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Question Number 212242 by York12 last updated on 08/Oct/24
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Help
Commented by York12 last updated on 08/Oct/24
Commented by Frix last updated on 08/Oct/24
Fibonacci Sequences:  a_n =(1/(2(√2)))((3+2(√2))^n −(3−2(√2))^n )  a_1 =0; a_2 =2; a_n =−a_(n−2) +6a_(n−1)     b_n =(1/2)((3+2(√2))^n +(3−2(√2))^n )  b_1 =1; b_2 =3; b_n =−b_(n−2) +6b_(n−1)     ⇒ gcd=1
FibonacciSequences:an=122((3+22)n(322)n)a1=0;a2=2;an=an2+6an1bn=12((3+22)n+(322)n)b1=1;b2=3;bn=bn2+6bn1gcd=1
Commented by York12 last updated on 08/Oct/24
Thank you
Thankyou
Commented by York12 last updated on 08/Oct/24
TBH sir I do not understand why it′s  neccesarliy means that the gcd =1  if possible can you elaborate please.
TBHsirIdonotunderstandwhyitsneccesarliymeansthatthegcd=1ifpossiblecanyouelaborateplease.
Commented by Frix last updated on 08/Oct/24
I tried the following:  a_1 =0; a_2 =2  b_1 =1; b_2 =1+2  a_3 =12; b_3 =5+12  a_4 =70; b_4 =29+70  a_5 =408; b_5 =169+408  ...  a_k ; b_k =a_k +n  gcd (a_k ; a_k +n) =q ⇒ q∣a_k ∧q∣n  n is odd ⇒ q is odd  I found no q≠1 up to k=100
Itriedthefollowing:a1=0;a2=2b1=1;b2=1+2a3=12;b3=5+12a4=70;b4=29+70a5=408;b5=169+408ak;bk=ak+ngcd(ak;ak+n)=qqakqnnisoddqisoddIfoundnoq1uptok=100

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