A228933 Optimal ascending continued fraction expansion of phi-1=1/phi=(sqrt(5)-1)/2 .

2, 4, -18, 322, 103682, 10749957122, 115561578124838522882, 13354478338703157414450712387359637585922, 178342091698891843163466683840822101223162205277179656650156983624835803932590082

Offset: 1

Author

Giovanni Artico, Sep 10 2013

Keywords

• sign
• cofr

Comments

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.

Examples

			phi = 1+1/2*(1+1/4*(1-1/18*(1+1/322*(1+1/103682*(1+1/10749957122*(1+...))))))
		

Crossrefs

Cf. A001622, A081459, A094214, A228929, A228931, A228932.

Programs

Maple

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)

Mathematica

Flatten[{2, 4, RecurrenceTable[{a[n] == a[n-1]^2 - 2, a[3] == -18}, a, {n, 3, 10}]}] (* Vaclav Kotesovec, Sep 20 2013 *)

Formula

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

a(n) = 2*A081459(n-1) for n>3. - Amiram Eldar, Apr 07 2023