Problem with determining squareroots in the progressions..

Annunaki · 2003-07-18, 18:02

Well may be i even can`t formulate my problem in short statement, because of my high-school level knowledge in math and my bad knowledge in mathematical terms in english, but i will try to describe my problem..
But first little bit about me.. Time by time i am returning to my ideas in factoring and trying to write simple algorithms in factoring and finding primes.. Because of lack of knowledge I can`t even estiminate how eficient they are but i know that they aren`t too eficient..
On one of my ideas i met the problem of determining when a sum of certain arithemtic progresiion will be equal to some natural n^2..
thus for example I know the first element in arithmetic progresion is n..
every next element is increased by 2.. the question is can modern math easily tell when the summ of hole chain will be some natural number n^2..
for example-
let the first element be 31
so the second will be 31+2=33
31+33=64 and that is 8^2..
in this example we are lucky and we need to do only one addtion and we have some natural n^2..
but some other examples can illustrate that we may need a lot of more...
i have some ideas how to do this, but every time it needs a lot of cycles(time) with large numbers...

Orgasmic Troll · 2003-07-18, 20:32

I don't know if you want a general case, but if the constant is 2 (such as the example you gave), then the sum of k terms is k^2 + k(n-1), if that helps any

Orgasmic Troll · 2003-07-18, 21:44

I've been playing around a bit with this .. for n where n==3 mod 4 (n leaves a remainder of 3 when divided by 4), when the sequence reaches (n-1)^2/4 then the sum of those terms is square. This isn't necessarily the soonest it sums to a square, but it always does

for n==1 mod 4, the sequence is a square when it reaches (n-1)^2/8 + 1, again, this isn't always the smallest answer.

So that takes care of the odd numbers .. haven't really looked into even numbers much

I sort of worked this out as a visual proof, if anybody is really interested, I can try to put it together up here, and if not, I sure won't mind not having to.

Orgasmic Troll · 2003-07-18, 21:53

for n even, the number of steps is (n-2)^2/4 - 1

ET_ · 2003-07-18, 23:19

It seems the opposite of the 3x-1 problem.

3x-1 chains seem to always decade into a 2^x loop.

Luigi

cheesehead · 2003-07-19, 00:25

Quote:

Originally Posted by Annunaki

my high-school level knowledge in math

How much algebra have you studied in school? (I ask because it could help us respond.)

Quote:

the question is can modern math easily tell when the summ of hole chain will be some natural number n^2..

Yes.

In order to figure out the efficiency of your algorithms, you will need to use the properties of, and methods of analyzing, mathematical series and sequences. In some schools this is included in first-year algebra; in other schools, later.

Maybe an online math instruction site could help you. Here is one: http://www.ilovemaths.com/3.htm

Annunaki · 2003-07-19, 09:01

About my level of math.. As i learned in trade technikal school the level of math was definitely not good.. However I have allways learned math for my self buy finding my own problems and trying to solve them..
In other words do not expect me to understand expressions fo high level math.. I just don`t know the syntax of that "high" math.. But i will try to understand whar you mean if you will explain me it.. For example I do`nt know if I understand correctly that mod expressions.. for example:
n==3 mod 4 does that meens that if you divide n by 4 you will allways get 3 in reminder or in other word your result will be some notural n + 0.75?..
.. and form your replies "when the sequence reaches (n-1)^2/4" i don`t think i understad this..
aslo i don`t understand what you mean with 3x-1 problem..
sorry i know my knowlegde in math is awfull.. i just like to write my algorithms and solve diferent problems.. I just haven`t read many abaout the math and problems in there.. may be it also good, because if your mind is filled with other ideas and thoughts, may be you can`t find yours anymore{just my opinion you may try to disagree with that}
Also I know how to get the sum of such progressions i described.. I just don`t know how te easely determinete when such sum will be some natural n^2..
More than that.. does any of you know have mathemathic still some problems with chains and calculating sum of chains..
for example i have a chain first eleent is 6.. diferential is 9 but it increases by 9 every time..
for example the sum of such chain if it consist of:
1 element = 6
2 elements = 6+6+9= 21
3 elements = 21+6+9+9=45
4 elements = 45+6+9+9+9=78 etc..
so question is can math easely calculate the sum of such chain wih n elements, because i think i can do it and i can even give a formula..
another example could be chain wich elements is all natural n^2 starting with 1.. and the sum of this chain would be the sum of all natural n^2 form 1 to some natural K.. of course summing all natural n together isn`t the most efficient way to calculate the sum of such chain.. if you are interested i can even give my formula.. but i doubt if my so called formulas is needed cause i assume math has gone far far beyond my imagination these days, and may be this what I describes is not a problem allready for n years..
However this is only additional question.. main question remains about the squareroots..
If you can tell me the way with which i can precisely tell when the
sum of chain where first element is natural n the second is n+2 the third n+4 etc. will be some natural n^2 then i think factoring of even large numbers can be done in few seconds.. But this is only if you can tell me the method with which you can precisely tell how much elements i need in that kind of chain (of course the element number will differ in diferent chains but..) to sum of this chain be some natural n^2..

Annunaki · 2003-07-19, 09:05

yes about my education may be I have mistaken.. I thought the high-school level corresponds to one level higher level than primary education.. Is it so? Well this year i hope i will already student of University of Latvia..

Annunaki · 2003-07-19, 17:02

I am sorry but I forgot to mention one thing about that progression..
In overal it is normal arithmethic progression with first element(natural n) and diferential 2, except that it has anoher integer(element) before other element that make arithmethic progression.. this (we could call it 0th element) is independent from other elements of progression and only thing that i can say about it is that it is allways less than first element of chain by at least 3..
here is an example of such chain..
14; 51; 53; 55; 57; 59; 61; etc..
here we can see that summ of first 6 elements is(or 5 if we count 14 as )th element) 289 which is 17^2 which is exactly what we needed to find in this case...

dsouza123 · 2003-07-20, 15:24

High school is usually grades 9,10,11,12
Some students instead go to a Vocational ( Technical ) High School also usually grades 9,10,11,12

University/College (Bachelors/Associate degree)/Technical school follows.

======= Back to the math. ================

The series consisting of the sum( addition ) of consecutive odd numbers
( starting with 1 ) results in (perfect) square numbers.

Since it is the sum consecutive odd numbers each number in the series is an increase of 2 from the previous number in the series.

Example
1+3= 4 1+3+5=9 1+3+5+7=16 etc.

If k = count of odd numbers in a section of the series
( k = count of numbers summed )
Then k^2 = sum(1..count)

Example
1+3+5+7=16 so k=4 and k^2=16

For mod it is the remainder after integer division

Using div to represent integer division

Example n = 23 n div 4 = 5 with a remaider 3 so n mod 4 = 3
for integer math numerator DIV denominator = result div

n = result div*denominator + remainder
23 = 5*4 + 3

if it wasn't integer math then
n = 23 n / 4 = 5.75

Annunaki · 2003-07-20, 17:49

I know that perfect squares or every natural n^2 is the sum of arithmethic progression of n elements where first element is 1 but every ext is increased by 2..
But my question is wether you can determinate when the sum of arithmethci progression is some natural n^2 or better how many elements you will need in arithmethic progression to it`s sum to be some natural n^2.. I know it is easy to tell this if first element is 1 and every next +2(because then in all cases sum offirst n elements of chain will be n^2), but in my case the fist element is never 1.. Although afetr the fisrt element every next is increased by 2(like in the case with that "perfect square string") it is not so easy to tell how many elements you will need in this chain to sum of this chain be some perfect sqare.. Actually i haven`t jet found a better way than just to add +1 element and check every time if now the sum of chain is some natural n^2.. but that is very slow and unefective.. i need better ways.. I have som ideas, but i just thought may be somene else has faced this problem and found some better solutions than me..

2003-07-18, 18:02	#1
Annunaki Jul 2003 31 Posts	Problem with determining squareroots in the progressions.. Well may be i even can`t formulate my problem in short statement, because of my high-school level knowledge in math and my bad knowledge in mathematical terms in english, but i will try to describe my problem.. But first little bit about me.. Time by time i am returning to my ideas in factoring and trying to write simple algorithms in factoring and finding primes.. Because of lack of knowledge I can`t even estiminate how eficient they are but i know that they aren`t too eficient.. On one of my ideas i met the problem of determining when a sum of certain arithemtic progresiion will be equal to some natural n^2.. thus for example I know the first element in arithmetic progresion is n.. every next element is increased by 2.. the question is can modern math easily tell when the summ of hole chain will be some natural number n^2.. for example- let the first element be 31 so the second will be 31+2=33 31+33=64 and that is 8^2.. in this example we are lucky and we need to do only one addtion and we have some natural n^2.. but some other examples can illustrate that we may need a lot of more... i have some ideas how to do this, but every time it needs a lot of cycles(time) with large numbers...

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2003-07-18, 20:32	#2
Orgasmic Troll Cranksta Rap Ayatollah Jul 2003 1201₈ Posts	I don't know if you want a general case, but if the constant is 2 (such as the example you gave), then the sum of k terms is k^2 + k(n-1), if that helps any

2003-07-18, 21:44	#3
Orgasmic Troll Cranksta Rap Ayatollah Jul 2003 641 Posts	I've been playing around a bit with this .. for n where n==3 mod 4 (n leaves a remainder of 3 when divided by 4), when the sequence reaches (n-1)^2/4 then the sum of those terms is square. This isn't necessarily the soonest it sums to a square, but it always does for n==1 mod 4, the sequence is a square when it reaches (n-1)^2/8 + 1, again, this isn't always the smallest answer. So that takes care of the odd numbers .. haven't really looked into even numbers much I sort of worked this out as a visual proof, if anybody is really interested, I can try to put it together up here, and if not, I sure won't mind not having to.

2003-07-18, 21:53	#4
Orgasmic Troll Cranksta Rap Ayatollah Jul 2003 641 Posts	for n even, the number of steps is (n-2)^2/4 - 1

2003-07-18, 23:19	#5
ET_ Banned "Luigi" Aug 2002 Team Italia 3²×5×107 Posts	It seems the opposite of the 3x-1 problem. 3x-1 chains seem to always decade into a 2^x loop. Luigi

2003-07-19, 09:01	#7
Annunaki Jul 2003 31 Posts	About my level of math.. As i learned in trade technikal school the level of math was definitely not good.. However I have allways learned math for my self buy finding my own problems and trying to solve them.. In other words do not expect me to understand expressions fo high level math.. I just don`t know the syntax of that "high" math.. But i will try to understand whar you mean if you will explain me it.. For example I do`nt know if I understand correctly that mod expressions.. for example: n==3 mod 4 does that meens that if you divide n by 4 you will allways get 3 in reminder or in other word your result will be some notural n + 0.75?.. .. and form your replies "when the sequence reaches (n-1)^2/4" i don`t think i understad this.. aslo i don`t understand what you mean with 3x-1 problem.. sorry i know my knowlegde in math is awfull.. i just like to write my algorithms and solve diferent problems.. I just haven`t read many abaout the math and problems in there.. may be it also good, because if your mind is filled with other ideas and thoughts, may be you can`t find yours anymore{just my opinion you may try to disagree with that} Also I know how to get the sum of such progressions i described.. I just don`t know how te easely determinete when such sum will be some natural n^2.. More than that.. does any of you know have mathemathic still some problems with chains and calculating sum of chains.. for example i have a chain first eleent is 6.. diferential is 9 but it increases by 9 every time.. for example the sum of such chain if it consist of: 1 element = 6 2 elements = 6+6+9= 21 3 elements = 21+6+9+9=45 4 elements = 45+6+9+9+9=78 etc.. so question is can math easely calculate the sum of such chain wih n elements, because i think i can do it and i can even give a formula.. another example could be chain wich elements is all natural n^2 starting with 1.. and the sum of this chain would be the sum of all natural n^2 form 1 to some natural K.. of course summing all natural n together isn`t the most efficient way to calculate the sum of such chain.. if you are interested i can even give my formula.. but i doubt if my so called formulas is needed cause i assume math has gone far far beyond my imagination these days, and may be this what I describes is not a problem allready for n years.. However this is only additional question.. main question remains about the squareroots.. If you can tell me the way with which i can precisely tell when the sum of chain where first element is natural n the second is n+2 the third n+4 etc. will be some natural n^2 then i think factoring of even large numbers can be done in few seconds.. But this is only if you can tell me the method with which you can precisely tell how much elements i need in that kind of chain (of course the element number will differ in diferent chains but..) to sum of this chain be some natural n^2..

2003-07-19, 09:05	#8
Annunaki Jul 2003 31 Posts	yes about my education may be I have mistaken.. I thought the high-school level corresponds to one level higher level than primary education.. Is it so? Well this year i hope i will already student of University of Latvia..

2003-07-19, 17:02	#9
Annunaki Jul 2003 31 Posts	I am sorry but I forgot to mention one thing about that progression.. In overal it is normal arithmethic progression with first element(natural n) and diferential 2, except that it has anoher integer(element) before other element that make arithmethic progression.. this (we could call it 0th element) is independent from other elements of progression and only thing that i can say about it is that it is allways less than first element of chain by at least 3.. here is an example of such chain.. 14; 51; 53; 55; 57; 59; 61; etc.. here we can see that summ of first 6 elements is(or 5 if we count 14 as )th element) 289 which is 17^2 which is exactly what we needed to find in this case...

2003-07-20, 15:24	#10
dsouza123 Sep 2002 2·331 Posts	High school is usually grades 9,10,11,12 Some students instead go to a Vocational ( Technical ) High School also usually grades 9,10,11,12 University/College (Bachelors/Associate degree)/Technical school follows. ======= Back to the math. ================ The series consisting of the sum( addition ) of consecutive odd numbers ( starting with 1 ) results in (perfect) square numbers. Since it is the sum consecutive odd numbers each number in the series is an increase of 2 from the previous number in the series. Example 1+3= 4 1+3+5=9 1+3+5+7=16 etc. If k = count of odd numbers in a section of the series ( k = count of numbers summed ) Then k^2 = sum(1..count) Example 1+3+5+7=16 so k=4 and k^2=16 For mod it is the remainder after integer division Using div to represent integer division Example n = 23 n div 4 = 5 with a remaider 3 so n mod 4 = 3 for integer math numerator DIV denominator = result div n = result divdenominator + remainder 23 = 54 + 3 if it wasn't integer math then n = 23 n / 4 = 5.75

2003-07-20, 17:49	#11
Annunaki Jul 2003 37₈ Posts	I know that perfect squares or every natural n^2 is the sum of arithmethic progression of n elements where first element is 1 but every ext is increased by 2.. But my question is wether you can determinate when the sum of arithmethci progression is some natural n^2 or better how many elements you will need in arithmethic progression to it`s sum to be some natural n^2.. I know it is easy to tell this if first element is 1 and every next +2(because then in all cases sum offirst n elements of chain will be n^2), but in my case the fist element is never 1.. Although afetr the fisrt element every next is increased by 2(like in the case with that "perfect square string") it is not so easy to tell how many elements you will need in this chain to sum of this chain be some perfect sqare.. Actually i haven`t jet found a better way than just to add +1 element and check every time if now the sum of chain is some natural n^2.. but that is very slow and unefective.. i need better ways.. I have som ideas, but i just thought may be somene else has faced this problem and found some better solutions than me..