Sunday, January 22, 2006 10:59 PM [GMT+1=CET],
Christian Goetz <[EMAIL PROTECTED]> escribió:
I searched the web for the starting numbers and found a japanese page
with a better description of the algorithm
(http://web2.incl.ne.jp/yaoki/2seqp2.htm)
Problem:
A(n+1)=A(n)+B(n)
B(n+1)=A(n)*B(n)
A(1)=B(1)=1
Problem 1: General term A (n), B (n) please ask. (dunno what this
one means)
Problem 2:
A(1)=1,A(2)=2,A(3)=3,A(4)=5
A(5)=11,A(6)=41,A(7)=371,A(8)=13901
B(1)=1,B(2)=1,B(3)=2,B(4)=6
B(5)=30,B(6)=330,B(7)=13530,B(8)=5019630
Prove for 'A (n) * B (n) +1' the fact that it becomes prime.
---
With Derive
w(n) :=
Prog
i := 1
v := [[1, 1, true]]
Loop
If i > n
RETURN v
v :=APPEND(v, [[v↓i↓1 + v↓i↓2, v↓i↓1·v↓i↓2, PRIME((v↓i↓1 +
v↓i↓2)·(v↓i↓1·v↓i↓2) + 1)]])
i :+ 1
====>
⎡ 1 1 true ⎤
⎢ ⎥
⎢ 2 1 true ⎥
⎢ ⎥
⎢ 3 2 true ⎥
⎢ ⎥
⎢ 5 6 true ⎥
⎢ ⎥
⎢ 11 30 true ⎥
⎢ ⎥
⎢ 41 330 false ⎥
⎢ ⎥
⎢ 371 13530 true ⎥
⎢ ⎥
⎣ 13901 5019630 false ⎦
For i = 9 to 20 also is false. The sequences grow very quickly: A(20) =
9.199334563·10^2157 and B(20) = 4.557263981·10^3491.
A(n) ---> http://www.research.att.com/~njas/sequences/A003686
What is true, and trivial, is that A(n) and B(n) are relatively primes.
Best regards,
Ignacio Larrosa Cañestro
A Coruña (España)
[EMAIL PROTECTED]
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