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Fun question :D #Bored

Mathematics
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Function F(n) defined over all positive integers n satisfies the following: \[f(1)=1,f(2)=2, and\] \[f(n+2)=f(n+1)+(f(n))^{2}+4022 \ for \ n \ge1\] How many of the 4022 integers f(1), f(2),....f(4022) are multiple of 7?
well I notice f(3) is a multiple of 7 but checking all of those will be crazy so there has to be a slick way of doing this
:)

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|dw:1436502484982:dw|
-1* mod 7
so thats the answer for you dan?
no just thinking
You will get it dan lol
hm
1
is the answer 1
Well nope
i think my arithmetic is wrong maybe, i got a repetition at f(9) ill put what i got
well going to the whole mod thing I guess we can really just look at: \[f(n+2)=f(n+1)+(f(n))^2+4 \text{ thinking in mod } 7\] these will make the outputs easier to look at but that is still a lot of f( ) thingys to check: \[f(1)=1,f(2)=2,f(3)=7=0,f(4)=15=1, f(5)=68=5\] hmmm... \[f(6)=f(5)+(f(4))^2+4 \\ f(6)=5+1^2+4=10=3 \\ f(7)=f(6)+(f(5))^2+4 \\ f(7)=3+(5)^2+4=25=4 \\ f(8)=f(7)+(f(6))^2+4 \\ f(8)=4+(3)^2+4 \\ f(8)=1+9=10=3 \\ f(9)=f(8)+(f(7))^2+4 \\ f(9)=3+4^2+4=4^2=2 \\ f(10)=f(9)+(f(8))^2+4 \\ f(10)+2+3^2+4=6+2=8=1 \\ f(11)=f(10)+(f(9))^2+4 \\ f(11)=1+2^2+4=9=2\] \[f(12)=f(11)+(f(10))^2+4 \\ f(12)=2+1^2+4=7=0\] I think f(3) and f(12) so far are multiples of 7
|dw:1436504022956:dw|
k ya i saw myin writing same thing
lol nope @myininaya
I didn't give an answer
Oh thought you did lol sorry
I'm showing my work so far hoping it can help anyone else trigger more thoughts but my thought actually came from dan's work above
i have a feeling we will run into a repeating pattern in mod 7 like every 9th maybe
I solved it by programming and I asked @jagr2713 for confirmation of my answer but he insisted me to post my answer here lol... ` numbers = [1, 2] total = 0 for x in range(3, 4023): NewNumber = (numbers[1] + (numbers[0])**2 + 4022)%7 if NewNumber%7 == 0: total +=1 numbers = [numbers[1], NewNumber] print(total) ` Output is 447
Lets conjecture \[7\mid f(3+9k)\]
:P 4022 / 9 = 446.8 pretty close hehe
based on @myininaya 'w work..
WOw great work
Great job :D
brute force cheattoorr
lol dan got right after geerky lol
I see pattern 1,2,0,1,5,3,4,3,2,1,2,0,1,5,3,4,3,2,1,2,0,1,5,3,4,3,2,1,2,0,1,5,3,4,3,2,... I wonder if this is the pattern we have a 0 at 3rd term,12th term, 21th term, oh and I see where @ganeshie8 get's his conjecture now
yep
ask another one now!
k let start typing it :D
the proof might be interesting
by induction it can be proven
I would love to see a proof on this but I think that is above me.
once you get a full repettiion it must be true
but there could be a neater proof this is a lame way to prove it
that is true, once the remainders start repeating, we're technically done!
sadly i can't stay for the fun @jagr2713 will post it is time for me to retire :(
good night myin
For all real numbers x, the function f(x) satisfies \[f(1+x)+xf(1-x)=x ^{2}+9x\] If the values of f(-7) can be expressed as \[\frac{ a }{ b }\] Where a and b are coprime positive integers/ What is a+b Here you go @dan815
see ya @myininaya D:
f(1+x)=9x f(1-x)=x?
x*f(1-x)= x^2+ax+b f(1+x)=cx+d
oh and since u applied x after there can be no constant term in both
therefore f(1-x)=x+b and f(1+x)=ax they must of this tform
f(1+x) cannot have a constant term
and f(1-x) cannot have a coefficient on the x
so we are restrited with the unknowns here
restricted*
hm
f(1-x)=x+b and f(1+x)=ax f(-7)=-7+b f(-7)=-7a and f(-7)= c/d, where c and d are co prime int hmm i am wondering if the first lines are really okay to assume
hm lol waiting lol falling asleep lol
f(-7)=-7+b f(-7)=-7a a+b=9 f(-7)=-7+ 9-a =2-a f(-7)=-7*(a) 2-a=-7a a=1/3
lol what is going on
I don't know but I'll guess 13 and then I think I'll try solving it to make sure.
\[\large a+b=217\]
howd u do it
Nope ganesh D: is that your answer @dan815
but your close
simply plugin \(x=\pm 8\), you will get two linear eqn's which you can solve
\[\large a+b=229\] ?
Correcto :D Finally now i can sleep lol
dont know who to give the stars D:
Ahhh good trick @ganeshie8 ! I had tried to turn this into a differential equation like yesterday and got something completely terrible.
Ahhh good trick @ganeshie8 ! I had tried to turn this into a differential equation like yesterday and got something completely terrible.
ah i liked it
It looks like you also get the answer to what f(9) is for free as well haha
I subed 1+x=u and 1-x=v and tried to get a differential eqn like yesterday but got stuck
@ganeshie8 What you could try instead is plug in 1-x=y+h and 1-x=y and that'll turn it into a differential equation.
this could be an interesting question For all real numbers x, the function f(x) satisfies \[f(1+x)+xf(1-x)=x ^{2}+9x\] find \(f(x)\) Here you go @dan815
Ahh okay I see that should make it easy..
Unfortunately (I might have done it wrong) the differential equation we get is not a very pretty one to solve but we can do it with an integrating factor it seems, I just think it ends up having the erf(x) function in there as part of the answer #_#
do you think this assumption is right f(1+x)=ux f(1-x)=x+v
Hmmm I don't know?
if we ignore the fourier series and other orthogonal representations to polynomial functions
f(1+x)+x*f(1-x) = x^2+-9x then can we say f(1+x) must be 1 degree lower than x*f(1-x) so the x^2 term must come from that term
Ahhh I see what you're looking at now and why you would say that but I'm sure if I know if we can do that or not. Interesting observation though. It looks like we should be able to solve for f(x)=ax+b by plugging in I think.
yeah at least we will get all linear functions, if not all
yea because if this is really true u get restrictions all over the place since then x*f(1-x) cannot have a constant term that means f(1-x) must have form ax+b and then f(1+x) cannot have a constnat term as it all equalys x^2+9x
f(1-x)= x + u f(1+x)=v*x u+v=9
I was also considering if we get anything by splitting it up into an even and odd function: f(x)=g(x)+h(x) g(x)=g(-x) and h(x)=-h(-x) since we can always do this, then make a pair of equations this way that will be easy to solve hopefully.
oo
Looks f(x) cannot be a polynomial
does this work f(1+x) + xf(1-x) = x^2+9x f(1-x) -xf(1+x) = x^2-9x ----------------------------- f(1-x)(1+x^2) = x^3+10x^2-9x f(1-x) = (x^3+10x^2-9x)/(1+x^2)
i think so, now u can rewrite the right side as functions of 1-x ?
wow yeah 1-x = u should do
Nice, that wasn't so bad after all haha
So if we look at f(x) as being a generating function, what's the sequence it generates? :O

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