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var est:extended;i:integer;

function factorial(v:integer):integer;
var i:integer;
begin
  for i := v downto 0 do
  begin
    result := result+i;
  end;
end;

begin
  est := 0
  while not iskeydown(VK_RCONTROL) do
  begin
    est := ((pow(-1,i)*factorial(6*i)*(13591409+(545140134*i)))/(pow(factorial(i),3)*factorial(3*i)*pow(pow(640320,3),i+(1/2))));
    i := i+1;
    writeln(12*est);
  end;
end.

Error:

Compiled succesfully in 31 ms.
Math error at line 16

I dont see what the problem is, here is the formula: =]

first of all, factorial is multiplication of the numbers consecutively.

Next, you are doing, once again, integer math when you want floats.

Also, you have a complex number in your first pow().

How accurately are you trying to calculate pi?

And in how much time?

your factorial should be:

function Factorial(int number): Integer;
begin
  if ((number = 0) or (number = 1)) then
  begin
    result := 1;
    exit;
  end else
    result := (number * Factorial(number - 1));
end;

OR

function Factorial(number: Integer): Integer;
var temp, i: integer;
begin
  temp := 1;
  for i := number downto 1 do
    temp := temp * i;
end;

I would opt for the 2nd option, the first one using recursion could take awhile with big numbers.

^ By the way, there's really no point in doing temp * 1.. for i := number downto 2 do would be slightly faster and do the same thing. :)

Just curious, why downto? I know it doesn't matter, I'm just wondering. I feel like most people would do for i := 2 to number do temp := temp * i;.

Not sure why i used downto, thats just how i thought of it in my head.

EDIT: nvm, the reason i used downto is because when i do a factorial in my head or on a normal calculator i start from the high number and go down to 1

EDIT: the result:= result+1 works, i tested it before i used it.

EDIT3: new code:

var est,ii:extended;

function factorial(v:integer):integer;
var i:integer;
begin
  for i := v downto 1 do
  begin
    result := result*i;
  end;
end;

begin
  est := 0
  while not iskeydown(VK_RCONTROL) do
  begin
    est := ((pow(-1,ii)*factorial(int(6*ii))*(13591409+(545140134*ii)))/(pow(factorial(int(ii)),3)*factorial(int(3*ii))*pow(pow(640320,3),ii+(1/2))));
    ii := ii+1;
    writeln(12*est);
  end;
end.

error:

[Error] (17:150): Type mismatch at line 16
Compiling failed.

Another edit:

using a different formula that finds the nth digit of pi:

ar P:string;k:integer;
begin
  k:= 1;
  while not iskeydown(VK_RCONTROL) do
  begin
    p := p+FloatToStr((pow(16,-1*k) * ( 4/((8*k)+1) - 2/((8*k)+4) - 1/((8*k)+5) - 1/((8*k)+6) ))) ;
    inc(k);
  end;
end.       

error:

Math error at line 7.

EDIT:btw, the formula is for base16(i think that is hexadecimal) but i am using decimal, so how do i change it to hexadecimal

EDIT: i found out ther is a formula to calculate the nth digit of pi in any base but i cannot find a straight-forward answer, only essays that dont seem to give the formula.

You're using way too complicated formula, check out the Leibniz' formula:
http://en.wikipedia.org/wiki/Leibniz_formula_for_pi

lol, the problem with that is that Leibniz's formula converges exponentially slower with each iteration, whereas mine exponentially increases in accuracy. i tested Leibniz's formula in excel and i almost got 3 digits of pi with 1000 iterations. his formula works, but its horrible for convergence

EDIT: i did try to make one using his formula though, it didnt work either:

program new;
var p:double;plus:boolean;i:integer;
begin
  i := 0;
  plus := true
  repeat
    if plus then
    begin
      plus := false;
    end else
      plus := true;
    if plus then
      p := p+(1/((2*i)+1))
    else
      p := p-(1/((2*i)+1));
    inc(i);
    writeln(p);
    wait(100);
  until iskeydown(VK_RCONTROL)
end.            

the 2i+1 is what i use to generate the denominator and plus is whether to add or subtract that iteration.


EDIT: omg how many times have u guys told me not to use integer division here =\. i did it again and fixed it and now the script works, lol. i still need the other formulas to work though because they are faster.

From: g0tp0t
lol, the problem with that is that Leibniz's formula converges exponentially slower with each iteration, whereas mine exponentially increases in accuracy. i tested Leibniz's formula in excel and i almost got 3 digits of pi with 1000 iterations. his formula works, but its horrible for convergence

EDIT: i did try to make one using his formula though, it didnt work either:

program new;
var p:double;plus:boolean;i:integer;
begin
  i := 0;
  plus := true
  repeat
    if plus then
    begin
      plus := false;
    end else
      plus := true;
    if plus then
      p := p+(1/((2*i)+1))
    else
      p := p-(1/((2*i)+1));
    inc(i);
    writeln(p);
    wait(100);
  until iskeydown(VK_RCONTROL)
end.            

the 2i+1 is what i use to generate the denominator and plus is whether to add or subtract that iteration.


EDIT: omg how many times have u guys told me not to use integer division here =\. i did it again and fixed it and now the script works, lol. i still need the other formulas to work though because they are faster.

The speed is not a problem in this case, as your computer can do several million iterations in matter of seconds, and your computer only supports float numbers up to like ~20 digits

well speed would become an issue depending on how accurate you want it to be. 10000 terms is accurate to approx 4 decimal places and i think every 0 is good for another decimal place. so 100000000 terms would be good for 8 decimal places…

lol, sped is exactly the issue, i want to get thousands of digits, not for any real reason. Just for fun.

also each iteration is not a calculation, ther are several calculations per iteration and the computer isnt devoted to the script so it is much slowr than that. i was getting a very slow output, ill add a timeing procedure and post my proggy

i made the script tell me how many calculations it made per second and i got an average of 60. 60 calculations per second is nothing, im assuming its because pascal is single-threaded.


i definetly need a better algorithm

http://pastebin.com/8Cg5ar3x