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Sum of successive remainders in computing Euclidean algorithm for (1, 1/sqrt(-n)) has real and imaginary parts equal.
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%I #11 Oct 11 2017 05:07:45

%S 1,4,5,6,8,9,16,17,18,20,24,25,36,37,38,39,40,42,48,49,64,65,66,68,72,

%T 78,80,81,100,101,102,104,105,110,117,120,121,144,145,146,147,148,150,

%U 152,155,156,164,168,169,196,197,198,200,203,210,220,222,224,225,256

%N Sum of successive remainders in computing Euclidean algorithm for (1, 1/sqrt(-n)) has real and imaginary parts equal.

%C Since the computation of the algorithm needs an extension of the integer part over a subset of C, the rule: floor(I*x) = i*floor(x) is used (which is what MuPAD does). The following program computes the exact value of the sum.

%C For all a(n) in the sequence, the relation: (2k)^2 <= a(n) <= (2k+1)^2 is true.

%e kappa(1/sqrt(-203)) = (1/2 + (1/2)i) - (1/29 + (1/29)i)*sqrt(203).

%o (MuPAD) kappa_1_over_comp_sqrt := proc(n) local a,b,i,p; begin if (a := sqrt(-n)-isqrt(-n)) = 0 then return(0) end_if: a := simplify(1/a,sqrt); i := a := simplify(1/(a - floor(a)),sqrt); p := 1; b := 0; repeat p := p*a; b := b*a+a-floor(a); until (a := simplify(1/(a-floor(a)),sqrt)) = i end_repeat: return(simplify((1-isqrt(n)/sqrt(n))*(1+b/(p-1)+1/a-floor(1/a)),sqrt)); end_proc:

%Y Cf. A086378, A087948.

%K nonn

%O 1,2

%A _Thomas Baruchel_, Sep 07 2003