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A228933
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Optimal ascending continued fraction expansion of phi-1=1/phi=(sqrt(5)-1)/2 .
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2
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2, 4, -18, 322, 103682, 10749957122, 115561578124838522882, 13354478338703157414450712387359637585922, 178342091698891843163466683840822101223162205277179656650156983624835803932590082
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OFFSET
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1,1
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COMMENTS
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See A228929 for the definition of "optimal ascending continued fraction".
Conjecture: The golden ratio (phi) expansion exhibits from the fourth term the recurrence relation a(n) = a(n-1)^2 - 2 described in A228931.
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LINKS
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FORMULA
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a(n) = a(n-1)^2 - 2 for n>3.
For n>3, a(n) = (sqrt(5)+2)^(2^(n-2)) + (sqrt(5)-2)^(2^(n-2)). - Vaclav Kotesovec, Sep 20 2013
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EXAMPLE
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phi = 1+1/2*(1+1/4*(1-1/18*(1+1/322*(1+1/103682*(1+1/10749957122*(1+...))))))
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MAPLE
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ArticoExp := proc (n, q::posint)::list; local L, i, z; Digits := 50000; L := []; z := frac(evalf(n)); for i to q+1 do if z = 0 then break end if; L := [op(L), round(1/abs(z))*sign(z)]; z := abs(z)*round(1/abs(z))-1 end do; return L end proc
# List the first 8 terms of the expansion of 1/phi
ArticoExp((sqrt(5)-1)/2, 8)
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MATHEMATICA
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Flatten[{2, 4, RecurrenceTable[{a[n] == a[n-1]^2 - 2, a[3] == -18}, a, {n, 3, 10}]}] (* Vaclav Kotesovec, Sep 20 2013 *)
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CROSSREFS
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KEYWORD
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sign,cofr
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AUTHOR
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STATUS
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approved
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