Brain-teaser

[botg]

1
11
21
1211
111221

What number comes next?

If you already know the solution, please don't spoil it for the others.

[Superracingteam]

i haven't found it

what is my IQ?? under 100 or above it??


P.S I'm only 15 years old

[botg]

You already know too much. The more you know, the less likely you are able to solve this puzzle. University students usually won't be able to solve this, but school kids (around 2nd grade) will sove it easily.

[TheAngryPenguin]

Ah! We adults tend to complicate things more than they need to be.


D(t+1) = (sigma(K=1,LOG(D(t)*10)-LOG(D(t)*10)%1,((D(t)-D(t)%10^(LOG(D(t))-
LOG(D(t))%1)+sigma(S=1,LOG(D(t))-LOG(D(t))%1,(((sigma(R=1,LOG(D(t)*10)
-LOG(D(t)*10)%1-1,ABS((D(t)*10-D(t)*10%10^(R+1))%10^(R+2)/10-(D(t)*10-D(t)*
10%10^R)%10^(R+1)))/10)-(sigma(R=1,LOG(D(t)*10)-LOG(D(t)*10)%1-1,ABS((D(t)*
10-D(t)*10%10^(R+1))%10^(R+2)/10-(D(t)*10-D(t)*10%10^R)%10^(R+1)))/10)
%10^(S-1))%10^S+1)%(((sigma(R=1,LOG(D(t)*10)-LOG(D(t)*10)%1-1,ABS((D(t)
*10-D(t)*10%10^(R+1))%10^(R+2)/10-(D(t)*10-D(t)*10%10^R)%10^(R+1)))/10)-(
sigma(R=1,LOG(D(t)*10)-LOG(D(t)*10)%1-1,ABS((D(t)*10-D(t)*10%10^(R+1))%10
^(R+2)/10-(D(t)*10-D(t)*10%10^R)%10^(R+1)))/10)%10^(S-1))%10^S+.5)*2*(D(t)
-D(t)%10^(S-1))%10^S))-(D(t)-D(t)%10^(LOG(D(t))-LOG(D(t))%1)+sigma(S=1,
LOG(D(t))-LOG(D(t))%1,(((sigma(R=1,LOG(D(t)*10)-LOG(D(t)*10)%1-1,ABS((D(t)
*10-D(t)*10%10^(R+1))%10^(R+2)/10-(D(t)*10-D(t)*10%10^R)%10^(R+1)))/10)
-(sigma(R=1,LOG(D(t)*10)-LOG(D(t)*10)%1-1,ABS((D(t)*10-D(t)*10%10^(R+1))%
10^(R+2)/10-(D(t)*10-D(t)*10%10^R)%10^(R+1)))/10)%10^(S-1))%10^S+1)%(((
sigma(R=1,LOG(D(t)*10)-LOG(D(t)*10)%1-1,ABS((D(t)*10-D(t)*10%10^(R+1))%
10^(R+2)/10-(D(t)*10-D(t)*10%10^R)%10^(R+1)))/10)-(sigma(R=1,LOG(D(t)*
10)-LOG(D(t)*10)%1-1,ABS((D(t)*10-D(t)*10%10^(R+1))%10^(R+2)/10-(D(t)*
10-D(t)*10%10^R)%10^(R+1)))/10)%10^(S-1))%10^S+.5)*2*(D(t)-D(t)%10^(S
-1))%10^S))%10^(K-1))%10^K/10^(K-1)*100^(2*sigma(N=1,K,(((D(t)-D(t)%
10^(LOG(D(t))-LOG(D(t))%1)+sigma(S=1,LOG(D(t))-LOG(D(t))%1,(((sigma(R=1,
LOG(D(t)*10)-LOG(D(t)*10)%1-1,ABS((D(t)*10-D(t)*10%10^(R+1))%10^(R+2)/10-
(D(t)*10-D(t)*10%10^R)%10^(R+1)))/10)-(sigma(R=1,LOG(D(t)*10)-LOG(D(t)*
10)%1-1,ABS((D(t)*10-D(t)*10%10^(R+1))%10^(R+2)/10-(D(t)*10-D(t)*10%10^R)%
10^(R+1)))/10)%10^(S-1))%10^S+1)%(((sigma(R=1,LOG(D(t)*10)-LOG(D(t)*10)%
1-1,ABS((D(t)*10-D(t)*10%10^(R+1))%10^(R+2)/10-(D(t)*10-D(t)*10%10^R)%
10^(R+1)))/10)-(sigma(R=1,LOG(D(t)*10)-LOG(D(t)*10)%1-1,ABS((D(t)*10-D(t)
*10%10^(R+1))%10^(R+2)/10-(D(t)*10-D(t)*10%10^R)%10^(R+1)))/10)%10^(S-1))
%10^S+.5)*2*(D(t)-D(t)%10^(S-1))%10^S))-(D(t)-D(t)%10^(LOG(D(t))-LOG(D(t))
%1)+sigma(S=1,LOG(D(t))-LOG(D(t))%1,(((sigma(R=1,LOG(D(t)*10)-LOG(D(t)*10)
%1-1,ABS((D(t)*10-D(t)*10%10^(R+1))%10^(R+2)/10-(D(t)*10-D(t)*10%10^R)%10^
(R+1)))/10)-(sigma(R=1,LOG(D(t)*10)-LOG(D(t)*10)%1-1,ABS((D(t)*10-D(t)*10%
10^(R+1))%10^(R+2)/10-(D(t)*10-D(t)*10%10^R)%10^(R+1)))/10)%10^(S-1))%10^
S+1)%(((sigma(R=1,LOG(D(t)*10)-LOG(D(t)*10)%1-1,ABS((D(t)*10-D(t)*10%10^
(R+1))%10^(R+2)/10-(D(t)*10-D(t)*10%10^R)%10^(R+1)))/10)-(sigma(R=1,
LOG(D(t)*10)-LOG(D(t)*10)%1-1,ABS((D(t)*10-D(t)*10%10^(R+1))%10^(R+2)/
10-(D(t)*10-D(t)*10%10^R)%10^(R+1)))/10)%10^(S-1))%10^S+.5)*2*(D(t)-D(t)%
10^(S-1))%10^S))%10^(N-1))%10^N+1)%(((D(t)-D(t)%10^(LOG(D(t))-LOG(D(t))%
1)+sigma(S=1,LOG(D(t))-LOG(D(t))%1,(((sigma(R=1,LOG(D(t)*10)-LOG(D(t)*
10)%1-1,ABS((D(t)*10-D(t)*10%10^(R+1))%10^(R+2)/10-(D(t)*10-D(t)*10%10^
R)%10^(R+1)))/10)-(sigma(R=1,LOG(D(t)*10)-LOG(D(t)*10)%1-1,ABS((D(t)*10-
D(t)*10%10^(R+1))%10^(R+2)/10-(D(t)*10-D(t)*10%10^R)%10^(R+1)))/10)%10^
(S-1))%10^S+1)%(((sigma(R=1,LOG(D(t)*10)-LOG(D(t)*10)%1-1,ABS((D(t)*10-
D(t)*10%10^(R+1))%10^(R+2)/10-(D(t)*10-D(t)*10%10^R)%10^(R+1)))/10)-
(sigma(R=1,LOG(D(t)*10)-LOG(D(t)*10)%1-1,ABS((D(t)*10-D(t)*10%10^(R+1))%
10^(R+2)/10-(D(t)*10-D(t)*10%10^R)%10^(R+1)))/10)%10^(S-1))%10^S+.5)*2*
(D(t)-D(t)%10^(S-1))%10^S))-(D(t)-D(t)%10^(LOG(D(t))-LOG(D(t))%1)+
sigma(S=1,LOG(D(t))-LOG(D(t))%1,(((sigma(R=1,LOG(D(t)*10)-LOG(D(t)*10
)%1-1,ABS((D(t)*10-D(t)*10%10^(R+1))%10^(R+2)/10-(D(t)*10-D(t)*10%10^
R)%10^(R+1)))/10)-(sigma(R=1,LOG(D(t)*10)-LOG(D(t)*10)%1-1,ABS((D(t)*
10-D(t)*10%10^(R+1))%10^(R+2)/10-(D(t)*10-D(t)*10%10^R)%10^(R+1)))/
10)%10^(S-1))%10^S+1)%(((sigma(R=1,LOG(D(t)*10)-LOG(D(t)*10)%1-1,
ABS((D(t)*10-D(t)*10%10^(R+1))%10^(R+2)/10-(D(t)*10-D(t)*10%10^R)%
10^(R+1)))/10)-(sigma(R=1,LOG(D(t)*10)-LOG(D(t)*10)%1-1,ABS((D(t)*
10-D(t)*10%10^(R+1))%10^(R+2)/10-(D(t)*10-D(t)*10%10^R)%10^(R+1)))
/10)%10^(S-1))%10^S+.5)*2*(D(t)-D(t)%10^(S-1))%10^S))%10^(N-1))%10^
N+.5))))/100)+(sigma(K=1,LOG(D(t)*10)-LOG(D(t)*10)%1,100^(1+sigma(N=
1,K-1,2*((((sigma(R=1,LOG(D(t)*10)-LOG(D(t)*10)%1-1,ABS((D(t)*10-D(t)*
10%10^(R+1))%10^(R+2)/10-(D(t)*10-D(t)*10%10^R)%10^(R+1)))/10)-(
sigma(R=1,LOG(D(t)*10)-LOG(D(t)*10)%1-1,ABS((D(t)*10-D(t)*10%10^(R+1))%
10^(R+2)/10-(D(t)*10-D(t)*10%10^R)%10^(R+1)))/10)%10^(N-1))%10^N/10^(N-
1)+1)%(((sigma(R=1,LOG(D(t)*10)-LOG(D(t)*10)%1-1,ABS((D(t)*10-D(t)*10%
10^(R+1))%10^(R+2)/10-(D(t)*10-D(t)*10%10^R)%10^(R+1)))/10)-(sigma(R=
1,LOG(D(t)*10)-LOG(D(t)*10)%1-1,ABS((D(t)*10-D(t)*10%10^(R+1))%10^(R+
2)/10-(D(t)*10-D(t)*10%10^R)%10^(R+1)))/10)%10^(N-1))%10^N/10^(N-1)+
.5)))))/10)

[botg]

What is that formula below your post? If you tried to interpolate the next number somehow, you can get it a lot easier with a simple polynom, but in any case the solution is wrong.

[TheAngryPenguin]

Apparently "D is a recursive function and t is the term number."

"By the way, % here is a certain non-integer remainder function. 2.1%0.1 would be 0, 2.1%0.2 would be 0.1, 2.1%0.3 would be 0 since 0.3 fits evenly into 2.1, etc.) If you really want conventional operators, you could define % with limits and modular arithmetic..."

From where I found the solution:

hxxp://dheera.net/sci/sequence_sol.php

...all of which is way beyond me

[BC-A]

1
11
21
1211
111221

What number comes next?

If you already know the solution, please don't spoil it for the others.

Nobody answered ?
I bet 312211

[grimboy]

A solution follows:
The number gets bigger each time.
That is all.

[BigDAS]

the numeral 1 is just one one

one one can be written as 1 1

1 1 is actually two ones.

two ones can be written as 2 1

2 1 is one two and one one

one one and one two is 1 2 1 1

1 2 1 1 is one one, one two, and two ones.

one one, one two, and two ones is 1 1 1 2 2 1.

1 1 1 2 2 1 is three ones, two twos, and one one.

So the next number is 3 1 2 2 1 1.

[Asiadeep]

Damn....thats pretty complicated if you don't actually "say it out loud in your head"

[ambrose8080]

312211,13112221,1113213211,31131211131221