1the last perfect square is m which is 2n+1 behind a0
made this assumption for obvious reasons ... dat if it is even, then twice of it is always even.
waise, twice of ANYTHING is even!
pls say if i am right.
Not the kind of convergence that appears in undergrad courses. So please dont be scared away. In fact if you apply your mind to this one, you may be rewarded.
For every integer n \ge 0 , let S(n) = n - m^2 where m is the greatest integer with m^2 \le n.
Define a sequence (a_k)_{k=0}^{\infty} by a0 = A, where A is a natural number and ak+1 = ak + S(ak).
For what values of A is the sequence eventually constant?
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42 Answers
62yes that is correct manipal.. expand it and see if it makes any sense.. and also look at the expression given in post 30
Those two together are just 1 step away from the final answer
11@prophet,
bhaiya i got my mistake... i was taking wrong S(ak) ... dats y din get constant series for perfect square.
tx.
62There is a slight mistake I think in what you have done sky...
you are onthe right directionto say the least...
1skygirl
:)
wat mistake ?
in sequence ?
Nishant
Reply to chat at 12:16
only a small mistake
see take the number to be n^2+k
skygirl
ok
but where ?
Nishant
where 0<=k<2n+1
skygirl
i cudn catch where
Nishant
now what will S() be?
skygirl
huh
mere upar se gaya
:(
wait wait
i took any arbitrary 'k'
Nishant
okie
skygirl
and n is also anything...
like should that limit of k dat u said matter ?
Nishant
ruko.. let me see the notation that is being used
skygirl
then i am thinking agn
ok
Nishant
let ak = m^2+k
where k <2m+1
becasue if k>2m+1
then ak > (m+1)^2
skygirl
haan
no but the thing is that i took that thing obvious dat woh jo diff hai that is i mean smaller than k ...
62Yaar you guys have to give it just a little shot
i will give one more hint
ak = n2+k where 0<=k<2n+1
S(ak) = k
ak+1=ak+S(ak) = n2+2k
now does this ring bell?
11no ringing of bells[3]
please help us out experts!!!!!!!!!!!!!
62One more hint
the 2 next Perfect square bigger than n2 will be ??
1in ak+1 = ak + S(ak) ,
let ak+1 be a perfect square.
say, the last perfect square is m which is 2n+1 behind a0
then S(ak) =2n+1
so finally we have ,
ak+1 = m + (2n+2)
=> ak+1 - m = even integer..... this is impossible because
difference between two perfect
squares is always odd.
hence if A = not a perfect square, the series will never converge.
62Prophet sir, no one seems to be answering.. so may be you should finally give the answer :)
Somehow i guess they believe this is not in JEE so not to be solved ;)
1seriously speaking i cudn add anything more logical to my last post...
though u explained me abt that k ... and in my expression it is 2n+2 .
but ... [12] [2]
62Ok i will only post the full solution....
ak = n2+k where 0<=k<2n+1
S(ak) = k
ak+1=ak+S(ak) = n2+2k
ak+1<= n2+4n (because of the size of k!)
so the only perfect squares greater than n2 and less than n2+2k is
n2+2n+1
which is never equal to ak+1
because ??? (just th elast 8-10 words left.. .atleast fill those :D)
39because
ak+1 = r2 +2S(ak) is greater than r2 but
less than (r +2)2, and not equal to (r+1)2 (its obvious)
it was a question of my birth year :P
9hmm this Q is easy only yaar (definitely not olympiad level)
Ok for
sequence to repeat ,
S(ak)=0 for some k
ie when A is a square (obvious)
when A isnt a square
let some other ak+1=α2 turn out to be a square first
so
α2= 2n - m2 with (ak=n)
now we know m2<n<m2+2m+1
so α2<2(m2+2m+1) -m2
α2< m2+4m+2 <(m+2)2
also α2>2(m2)-m2> m2
hence α2 =(m+1)2
now
(m+1)2=2n - m2
ie 2n =m2 + (m+1)2 LHS is even
RHS is cumpolsorily even + odd = odd
hence situation impossible
39celestine , good judgement that it s not olympiad level.
its a putnam q. now wat u call putnam i donno.
9dint know putnam level was so low!!!!
some Qs asked in jee are abv this level man
9b555 i wud call Qs posted by u only as olympiad level (for some i cant solve even if i take hrs)
prophets Q are simple only
62yes celestine only after seeing the putnam questins i realised that the questions were so easy
most of them are easier than JEE questions ... only that they are a bit different from the jee syllabus.
and yes your proof is good :)
62bhargav, you are only repeating yourself..
this i had written in the first post that i deleted...
This is the first step to that proof.
the next step is the most obvious step... pls see if you can fill it!
I dont know if prophet could help you in understanding what i am trying to say...
62Eureka..
trust me this is not the oLympiad type
it can be solved by the normal maths that you know
May be it is not useful for XII guys giving jee in 1 month
but is more than userful for a XI guy ...
24thats why I said.....its not worth soving these at this point of time
62but does that prove that it cannot go on to become a perfect square later
According to your claim,
You have to prove that if A is not a perfect square then none of ak's become a perfect square...
39that is what i meant, if A is not a perfect square then none of ak's become a perfect square...
39its a kind of obvious na bhaiyan, if A is not a perfect square then s(a0)wont be equal to zero, there upon others also non zero and the sequence can no more be a constant ya got now watever u said bhaiyan
