A228931 Optimal ascending continued fraction expansion of sqrt(2)-1.

2, -6, 34, 1154, 1331714, 1773462177794, 3145168096065837266706434, 9892082352510403757550172975146702122837936996354

Offset: 1

Author

Giovanni Artico, Sep 09 2013

Keywords

• sign
• cofr
• easy

Comments

See A228929 for the definition of "optimal ascending continued fraction".

Conjecture: The terms from a(3) are all positive and can be generated by the recurrence relation a(k+1) = a(k)^2 - 2.

This relation was studied by Lucas with reference to Engel expansion.

This recurrence is not peculiar of sqrt(2) but is present in the expansion of the square root of many other numbers, starting from some term onward, but not for all numbers. Here is a list of the numbers in range 1..200 having the recurrence: 2, 3, 5, 6, 7, 8, 10, 11, 12, 13, 14, 15, 17, 18, 19, 20, 21, 22, 23, 24, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 37, 38, 39, 40, 41, 42, 45, 47, 48, 50, 51, 52, 54, 55, 56, 57, 59, 60, 62, 63, 65, 66, 68, 69, 70, 71, 72, 74, 75, 77, 78, 79, 80, 82, 83, 84, 87, 88, 90, 92, 93, 95, 96, 98, 99, 101, 102, 104, 105, 107, 110, 111, 112, 114, 117, 119, 120, 122, 123, 124, 126, 128, 130, 132, 133, 135, 136, 138, 140, 141, 142, 143, 145, 146, 147, 148, 150, 152, 155, 156, 158, 162, 164, 165, 167, 168, 170, 171, 174, 175, 178, 180, 182, 183, 185, 187, 188, 189, 192, 194, 195, 197, 198, 200

Essentially the same as A003423. - R. J. Mathar, Sep 21 2013

Examples

			sqrt(2)=1+1/2*(1-1/6*(1+1/34*(1+1/1154*(1+1/1331714*(1+1/1773462177794*(1+.....))))))
		

Links

• P. Bala, A modified Engel expansion for certain quadratic irrationals
• Giovanni Artico, Proof of the conjecture

Crossrefs

Cf. A228929, A220335.

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 sqrt(2)-1
ArticoExp(sqrt(2),8)

Mathematica

Flatten[{2, RecurrenceTable[{a[n] == a[n-1]^2 - 2, a[2] == -6}, a, {n, 2, 10}]}] (* Vaclav Kotesovec, Sep 20 2013 *)

Formula

a(n) = a(n-1)^2 - 2, for n > 2.

For n>2, a(n) = (sqrt(2)+1)^(2^(n-1)) + (sqrt(2)-1)^(2^(n-1)). - Vaclav Kotesovec, Sep 20 2013

Extensions

Added a pdf file with a proof of the conjecture by Giovanni Artico