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anonymous
 one year ago
a12 and a20 are the 12th and 20th terms of Fibonacci sequence.find (a12,a20)
anonymous
 one year ago
a12 and a20 are the 12th and 20th terms of Fibonacci sequence.find (a12,a20)

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0first i wrote the first terms of Fibonacci sequence. 1,1,2,3,5,8,13,21,34,55,89,144,233 then i solved these examples : (a2,a4)=(1,3)=1=a2 (a3,a6)=(2,8)=2=a3 (a4,a6)=(3,8)=1=a2 so i guessed (am,an)=a(m,n)

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2what does: find (a12,a20) mean?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0(a,b) means Greatest common divisor of a and b

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2hmm, you only wroted 13 terms .... any reason you didnt just compute it with the actual values?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2not sure if there is a way i can think of to prove your conjecture tho ... unless we develop the closed form of a fibber

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2x^(n+2) = x^(n+1) + x^n x^(n+2)  x^(n+1)  x^n = 0 x^n (x^2  x  1) = 0 x = 1/2 (1 + sqrt(5)) ................................. setting a system we get Ax1^0 + Bx2^0 = 1 , since f(0) = 1 Ax1^1 + Bx2^1 = 1 , since f(1) = 1  A + B = 1 Ax1 + Bx2 = 1 Ax1  Bx1 = x1 Ax1 + Bx2 = 1  B(x2x1) = 1x1 B = (1x1)/(x2x1) B = (1(1 + sqrt(5))/2)/((1  sqrt(5))/2(1 + sqrt(5))/2) B = (5sqrt(5))/10 and A = 1 B A = (5+sqrt(5))/10 the closed form is therefore: Ax1^n + Bx2^n

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2of course im starting at n=0 .. so we would have to adjust for index

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2not sure if this would be useful in a GCD setup tho ... since its 2terms

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2well, it would at least be useful in defining the nth terms for large values of n, but then so could coding a computer script to work the manual stuff for you.
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