Jouer avec c = (a*b)
mod (a+b);
Itérer.
J’ai envoyé récemment
ce message aux listes MathFun et SeqFans:
______________________________________________
Date: lun.
07/07/2008 19:03
À: math-fun;
seqfan@ext.jussieu.fr
Objet : Playing
with c = (a*b) mod (a+b)
Hello MathFun
& SeqFans,
Start S = 2, 9,...
Next term is (2*9)mod(2+9)
18 mod 11 = 7
S = 2, 9, 7,...
Next term is (9*7)mod(9+7)
63 mod 16 = 15
S = 2, 9, 7, 15,...
Next term is (7*15)mod(7+15)
105 mod 22 = 17
S = 2, 9, 7, 15, 17,...
Next term is (15*17)mod(15+17)
255 mod 32 = 31
S = 2, 9, 7, 15, 17, 31,...
Next term is (17*31)mod(17+31)
527 mod 48 = 47
S = 2, 9, 7, 15, 17, 31, 47,...
Does this end in a loop?
If yes, is there another
start
which doesn't?
Best,
É.
______________________________________________
Trois réponses intéressantes ont fusé :
[Chuck
Seggelin] :
Hi Eric,
(...)
Looks like this sequence rapidly enters
into repetition:
2, 9, 7, 15, 17, 31, 47, 53, 91, 71, 143,
95, 19, 95, 95, 95, ...
All remaining terms are 95. Many sequences
of this variety (based on different starting terms) seem to "bottom
out" fairly quickly. An interesting related seq
might be if the first term is 2, what successive second terms give longer and
longer non-repeating sequences?
2,3 goes 4 terms.
2,5 goes 6 terms.
2,9 goes 14 terms without repeating.
2,25 goes 21 terms which is the next record (25 is also the
next odd square!)
2,27 goes 13.
2,29 goes 23 terms which is the new record.
The next record is at the next odd square 49, yielding 33 terms.
Then 99, yielding 36 terms.
...
So the first records are 3, 5, 9, 25, 29,
49, 99 (corresponding to 4, 6, 14, 21, 23, 33, 36).
Incidentally 2,61
gives the repeating loop with the most number of elements I've seen yet (six
elements: 61, 59, 119, 79, 95, 23).
-- Chuck PlasteredDragon
[Jack
Brennen] :
That one ends in a loop fairly quickly:
2, 9, 7, 15, 17, 31, 47, 53, 91, 71, 143, 95, 19, 95, 95, 95, ...
Playing around with different starting
pairs, it looks like all starting pairs may end up in a loop, or by going to
the pair (0,0) which is either a loop or a singularity, depending on whether
you take 0 mod 0 as being 0 or being undefined.
Furthermore, you can go an arbitrarily long time without looping. Take for large a:
.., 6*a+6, 6*a, ...
The next term is 6*a-6, and progressively
it goes down the ladder by steps of 6 until it terminates at ..., 18, 12, 6, 0,
0.
In a few minutes of searching, the longest
sequence I could find which eventually loops without going to zero was the
sequence starting with (29,574), which hits 855 at the 79th term and then stays
at 855.
Jack
Jack a envoyé cette
suite à l’OEIS de Neil Sloane, c’est désormais
la A137453.
[Maximilian
Hasler] :
> Next term is (17*31)mod(17+31)
>
527 mod 48 = 47
> S = 2, 9, 7, 15, 17, 31, 47,...
>
> Does this end in a loop?
rather in a fixed
point :
2, 9, 7, 15, 17, 31, 47, 53, 91, 71, 143,
95, 19, 95, 95, 95,
> If yes, is there another start which
doesn't?
2,7
ends in a true loop:
EA([2,7])
[2, 7, 5, 11, 7, 5, 11, 7,
5, 11, 7, 5, 11, 7, 5, 11, 7, 5, 11,]
At a first glance, I can't
find a starting value that would not end in a fixed point or a loop.
Many end in a loop
consisting of {5,7,11} (in some order), sometimes
reached after a long transition :
32, 51, 55, 49, 95, 47, 63,
101, 131, 7, 89, 47, 103, 41, 47, 79, 59,
107, 5, 87, 67, 131, 65, 87,
31, 101, 95, 187, 281, 131, 143, 101, 47,
11, 53, 7, 11, 5, 7, 11,...
But there are other loops
possible, the first I found is
12, 99, 78, 111, 153, 87,
111, 153, 87, 111, 153, 87, 111, ...
There are longer loops :
14, 68, 50, 96, 128, 192,
256, 320, 128, 192, 256, 320, 128, 192, 256, 320
58, 79, 61, 59, 119, 79, 95,
23, 61, 59, 119, 79, 95, 23, 61, 59, 119, 79, 95, ...
32, 51, 55, 49, 95, 47, 63,
101, 131, 7, 89, 47, 103, 41, 47, 79, 59, 107, 5, 87, 67, 131, 65, 87, 31, 101,
95, 187, 281, 131...
One could imagine 2 seq :
A1 : square array T[i,j] = length of the cycle the seq. (b1,b2,(b1*b2) mod
(b1+b2), (b2*b3) mod (b2+b3),...) will end in
A2 : square array T[i,j] = transition time before reaching the loop
One could also imagine a
seq. where b[k]
+/* b[k-1] is replaced by sum/product of more (or all) terms.
As to the latter, will it
always end in 0 ?
[3, 19, 13, 6, 18, 24, 52,
54, 54, 0 .....
[3, 17, 11, 3, 17, 0, 0, 0,
0, 0, 0.....
[3, 16, 10, 16, 30, 0, 0, 0,
0...
[3, 14, 8, 11, 24, 24, 0, 0,
0,.....
Then there is :
A3 : square array T[i,j] = transition time before reaching 0
which might be related to some gcd
(euclid's algorithm) properties
Maximilian
... et voici une quatrième réponse
somptueuse de [Jacques
Tramu] (citant par ailleurs Georges Brougnard) :
The following describes the behaviour of the sequence u(n) = (u(n-2)*u(n-1)) mod(u(n-1) + u(n-2)) for large integers:
10^1000
3.7125500000 4706 43
57
We take 100 random pairs of starters (u(0) =a, u(1)= b) < 10^1000
(1000 digits), and see that:
- the average sequence length (AVL) is 1000* 3.71
- the
largest
sequence found has length 4706,
- 43% of sequences
end in (0,0),
- 57 % end in a loop of fixed point.
The same for
10^p, 200 <= p <= 5000, each time 100 random pairs of starters.
One can see that the ratio R= AVL/p is close to 3.6 for all p in this
sample.
10^200
3.7440500000 1009
50 50
10^400
3.7919000000 1960 43
57
10^600
3.6428000000 2863 52
48
10^800
3.6871250000 3787 47
53
10^1000
3.7125500000 4706 43
57
10^1200
3.5687666667 5463 54
46
10^1400
3.6217214286 6151 48
52
10^1600
3.5595187500 7125 52
48
10^1800
3.5865500000 7852 50
50
10^2000
3.5276800000 8875 55
45
10^2200
3.4119636364 9566 63
37
10^2400
3.5224416667 10469 55
45
10^2600
3.5431000000 11399 52
48
10^2800
3.6714142857 12160 42
58
10^3000
3.5366266667 12997 52
48
10^3200
3.4938656250 13873 56
44
10^3400
3.4834970588 14798 56
44
10^3600
3.6956305556 15528 39
61
10^3800
3.6228578947 16425 45
55
10^4000
3.4844000000 16984 55
45
10^4200
3.4705285714 18003 57
43
10^5000
3.5578060000 21355 49
51
Question : does the
ratio R have a limit, have upper and lower bounds ?
My friend Georges Brougnard conjectured
that lim(R) = PI/2 * log(10) = 3.616892206, but offered no proof.
Regards,
JT
No divergent sequence
was found,
No animal was harmed in the making of this experiment.
______________________________________________
Thanks to everyone !
(merci à tous)
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