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%I #11 Nov 05 2018 21:17:38
%S 0,2,1,0,1,2,0,1,2,0,2,1,0,2,1,0,1,2,0,2,1,0,2,1,0,1,2,0,2,1,0,1,2,0,
%T 1,2,0,2,1,0,1,2,0,1,2,0,2,1,0,2,1,0,1,2,0,2,1,0,1,2,0,1,2,0,2,1,0,1,
%U 2,0,1,2,0,2,1,0,2,1,0,1,2,0,2,1,0,1,2,0,1,2,0,2,1,0,2,1,0,1,2,0,2,1,0,2,1,0,1,2,0,2,1,0,1,2,0,1,2,0,2,1,0,2,1,0,1,2,0,2,1,0,2,1,0,1,2,0,2,1,0,1,2,0,1,2,0,2,1,0,1,2,0,1,2,0,2,1,0,2,1,0,1,2,0,2,1,0,1,2,0,1,2,0,2,1,0,2,1,0,1,2,0,2,1,0,2,1,0,1,2,0,2,1,0,1,2,0,1,2,0,2,1
%N Sequence {a(n), n>=0} satisfying the continued fraction relation: if z = [a(0) + 1; a(1) + 1, a(2) + 1, a(3) + 1, ..., a(n) + 1, ...], then 7*z = [a(0) + 9; a(1) + 11, a(2) + 11, a(3) + 11, ..., a(n) + 11, ...].
%C a(n) = 2 - A321090(n), for n >= 0.
%F CONTINUED FRACTION RELATION - this sequence {a(n), n>=0} satisfies:
%F If y(k,n) = [k*a(0) + n; k*a(1) + n, k*a(2) + n, k*a(3) + n, ...],
%F then (n^2 + 3*k*n + 2*k^2 + 1)*y(k,n) = [k*a(0) + m-k-n; k*a(1) + m, k*a(2) + m, k*a(3) + m, ...], where m = n^3 + 3*k*n^2 + (2*k^2 + 3)*n + 2*k, for n >= 0, k >= 0.
%F FORMULA FOR TERMS: for n >= 0,
%F (1) a(3*n) = 0,
%F (2) a(3*n+2) = 3 - a(3*n+1),
%F (3) a(9*n+1) = 2,
%F (4) a(9*n+7) = 1,
%F (5) a(9*n+4) = 3 - a(3*n+1).
%e ILLUSTRATION OF CONTINUED FRACTION PROPERTY.
%e Define y(k,n) = [k*a(0) + n; k*a(1) + n, k*a(2) + n, k*a(3) + n, ...],
%e then
%e (n^2 + 3*k*n + 2*k^2 + 1) * X(k,n) = [k*a(0) + m-k-n; k*a(1) + m, k*a(2) + m, k*a(3) + m, ...], where m = n^3 + 3*k*n^2 + (2*k^2 + 3)*n + 2*k, for n >= 0, k >= 0.
%e EXAMPLES of constants y(k,n) and respective continued fractions for initial k and n are as follows.
%e CASE k = 1, n = 1.
%e y(1,1) = 1.29663382206594201985347001536274116440601452468308746847...
%e y(1,1) = [1; 3, 2, 1, 2, 3, 1, 2, 3, 1, 3, 2, 1, 3, ..., a(n) + 1, ...],
%e 7*y(1,1) = [9; 13, 12, 11, 12, 13, 11, 12, 13, 11, ..., a(n) + 11, ...].
%e CASE k = 1, n = 2.
%e y(1,2) = 2.23302966146823013079630091558411348943843779308298734989...
%e y(1,2) = [2; 4, 3, 2, 3, 4, 2, 3, 4, 2, 4, 3, 2, 4, ..., a(n) + 2, ...].
%e 13*y(1,2) = [29; 34, 33, 32, 33, 34, 32, 33, 34, 32, ..., a(n) + 32, ...].
%e CASE k = 1, n = 3.
%e y(1,3) = 3.19112838213609195362054670452820227524052071087999217614...
%e y(1,3) = [3; 5, 4, 3, 4, 5, 3, 4, 5, 3, 5, 4, 3, 5, ..., a(n) + 3, ...].
%e 21*y(1,3) = [67; 73, 72, 71, 72, 73, 71, 72, 73, 71, ..., a(n) + 71, ...].
%e CASE k = 2, n = 1.
%e y(2,1) = 1.18990000724532672619738638935609891545233786934727750160...
%e y(2,1) = [1; 5, 3, 1, 3, 5, 1, 3, 5, 1, 5, 3, 1, 5, ..., 2*a(n) + 1, ...].
%e 16*y(2,1) = [19; 26, 24, 22, 24, 26, 22, 24, 26, 22, ..., 2*a(n) + 22, ...].
%e CASE k = 3, n = 1.
%e y(3,1) = 1.13873249345174370130452490021023011324120719384639850933...
%e y(3,1) = [1; 7, 4, 1, 4, 7, 1, 4, 7, 1, 7, 4, 1, 7, ..., 3*a(n) + 1, ...].
%e 29*y(3,1) = [33; 43, 40, 37, 40, 43, 37, 40, 43, 37, ..., 3*a(n) + 37, ...].
%e CASE k = 3, n = 2.
%e y(3,2) = 2.12220007282539436078116266359839811814939013508321100093...
%e y(3,2) = [2; 8, 5, 2, 5, 8, 2, 5, 8, 2, 8, 5, 2, 8, ..., 3*a(n) + 2, ...].
%e 41*y(3,2) = [87; 98, 95, 92, 95, 98, 92, 95, 98, 92, ..., 3*a(n) + 92, ...].
%o (PARI) /* Generate over 5000 terms */
%o {CF=[1]; for(i=1, 8, M = contfracpnqn( CF + vector(#CF, i, if(i==1,8,10)) ); z = (1/7)*M[1, 1]/M[2, 1]; CF = contfrac(z) )}
%o for(n=0, 200, print1(CF[n+1] - 1, ", "))
%o (PARI) /* Using formula for terms */
%o {a(n) = if(n%3==0, 0,
%o if(n%3==2, 3 - a(n-1),
%o if(n%9==1, 2,
%o if(n%9==7, 1,
%o if(n%9==4, 3 - a((n-1)/3) )))))}
%o for(n=0, 200, print1(a(n), ", "))
%Y Cf. A321090.
%K nonn,cofr
%O 0,2
%A _Paul D. Hanna_, Nov 03 2018
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