!]
-(k+1)!/[(k-2)!*3!]*1/(k+1)
-k!/[(k-2)!*2!]*1/2
=
k(k-1)/2
-k(k-1)/6
-k(k-1)/4
=
k(k-1)/12

(k-1)*ak-1 = k(k-1)/12

ak-1 = k/12   ---eqn.AA227

<a name="docB87">
Up to here
Proved Coefficient triangle
column 03 coefficient = k/12

Other coefficients can be build with same
method.
2009-09-05-13-59 stop

<a name="docB88">
"Update 2009-09-06" add following few more
calculations and add first9 coefficients.

2009-09-06-12-20 document start
Few more calculations.

Next is Coefficient triangle column 04
expect get all zero.

Equations below is 2009-09-05 draft work.
On 2009-09-06 include draft work to web
page and put explanation in web page.

<a name="docB89">
2009-09-05-21-04
From general equation eqn.AA209
or from expanded whole eqn.AA211
or from detailed whole eqn.AA217
Collect n^(k-3) terms. Need its coefficient,
set coef to zero to find ak-2

<a name="docB90">
pow(n,k-3)* ---eqn.AA228
[
+bino(k+1,k-3)*ak+1
+bino(k,k-3)*ak
+bino(k-1,k-3)*ak-1
+bino(k-2,k-3)*ak-2
+bino(k-3,k-3)*ak-3
-bino(k,k-3)
-ak-3
]

<a name="docB91">
proof, program, Conjecture, triangle
All 'bino()' come from (n+1)^k expansion.
Other side of general equation use n^k,
Other side n^k no expansion and no 'bino()'

Next is bino(s,t) evaluation.

<a name="docB92">
2009-09-05-21-11
pow(n,k-3)* ---eqn.AA229
[
(k+1)*(k)*(k-1)*(k-2)*ak+1/4!
(k)*(k-1)*(k-2)*ak/3!
(k-1)*(k-2)*ak-1/2!
(k-2)*ak-2/1!
1*ak-3/0!
-(k)*(k-1)*(k-2)/3!

]

<a name="docB93">
bino(k-3,k-3)*ak-3 -ak-3 = zero
Use ak+1 value
Use ak value
Use ak-1 value
Above equation change to next.

<a name="docB94">
2009-09-05-21-16
pow(n,k-3)* ---eqn.AA230
[
(k+1)*(k)*(k-1)*(k-2)*1/(k+1)/4!
+(k)*(k-1)*(k-2)/2/3!
+(k-1)*(k-2)*k/12/2!
+(k-2)*ak-2
-(k)*(k-1)*(k-2)/3!
]

<a name="docB95">
Write clearly factor '(k)*(k-1)*(k-2)'
as next

2009-09-05-21-18
pow(n,k-3)* ---eqn.AA231
[
(k)*(k-1)*(k-2)/4!
+(k)*(k-1)*(k-2)/2/3!
+ k *(k-1)*(k-2)/12/2!
+(k-2)*ak-2
-(k)*(k-1)*(k-2)/3!
]

<a name="docB96">
proof, program, Conjecture, triangle
Collect coefficient for '(k)*(k-1)*(k-2)'

2009-09-05-21-20
pow(n,k-3)* ---eqn.AA232
[
(k)*(k-1)*(k-2)
*[1/4! + 1/2/3! + 1/12/2! -1/3!]
+(k-2)*ak-2
]

<a name="docB97">
Change factorial to numbers

2009-09-05-21-21
pow(n,k-3)* ---eqn.AA233
[
(k)*(k-1)*(k-2)
*[1/24 + 1/12 + 1/24 -1/6]
+(k-2)*ak-2
]

<a name="docB98">
Calculate

2009-09-05-21-22
pow(n,k-3)* ---eqn.AA234
[
(k)*(k-1)*(k-2)
*[0]
+(k-2)*ak-2
]

<a name="docB99">
Terms in square bracket is pow(n,k-3)
coefficient. Since equation is valid
for ANY n. MUST n's coefficient be zero

demand
(k)*(k-1)*(k-2)*[0] ---eqn.AA235
+(k-2)*ak-2 =0

get ak-2=0 ---eqn.AA236

<a name="docB100">
Up to here
Proved Coefficient triangle
column 04 coefficient = 0
2009-09-05-21-23

<a name="docB101">
proof, program, Conjecture, triangle
2009-09-06-13-05 document to here
Next is Coefficient triangle column 05

When repeat same work few times, it is easy
to find out next new coefficient equation.
1. change all ",k-3)" to ",k-4)"
2. add one new line "+bino(k-4,k-4)*ak-4"
3. change "-ak-3" to "-ak-4"
then start new calculation.
Must carry out point 2. add new line.
In current calculation, point 2. and 3.
two terms cancel each other.
In next iteration, point 2. added new line
provide next iteration with a chance to
change from "+bino(k-4,k-4)*ak-4"
to "+bino(k-4,k-5)*ak-4". so next equation
is complete.

<a name="pownk">
2009-09-06-18-04
What is "pow(n,k-4)" ?
If k=7, "pow(n,k-4)" = "pow(n,3)" = n*n*n

<a name="docB102">
2009-09-05-21-24
pow(n,k-4)*
[
+bino(k+1,k-4)*ak+1
+bino(k  ,k-4)*ak
+bino(k-1,k-4)*ak-1
+bino(k-2,k-4)*ak-2
+bino(k-3,k-4)*ak-3
+bino(k-4,k-4)*ak-4
-bino(k,k-4)
-ak-4
]

<a name="docB103">
Above equation has cancellation.
+bino(k-4,k-4)*ak-4-ak-4
=1*ak-4-ak-4
=0
This type cancellation occurs in each
calculation.

2009-09-05-21-29
pow(n,k-4)*
[
+(k+1)*(k+0)*(k-1)*(k-2)*(k-3)*ak+1/5!
+(k+0)*(k-1)*(k-2)*(k-3)*ak/4!
+(k-1)*(k-2)*(k-3)*ak-1/3!
+(k-2)*(k-3)*ak-2/2!
+(k-3)*ak-3
-k*(k-1)*(k-2)*(k-3)/4!
]

<a name="docB104">
2009-09-05-21-37
Use previous calculated values ak+1, ak
ak-1, ak-2. Please see List. Change above
equation to next equation.
pow(n,k-4)*
[
+(k+1)*(k+0)*(k-1)*(k-2)*(k-3)/(k+1)/5!
+(k+0)*(k-1)*(k-2)*(k-3)/2/4!
+(k-1)*(k-2)*(k-3)*k/12/3!
+(k-2)*(k-3)*0/2!
+(k-3)*ak-3
-k*(k-1)*(k-2)*(k-3)/4!
]

<a name="docB105">
2009-09-05-21-39
pow(n,k-4)*
[
(k+0)*(k-1)*(k-2)*(k-3)/5!
+(k+0)*(k-1)*(k-2)*(k-3)/2/4!
+  k  *(k-1)*(k-2)*(k-3)/12/3!
+(k-3)*ak-3
-k*(k-1)*(k-2)*(k-3)/4!
]

<a name="docB106">
proof, program, Conjecture, triangle
2009-09-05-21-41
pow(n,k-4)*
[
(k+0)*(k-1)*(k-2)*(k-3)*
[1/5! + 1/2/4! + 1/12/3! -1/4!]
+(k-3)*ak-3
]

2009-09-05-21-42
pow(n,k-4)*
[
(k+0)*(k-1)*(k-2)*(k-3)*
[ 1/120 + 1/48 + 1/72 -1/24 ]
+(k-3)*ak-3
]

<a name="docB107">
2009-09-05-21-44
pow(n,k-4)*
[
(k+0)*(k-1)*(k-2)*(k-3)*
[ 1/(3*4*2*5) + 1/(3*4*4) + 1/(2*2*2*3*3) -1/(2*2*2*3) ]
+(k-3)*ak-3
]

2009-09-05-21-48
pow(n,k-4)*
[
(k+0)*(k-1)*(k-2)*(k-3)*
(6+15+10-30)/(3*3*4*4*5)
+(k-3)*ak-3
]

<a name="docB108">
2009-09-05-21-50
pow(n,k-4)*
[
(k+0)*(k-1)*(k-2)*(k-3)*
(+1)/(3*3*4*4*5)
+(k-3)*ak-3
]

require
(k+0)*(k-1)*(k-2)*(k-3)*
(+1)/(3*3*4*4*5)
+(k-3)*ak-3
=0

<a name="docB109">
ak-3=
-k*(k-1)*(k-2)/(3*3*4*4*5)

Up to here
Proved Coefficient triangle
column 05 coefficient = -k*(k-1)*(k-2)/(3*3*4*4*5)
2009-09-05-21-52

The following work is routine and longer.
Omit here. Few more results are listed here.

Below is a summary of coefficients up to
ak-7.

<a name="docB110">
2009-09-05-21-58
summary

ak+1=1/(k+1)
ak  =1/2
ak-1=k/12
ak-2=0
ak-3=-k*(k-1)*(k-2)/(5*4*3*2*3*2)
ak-4=0
ak-5=k*(k-1)*(k-2)*(k-3)*(k-4)/(7*6*5*4*3*2*3*2)
ak-6=0
ak-7=
-k*(k-1)*(k-2)*(k-3)*(k-4)*(k-5)*(k-6)/(8*7*6*5*4*3*2*3*2*5)

2009-09-06-13-21 document stop
proof, program, Conjecture, triangle

<a name="docB111">
2009-09-07-14-48
"Update 2009-09-07" add link to second page
Please visit second page
http://freeman2.com/jspico2e.htm
If you saved a copy, click jspico2e.htm
2009-09-07-14-51
```

<a name="Arndt01">
linear sets of equations compute
2009-05-28-23-33 LiuHH access
http://www.arndt-bruenner.de/mathe/scripts/gleichungssysteme.htm

Below is Arndt Brünner's work

<a name="Arndt02">

## Computer for the release of linear sets of equations

With interactive explanations to the Gauss solution procedure

This Javascript solves linear sets of equations up to 26 variables and homogeneous sets of equations, whose solutions depend all on exactly a free parameter. The solution method is represented when desired in detail on the basis the entered set of equations. The Script counts recently on breaks, i.e. the results are exactly, so far counters and denominators of inputs, intermediate steps or results the border of 10^15 do not exceed.
<a name="Arndt03">

In the left text window the equations are line by line entered, so many, as altogether variables are contained. It is not necessary that in each equation all variable emerge, also is all the same the sequence. Necessary it is only that possibly on the left and on the right from the equals sign a linear sum of variables (with signs and/or factors) stands and absolute member (number without variable).
Again: Also break numbers can be entered (e.g.: 2/3x).
Clasped terms, break terms or powers cannot be processed.
The variable names must be individual letters. The entire alphabet (without umlauts) is available, large and lower case is not differentiated. Multiplication symbol (* or ·) are not necessary.

<a name="Arndt04">   program0   sum_i_from_1_to_n
 Here equations (or coefficient matrix) enter Solutions (computed) Expenditure as breakas decimal number produce explanations always immediately
<a name="Arndt05">
[[
LiuHH add "try5" to generate ∑ i^(5) from i=1 to i=7 input data. get answer
∑ i^(5)= n*n*n*n*n*n/6+n*n*n*n*n/2+5*n*n*n*n/12-n*n/12
2009-09-01-14-32 LiuHH made above notes.
]]
<a name="Arndt06">

The solutions of homogeneous sets of equations are indicated in dependence of a free parameter. If a not-homogeneous set of equations has infinitely many solutions, the message “no clear solution is indicated found”.
A homogeneous set of equations does not possess (after simplification) absolute members.

Javascript uses the Gauss algorithm, which is called also Gauss elimination procedure, since successively in the equations systematically variables are eliminated. In addition only two methods are used: Multiply an equation by a number, an adding (a multiples) of an equation to another.

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Over the following switching surface an explanation of the solution method can be produced for the sets of equations or coefficient stencils entered above:

 Random example produce Coefficient matrix Equations integral count

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#### To practicing:

The solutions are smallprinted provided.

Minimum dimension: Maximum dimension:

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Notes:
The attitudes for minimum and maximum dimension are effective also for the example production on this side.
All examples are solved to the sample by the algorithm, although the solutions are first produced with the production of the tasks and for the program to that extent already admit are. If solutions are thus indicated, then that means that the task is also clearly solvable.

Possibly helpfully: ® basic operations of arithmetic of the Bruchrechnung

Sample applications for the algorithm are on these three sides:
Linear involutionCircle by 3 pointsBall by 4 points

Above is Arndt Brünner's work
proof, program, Conjecture, triangle

Javascript index
http://freeman2.com/jsindex2.htm
Space Curve Projector
http://freeman2.com/curve3d2.htm
foot of perpendicular
http://freeman2.com/eyefoot2.htm
Gram-Schmidt Process
http://freeman2.com/gramsch2.htm
complex polynomial root
http://freeman2.com/polyroot.htm
complex variable functions
http://freeman2.com/complex2.htm
Hilbert's Inequality and Schur Constant
http://freeman2.com/tute0009.htm

This page, Positive Integer power sum COefficient equation.
http://freeman2.com/jspico_e.htm
First upload 2009-09-02

Program name in Chinese: (2009-09-03-11-36)
Brief name: 指和函數
Full name: 正整數指數和積係數公式函數

Thank you for visiting Freeman's page.
Freeman 　2009-09-01-23-36

Please visit second page freeman2.com/jspico2e.htm