I was trying to prove that for all n∈N there are integer numbers {a1,a2,…,an,bn} s.t. a21+a22+⋯+a2n=b2n.
I founded that if {a1,a2,…,an,bn} have the property then {a1,a2,…,an,an+1=b2n−12,bn+1=b2n+12} are the n+1 numbers which have the property too. I start with a1=b1=3 and get the sequence 3,4,12,84,… and now try found an in terms of n. I also prove that solving bn+1=b2n+12,b1=3 is equivalent to solving cn+1=c2n+cn+1,c1=1. Thanks for reading the text and any helps.
NOTE: It's eaasy to check that for b1=3 all ai and bi will be integer.
Answer
Let us tackle cn+1=c2n+cn+1
Set xn=cn+12.
We get
xn+1=x2n+54
This paper: http://www.fq.math.ca/Scanned/11-4/aho-a.pdf (see page 434) shows that xn is of the form: (nearest integer to k2n+12) - 12 for all large enough n, for some constant k (defined in terms of xn itself).
and thus cn=⌈k2n+12⌉
It is unlikely you will get a "neater" closed form.
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