Question

Conway series, can someone explain the formula?

Answer 1

a(1)=a(2)=1, a(n) = a(a(n-1)) + a(n-a(n-1)) for n>2.

Answer 2

I have two version of the program:

1)conway:=proc(n)
description`Conway with remember table`:
option remember:
if n=1 then
return 1;
elif n=2 then
return 1;
else
return conway(a(a[n-1]))+conway(a(n-a[n-1]));
end if;
end proc;

Answer 3

2)conway2:=proc(n)
local A,k;
A[1],A[2]:=1,1;
for k from 3 to n do
A[n]:=A[A[k-1]]+A[k-A[k-1]];
end do;
return A;
end proc;

Answer 4

both are not returing the sequence

Answer 5

what is the deal with the conway series? can somone explain the formula?

Answer 6

man, i just read this paper on recursive programing and i think this might be more subtle than i thought

Answer 7

but give me your thoughts on it

Answer 8

i might just be missing somehting obvious

Answer 9

I think it's this
\[\huge a _{n} = a _{a _{n-1}} + a _{n-a _{n-1}}\]

Answer 10

yeah, that what i am thinking, thats why i put in maple a[a[n-1]], the brackets are suppose to denote subscripts

Answer 11

you supposed to be implementing recursively?

Answer 12

i guess so, its not explicetly mentioned in the problem(the problem only says go to internet and search conway, then right a program on it), but upon googling conway series it says that its a recursive formula

Answer 13

sec

Answer 14

check the sixth link
https://www.google.com/webhp?hl=en&tab=Xw#hl=en&safe=off&sclient=psy-ab&q=%22conway+series%22+math&oq=%22conway+series%22+math&gs_l=serp.3...11031.13975.0.14204.2.2.0.0.0.0.98.182.2.2.0.les%3B..0.0...1c.1.xljvqXLccU4&pbx=1&bav=on.2,or.r_gc.r_pw.r_cp.r_qf.&fp=a13ecffab696db35&biw=1024&bih=655

RECURSION

Answer 15

so what site are you looking at?

Answer 16

http://www.skrause.org/math/conway.shtml

Answer 17

thats basically it on info about conway, there are someothers like wolfram that has info on conway series, but its all the same

Answer 18

we should click your own link a bunch of times and then you should ask the question: "Does anyone know any know of any good links with info about the Conway Series?? plz hapl"

Answer 19

eventually, due to recursion, we'll get a good link with all the info we need.

Answer 20

you sent me back to my own problem, your a genius :)

Answer 21

i might have found something broski, let me try it

Answer 22

stand back everybody

summoning the math gods...
@UnkleRhaukus
@experimentX
@estudier
@ganeshie8
@ByteMe
@algebraic!

Answer 23

I summoned myself only for the sake of recursion.

Answer 24

like i am using this program:
conway:=proc(n)
description`Conway with remember table`:
option remember:
if n=1 then
return 1;
elif n=2 then
return 1;
else
return A(A(n-1))+A(n-A(n-1));
end if;
end proc;


but when i input in maple i get this:


conway(1);
1
conway(2);
1
conway(3);
A(A(2)) + A(3 - A(2))

Answer 25

hmm yeah, maple probably has some special notation for terms in a series with sub scripts

Answer 26

that's not "A[ ] "

Answer 27

now i changed it to:A[A(n-1)]+A[n-A(n-1)];

and got:
conway(3);
A[A(2)] + A[3 - A(2)]

Answer 28

what are you trying to get?

Answer 29

yeah I don't think " A[ ]" is going to do it... we need something like:

Answer 30

i am trying output the conway series

Answer 31

he's trying to find a recursion relation for the conway series so he can implement in Maple

Answer 32

Error, unable to match delimiters

could you edit the program?

Answer 33

I find it easy to program in Matlab.

Answer 34

Need something like \[\huge Conway _{n} = Conway _{Conway _{n-1}} +Conway _{n-Conway _{n-1}}\]

Answer 35

wow that totally worked

Answer 36

\[ Conway _{n} = Conway _{Conway _{n-1}} +Conway _{n-Conway _{n-1}}\]

Answer 37

\[\huge Conway _{n} = Conway _{Conway _{n-1}} \]

\[\huge +Conway _{n-Conway _{n-1}}\]

Answer 38

check this out, its different versions of what i have tried:

Answer 39

see what I'm saying?

Answer 40

if you can find the syntax in Maple for 'series terms with subcripts' it would make it easier...

Answer 41

did you see the first part of that pdf i sent you

Answer 42

i have made a matlab code that seems working

Answer 43

please post it, thank you

Answer 44

the thing with mine, is that it gives me the subscripts but it doesnt want to evauate or give me the value of that term, if i could figure out how to do that, it would work

Answer 45

% conway %
L=20; % lenght of series
a=1:L; % set vector a
a(1)=1; % initial value
a(2)=1; % second value
for n=3:L;
a(n) = a(a(n-1)) + a(n-a(n-1)); % other values
end
a % dipsplay

Answer 46

>> conway

a =

Columns 1 through 20

1 1 2 2 3 4 4 4 5 6 7 7 8 8 8 8 9 10 11 12

Answer 47

yeah so, find out the Maple syntax for terms in a series with sub scripts
it's evidently not " A [ ]"

Answer 48

yup thats the series

Answer 49

To select members of a sequence, use index notation: s[i]refers to the ith element of the sequence s.

Answer 50

or figure out how to write a Maple proc that opens matlab and executes that program and then returns the result, all the while pretending that it did the work... *whistles innocently* might help

Answer 51

there ya go

Answer 52

but isnt that notation what i have used in my program?

Answer 53

I took it to mean you have to use 's' specifically... maybe A is some notation used for a matrix? I dunno...

Answer 54

ah, part of the problem is that in your program I don't think you tell it that A[1] =1 and A[2] =1 ...?

Answer 55

you just tell it what to return...

Answer 56

@UnkleRhaukus in matlab is () suppose to denote a subscript?

Answer 57

i do define a[1] and a[2]

Answer 58

yeah

Answer 59

ok, in the second program..?
but you have the brackets wrong in that one...

Answer 60

s[n] := s[s(n-1)]+s[n-s(n-1)] <---you mean here?

Answer 61

can't copypasta from pdf but... in " conway2:= "
you have the wrong brackets for some of the things that *should* be subscripts....

Answer 62

I changed that and look what i got:


seq(G[i],i=3..30);


s[3], s[4], s[5], s[6], s[7], s[8], s[9], s[10], s[11], s[12],

s[13], s[14], s[15], s[16], s[17], s[18], s[19], s[20], s[21],

s[22], s[23], s[24], s[25], s[26], s[27], s[28], s[29],

s[s[29]] + s[30 - s[29]]

Answer 63

" A[A(n-1)] "
try it with A[A[n-1]]

Answer 64

whats the program is not doing is this:

i defined the terms s[1] and s[2], they should both be 1, so when i plug in n=3 meaning i want the s[3] term, it should automatically say okay s[3-1] is s[2] and i know s[2] is =1 , so then it should use 1 and the subscript for the outer s

Answer 65

in matlab is () suppose to denote a subscript? Yes

Answer 66

I'd try the exact code you have in conway2 , but with the right brackets... if that doesn't work... then, one problem down.

Answer 67

I switched it to this:s[n]:= s[s[n-1]]+s[n-s[n-1]];

Answer 68

ok

Answer 69

with s[1] = s[2] =1 ?

Answer 70

and it gave me this:seq(G[i],i=3..30); s[3], s[4], s[5], s[6], s[7], s[8], s[9], s[10], s[11], s[12], s[13], s[14], s[15], s[16], s[17], s[18], s[19], s[20], s[21], s[22], s[23], s[24], s[25], s[26], s[27], s[28], s[29], s[s[29]] + s[30 - s[29]]

Answer 71

yup i said:s[1], s[2] := 1, 1

Answer 72

try i=1..30 , looks like it's not including s[1], s[2]

Answer 73

all it does is include 1,1 but the rest of them stayed the same

Answer 74

seq(G[i]:=s[s[n-1]]+s[n-s[n-1]],i=1..30) ?

Answer 75

dead end

Answer 76

don't know... syntax thing.

Answer 77

conway2 looks closest to me... I'd play with it.

Answer 78

thanks for wokring with me, i'll try and keep working on it to see what happens

Answer 79

So check out this link, scroll down to where it says Maple:

http://oeis.org/A004001

Answer 80

you wanna know somethin crazy, i just called Neil Sloane, the guy that wrote that page. Well he didnt answer but i left my number.

Answer 81

anyway, can you believe that all i did was copy and paste that code into maple, and its acutally does return each indivdual conway term

Answer 82

rfl

Answer 83

but what i am trying to figure out is does A00400l have a deeper meaning in maple or is that just a name for soomething?

Answer 84

"option remember" maybe you have to turn recursion on in maple. I like how the function name is a link to the page. just keep recurring.

Answer 85

right, but in maple, does A004001 actually have a meaning or does it just name the program

Answer 86

name it I think, it's a joke...I'm guessing.

Answer 87

what does rfl stand for?

Answer 88

like, "hey, if I click this link, it'll show me what the function is!"

Answer 89

so, lets say that i edit my program and instead of A004001 i write conway, could it still work?

Answer 90

Yeah, I think so.

Answer 91

or maybe he just names his functions on each page after each page address. if they're* all recursive functions, that's kind of funny. when he calls back, ask him for me.

Answer 92

yea i will, you know whats funny? is that his wife or daughter answers, and when i ask to speak to mister sloan, she is like" sure, can he call you back", really politely, she dosent even who is. I think i am not the first student to call him.

Answer 93

she doesent even ask who it is *

Answer 94

lol. sounds like a family with a sense of humor.

Answer 95

now let me ask you, how does this program work?

Answer 96

just like we've been doing all along, I guess the brackets don't matter...
it's the same thing I wrote above. also I don't know what 'option remember' means in Maple, maybe that needs to be enabled so it can call a function inside itself (it remembers where it is in 'n')

Answer 97

oh man, the program rembers the values of the conway(1) and conway(2)