[VA] SRC #017 - April 1st, 2024 Spring Special

[Juan14]

For the function S(X); notice that:
1). 2^(12N+4) = 2^(12N)*16, so 16 can go outside the summation.
2). 2^(12(N+1))/2^(12N) = 2^12 = 4096.
3). COMB(2N, N) = (2*N)!/(N!^2).
COMB(2(N+1), N+1)/COMB(2N, N) = ((2*(N+1))!/((N+1)!^2))/((2*N)!/(N!^2))
= (4*N+2)/(N+1) = 2*(2-1/(N+1)).
Having (COMB(2N, N)^3)/(2^(12N)) stored in register 03, you can get to (COMB(2(N+1), N+1)^3)/(2^(12(N+1)) by multiplying register 03 by ((2*(2-1/(N+1)))^3)/4096
= ((2-1/(N+1))^3)/512.
With X in register 01, N+1 in register 02 and summation in register 04, we have the next program for the free42:

PHP Code:

01 LBL “S”
02 STO 01
03 0
04 STO 02
05 1
06 STO 03
07 5
08 STO 04
09 LBL 00
10 2
11 1
12 RCL+ 02
13 STO 02
14 1/X
15 -
16 3
17 Y↑X
18 512
19 ÷
20 RCL× 03
21 STO 03
22 RCL 01
23 RCL× 02
24 5
25 +
26 ×
27 STO+ 04
28 1E-34
29 X<Y?
30 GTO 00
31 RCL 04
32 1/X
33 16
34 ×
35 END 


S(42) = 3.141592653589793238462643383279504
wow, pi again!