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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,905

Growing...

{1, 3, 6, 10, 15, 21, 4, 5, 11, 14, 2, 7, 9, 16, 20, 29, 35, 46, 18, 31, 33, 48, 52, 12, 13, 23, 26, 38, 43, 57, 24, 25, 39, 42, 22, 27, 37, 44, 56, 8, 17, 19, 30, 34, 47, 53, 28, 36, 45, 55, 66, 78, 91, 105, 64, 80, 41, 40, 60, 61, 83, 86, 58, 63, 81, 88, 108, 117, 79, 65, 104, 92, 77, 67, 54, 90, 106, 119, 50, 71, 73, 96, 100, 69, 75, 94, 102, 123, 133, 156, 168}

Length-91

MaxNumber-168

IPBLE: Increasing Performance By Lowering Expectations.

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**seerj****Member**- Registered: 2005-12-21
- Posts: 42

Ehy seems to be cool.

If u want can u send me the algorithm? [if u want for sure!]

on leopardus@inwind.it

However thank u...

We need to understand which is this minimum n....

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**krassi_holmz****Real Member**- Registered: 2005-12-02
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I generalized my algoritm and inproved my program. Now it the first member don't have to be 1.

IPBLE: Increasing Performance By Lowering Expectations.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,905

here's my algoritm:

Let sqsei(x) give the smallest perfect square, greater than x.

a[1]=k-arbitary

How to find a[f]?

1.We calculate sqsei[a[f-1]]-a[f-1]. Let this be k.

2.If k is different with all numbers a[1],a[2],...,a[f-1] then a[f]=k.

3.If not, the new value of k must be sqsei[k+a[f-1]]-a[f-1] and we repeat step 3.

If a[f]+k=sqsei(a[f]) then the square chain is also circular.

*Last edited by krassi_holmz (2005-12-29 03:21:20)*

IPBLE: Increasing Performance By Lowering Expectations.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,905

I wrote my program on Mathematica language and it has some extra properties:

*Last edited by krassi_holmz (2005-12-29 03:53:28)*

IPBLE: Increasing Performance By Lowering Expectations.

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**seerj****Member**- Registered: 2005-12-21
- Posts: 42

krassi_holmz wrote:

I wrote my program in Mathematica language and it has some extra properties:

Oh Mathematica! I thought something like C/C++. Coz I dunno how to check if the sum of 2 consecutive numbers is perfect square??!

When the sum of 2 numbers is a perfect square?

if (a+b) = ?!?!

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,905

I'll tell you. I asked myself same thing when I started doing the program. But my mistake was that i started writting on Visual Basic. And I had to define some functions:

```
------------------------------------------------------------------------------------------------------------------
Function IsPerfectSquare(x) 'it gives 1 if x is square and 0 if not.
If Sqr(x) = Math.Round(Sqr(x)) Then
IsPerfectSquare = 1
Else
IsPerfectSquare = 0
End If
End Function
Function Ceiling(x) 'gives the smallest integer, greater or equal to x
If x > Math.Round(x) Then
Ceiling = Math.Round(x) + 1
ElseIf x <= Math.Round(x) Then
Ceiling = Math.Round(x)
End Function
Function PerfectSquareCeiling(x) 'gives the smallest perfect square that is greater than x
PerfectSquareCeiling = Ceiling(Sqr(x + IsPerfectSquare(x))) ^ 2
End Function
------------------------------------------------------------------------------------------------------------------
```

The sum of two numbers is perfect square if

IsPerfectSquare(a+b)==1.

IPBLE: Increasing Performance By Lowering Expectations.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,905

The third function is correct.

IPBLE: Increasing Performance By Lowering Expectations.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,905

I know C very little.

Sorry, but I'm young and the first language i've learned was VBS.

If someone can translate the functions to C I'll be grateful.

IPBLE: Increasing Performance By Lowering Expectations.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,905

Here's the Mathematica code:

```
(* PROGRAM CALCULATING THE SQUARE SUM CHAINS AND TESTING THEIR CIRCULARITY
Based on Georgiev's square sum chain creating algoritm
By Krasimir Geogiev
December, 2005 *)
(*____________________________________________________________*)
(* data *)
n := 17
k := 2
(* /data *)
(*____________________________________________________________*)
(* function defining *)
sq[x_] := If[x^(1/2) == Floor[x^(1/2)], 1, 0]
(* this function gives 1 when x is perfect square and 0 when it isn't. *)
sqcei[x_] := Ceiling[(x+sq[x])^(1/2)]^2
(* this function gives the smallest perfect square which is greater than x. *)
(* /function defining *)
(*____________________________________________________________*)
(* calculations *)
a[1] := k
Do[b[i] = 0, {i, 1, n^3}]
b[k] := 1
Do[
p = sqcei[a[i - 1]] - a[i - 1];
j = 0;
While[j == 0,
If[b[p] == 0,
b[p] = 1;
a[i] = p;
j = 1,
p = sqcei[p + a[i - 1]] - a[i - 1]
]], {i, 2, n}]
(* /calculations *)
(*____________________________________________________________*)
(* output *)
t=Table[a[i],{i,1,n}]
max = Max[t]
iscir=If[sq[a[n] + k] == 1, 1, 0]
(* /output*)
(*____________________________________________________________*)
```

here you choose n and k.

t is the chain

max is max number in the chain

iscir is 1 if the chain sequence is circular and 0 if not.

I've simplified the output. In my program actually it is:

```
(* output *)
Print["The chain:"]
t = Table[a[i], {i, 1, n}]
Print["Length:"]
n
Print["Maximal number:"]
Max[t]
Print["Graphical plot:"]
ListPlot[t, PlotJoined -> True]
isO[x_] := If[sq[a[x] + k] == 1, 1, 0]
Print["if j <= n chain is circular:"]
tt = Table[isO[i], {i, 1, n}]
Print["Graphical plot of the circular test:"]
Plot[isO[Floor[x]], {x, 1, n + 1}]
(* /output*)
```

IPBLE: Increasing Performance By Lowering Expectations.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,905

I just have found something very interesting:

The biggest number in the square sum chain sq(2,17)={2, 7, 9, 16, 20, 5, 4, 12, 13, 3, 1, 8, 17, 19, 6, 10, 15} with length 17 is 20!

Unfortunately sq(2,17) is not circular.

*Last edited by krassi_holmz (2005-12-29 07:40:52)*

IPBLE: Increasing Performance By Lowering Expectations.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
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Oh. I forgot!

sq(x,y) - x means the first number in the chain, y-the length of the chain.

*Last edited by krassi_holmz (2005-12-29 07:41:53)*

IPBLE: Increasing Performance By Lowering Expectations.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
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Conjection:

The greatest number in the chain sq(1, n) is >n and ≈2n.

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**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,588

1 3 6 10 15 21 4 12 24 25

11 5 31 18 7 9 40 41 8

All less than fifty. Length=19

1 8 17 19 6 10 15 21 4 5

11 14 2 23 13 12 24

All less than 50, Length = 17

1 3 6 10 15 21 4 12 13 23

2 34 30 19 45 36 28 8

All less than 50, Length = 18

1 3 6 10 15 21 4 5 20 16

9 27 22 14 2 23 13 36 28 8

41 40 24

All less than 50, Length = 23

1 3 6 10 15 21 4 12 24 25

11 5 44 20 16 33 31 18 46 35

All less than 50, Length = 20

*Last edited by John E. Franklin (2005-12-29 09:48:08)*

**igloo** **myrtilles** **fourmis**

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**John E. Franklin****Member**- Registered: 2005-08-29
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Length=25, #'s < 50

1 8 17 19 6 3 22 14 11 5

31 18 7 2 23 26 10 39 25 24

40 9 16 33 48

Length=21, #'s < 50

1 8 17 19 6 10 15 21 4 5

11 14 2 7 9 16 20 29 35 46

3

*Last edited by John E. Franklin (2005-12-29 09:54:32)*

**igloo** **myrtilles** **fourmis**

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,905

sq(3,24)={3, 1, 8, 17, 19, 6, 10, 15, 21, 4, 5, 11, 14, 2, 7, 9, 16, 20, 29, 35, 46, 18, 31, 33}

Length-24

#<50

*Last edited by krassi_holmz (2005-12-29 09:59:15)*

IPBLE: Increasing Performance By Lowering Expectations.

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**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,588

Does "correspond" mean they came out the same as yours?

Does sq(1,23) mean Length 23, starts on 1?

**igloo** **myrtilles** **fourmis**

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**John E. Franklin****Member**- Registered: 2005-08-29
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Another Length 25, all #'s < 50

1 3 6 10 15 21 4 32 17 19

30 34 2 7 42 22 14 35 29 20

44 5 11 25 24

A twenty long, below 50

1 3 6 10 15 21 4 32 17 8

28 36 13 23 41 40 9 16 33 48

*Last edited by John E. Franklin (2005-12-29 10:03:51)*

**igloo** **myrtilles** **fourmis**

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,905

sq(1,23) means with length 23, tarting on 1 that is generated by my algoritm. My algoritm gives it too.

IPBLE: Increasing Performance By Lowering Expectations.

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**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,588

Awesome! My algorithm uses random #'s, kind of cheating.

Everytime I run, I get a different answer, usually.

a twelve long, max #21

1 8 17 19 6 10 15 21 4 12

13 3

*Last edited by John E. Franklin (2005-12-29 10:07:19)*

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**krassi_holmz****Real Member**- Registered: 2005-12-02
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Then you have got much luck!

IPBLE: Increasing Performance By Lowering Expectations.

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**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,588

Well, I pick a random square and subtract last number from it

and check if number is used yet.

Here's 22 long (my favorite #), Max#41

1 3 6 10 15 21 4 12 13 23

2 7 9 27 22 14 11 25 24 40

41 8

a 19 long, max 46

1 3 6 10 15 21 4 12 13 23

2 7 29 20 5 31 18 46 35

*Last edited by John E. Franklin (2005-12-29 10:13:53)*

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**John E. Franklin****Member**- Registered: 2005-08-29
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My longest under 50.

Length=28, Max=41

1 3 6 10 15 21 4 12 13 23

2 14 22 27 9 16 20 29 7 18

31 5 11 25 24 40 41 8

**igloo** **myrtilles** **fourmis**

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**krassi_holmz****Real Member**- Registered: 2005-12-02
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A 28 long, MaxNumber=52(too close ):

sq(13,28)={13, 3, 1, 8, 17, 19, 6, 10, 15, 21, 4, 5, 11, 14, 2, 7, 9, 16, 20, 29, 35, 46, 18, 31, 33, 48, 52, 12}

IPBLE: Increasing Performance By Lowering Expectations.

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**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,588

a 9 long, max=24

1 3 6 10 15 21 4 12 24

an 11 long, max=34

1 8 17 19 6 10 26 23 2 34

15

a 14 long, max=40

1 3 6 10 15 21 4 5 31 18

7 9 40 24

a 15 long, max=36

1 3 6 10 15 21 4 32 17 8

28 36 13 12 24

*Last edited by John E. Franklin (2005-12-29 10:33:36)*

**igloo** **myrtilles** **fourmis**

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