OEIS A229580 mini challenge (RPL)

[Gerson W. Barbosa]

(05-02-2018 04:59 PM)Valentin Albillo Wrote:  You're welcome, thanks for the explanation. Dealing with continued fractions and their theory is a math topic I find remarkably interesting so I welcome any challenge or problem having to do with them.

However I think you were right in the first place, a recursive formula could have indeed been of help for my particular application. In the program below I have used a closed-form formula to compute the last denominator of the continued fraction, but I might as well have computed it using a recursive formula, every term being computed at each step of the first FOR-NEXT loop.

Regards,

Gerson.

===============

>list

10 DESTROY ALL
15 INPUT N
20 S1=0 @ S2=0 @ P=1
25 FOR I=1 TO N
30 P=4*P @ S1=S1+1/(2*I-1)^2
35 NEXT I
40 P=P/4 @ A=2*N-1 @ B=8*N+4 @ D=(2*N+1)*P
45 FOR I=N TO 1 STEP -1
47 PRINT D*A/B
50 S2=I^4/(D*N+S2) @ D=D*A/B @ A=A-2 @ B=B-8
55 NEXT I
60 S2=1/(4*N+S2)
65 PRINT 8*(S1+S2)

>run

? 12
 24117248
 5505024
 1245184
 278528
 61440
 13312
 2816
 576
 112
 20
 3
 .25
 9.86960440168