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%I #81 Nov 07 2018 09:33:07
%S 2,0,1,2,1,0,2,1,0,2,0,1,2,0,1,2,1,0,2,0,1,2,0,1,2,1,0,2,0,1,2,1,0,2,
%T 1,0,2,0,1,2,1,0,2,1,0,2,0,1,2,0,1,2,1,0,2,0,1,2,1,0,2,1,0,2,0,1,2,1,
%U 0,2,1,0,2,0,1,2,0,1,2,1,0,2,0,1,2,1,0,2,1,0,2,0,1,2,0,1,2,1,0,2,0,1,2,0,1,2,1,0,2,0,1,2,1,0,2,1,0,2,0,1,2,0,1,2,1,0,2,0,1,2,0,1,2,1,0,2,0,1,2,1,0,2,1,0,2,0,1,2,1,0,2,1,0,2,0,1,2,0,1,2,1,0,2,0,1,2,1,0,2,1,0,2,0,1,2,0,1,2,1,0,2,0,1,2,0,1,2,1,0,2,0,1,2,1,0,2,1,0,2,0,1
%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 3*z = [a(0) + 9; a(1) + 11, a(2) + 11, a(3) + 11, ..., a(n) + 11, ...].
%C a(n) = 2 - A321100(n) for n >= 0.
%C a(3*n+1) = A189706(n+1) for n >= 0.
%C A321091(n) = a(n) + 1 for n >= 1.
%C A321093(n) = 2*a(n) + 1 for n >= 1.
%C A321095(n) = 3*a(n) + 1 for n >= 1.
%C A321097(n) = 4*a(n) + 1 for n >= 1.
%H Paul D. Hanna, <a href="/A321090/b321090.txt">Table of n, a(n) for n = 0..10000</a>
%F CONTINUED FRACTION RELATION - this sequence {a(n), n>=0} satisfies:
%F (1) If X(k,n) = [k*a(0) + n; k*a(1) + n, k*a(2) + n, k*a(3) + n, ...],
%F then (n^2 + k*n + 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.
%F (2) If y(n) = [a(0) + n; a(1) + n, a(2) + n, a(3) + n, ...]
%F then (n^2 + n + 1)*y(n) = [a(0) + m-n-1; a(1) + m, a(2) + m, a(3) + m, ...], where m = n^3 + 3*n^2 + 5*n + 2, for n >= 0 (note special case at n=0).
%F (3) If z(n) = [n*a(0) + 1; n*a(1) + 1, n*a(2) + 1, n*a(3) + 1, ...],
%F then (n + 2)*z(n) = [n*a(0) + m-n-1; n*a(1) + m, n*a(2) + m, n*a(3) + m, ...], where m = 2*n^2 + 5*n + 4, for n >= 0.
%F FORMULA FOR TERMS: for n >= 0,
%F (1) a(3*n) = 2,
%F (2) a(3*n+2) = 1 - a(3*n+1),
%F (3) a(9*n+1) = 0,
%F (4) a(9*n+7) = 1,
%F (5) a(9*n+4) = 1 - a(3*n+1).
%e ILLUSTRATION 1 OF CONTINUED FRACTION PROPERTY.
%e Define y(n) = [a(0) + n; a(1) + n, a(2) + n, a(3) + n, ...]
%e then (n^2 + n + 1)*y(n) = [a(0) + m-n-1; a(1) + m, a(2) + m, a(3) + m, ...], where m = n^3 + 3*n^2 + 5*n + 2, for n >= 0 (note special case at n=0).
%e EXAMPLES of constants y(n) and respective continued fractions for initial n are as follows.
%e CASE n = 0.
%e y(0) = 3.43303149449604468606582652632993270180568661743523717776168...
%e Allowing for zero partial denominators in the continued fraction,
%e y(0) = [2; 0, 1, 2, 1, 0, 2, 1, 0, 2, 0, 1, 2, 0, ..., a(n) + 0, ...];
%e the simple continued fraction expansion of y(0) rewrites this as
%e y(0) = [3; 2, 3, 4, 3, 2, 4, 3, 2, 4, 2, 3, 4, 2, ..., a(n) + 2, ...].
%e CASE n = 1.
%e y(1) = 3.69674328597002790903797135061489969596768903498266397449760...
%e y(1) = [3; 1, 2, 3, 2, 1, 3, 2, 1, 3, 1, 2, 3, 1, ..., a(n) + 1, ...],
%e 3*y(1) = [11; 11, 12, 13, 12, 11, 13, 12, 11, 13, ..., a(n) + 11, ...].
%e CASE n = 2.
%e y(2) = 4.43303149449604468606582652632993270180568661743523717776168...
%e y(2) = [4; 2, 3, 4, 3, 2, 4, 3, 2, 4, 2, 3, 4, 2, ..., a(n) + 2, ...],
%e 7*y(2) = [31; 32, 33, 34, 33, 32, 34, 33, 32, 34, ..., a(n) + 32, ...].
%e CASE n = 3.
%e y(3) = 5.30877551945535750122355554493187782342126930062870320137262...
%e y(3) = [5; 3, 4, 5, 4, 3, 5, 4, 3, 5, 3, 4, 5, 3, ..., a(n) + 3, ...],
%e 13*y(3) = [69; 71, 72, 73, 72, 71, 73, 72, 71, 73, ..., a(n) + 71, ...].
%e CASE n = 4.
%e y(4) = 6.23845058448006611462883411378241316742379547477181191240204...
%e y(4) = [6; 4, 5, 6, 5, 4, 6, 5, 4, 6, 4, 5, 6, 4, ..., a(n) + 4, ...],
%e 21*y(4) = [131; 134, 135, 136, 135, 134, 136, 135, ..., a(n) + 134, ...].
%e etc.
%e ILLUSTRATION 2 OF CONTINUED FRACTION PROPERTY.
%e Define z(n) = [n*a(0) + 1; n*a(1) + 1, n*a(2) + 1, n*a(3) + 1, ...],
%e then (n + 2)*z(n) = [n*a(0) + m-n-1; n*a(1) + m, n*a(2) + m, n*a(3) + m, ...], where m = 2*n^2 + 5*n + 4, for n >= 0;
%e EXAMPLES of constants z(n) and respective continued fractions for initial n are as follows.
%e CASE n = 0.
%e z(0) = 1.6180339887498948482045868343... = (1 + sqrt(5))/2 ;
%e z(0) = [1; 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, ..., 0*a(n) + 1, ...];
%e 2*z(0) = [3; 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, ..., 0*a(n) + 4, ...].
%e CASE n = 1.
%e z(1) = 3.696743285970027909037971350614899695967689034... (A321092 - 1)
%e z(1) = [3; 1, 2, 3, 2, 1, 3, 2, 1, 3, 1, 2, 3, 1, ..., 1*a(n) + 1, ...];
%e 3*z(1) = [11; 11, 12, 13, 12, 11, 13, 12, 11, 13, ..., 1*a(n) + 11, ...].
%e CASE n = 2.
%e z(2) = 5.761342189260052577009778722181649926352515102... (A321094 - 1)
%e z(2) = [5; 1, 3, 5, 3, 1, 5, 3, 1, 5, 1, 3, 5, 1, ..., 2*a(n) + 1, ...];
%e 4*z(2) = [23; 22, 24, 26, 24, 22, 26, 24, 22, 26, ..., 2*a(n) + 22, ...].
%e CASE n = 3.
%e z(3) = 7.805401757688663373476939995437639876411562871... (A321096 - 1)
%e z(3) = [7; 1, 4, 7, 4, 1, 7, 4, 1, 7, 1, 4, 7, 1, ..., 3*a(n) + 1, ...];
%e 5*z(3) = [39; 37, 40, 43, 40, 37, 43, 40, 37, 43, ..., 3*a(n) + 37, ...].
%e CASE n = 4.
%e z(4) = 9.836308638532504943187035876427127876597685953... (A321098 - 1)
%e z(4) = [9; 1, 5, 9, 5, 1, 9, 5, 1, 9, 1, 5, 9, 1, ..., 4*a(n) + 1, ...];
%e 6*z(4) = [59; 56, 60, 64, 60, 56, 64, 60, 56, 64, ..., 4*a(n) + 56, ...].
%e etc.
%e EXTENDED TERMS.
%e The initial 1020 terms of this sequence are as follows.
%e A = [2,0,1,2,1,0,2,1,0,2,0,1,2,0,1,2,1,0,2,0,1,2,0,1,2,1,0,2,0,1,
%e 2,1,0,2,1,0,2,0,1,2,1,0,2,1,0,2,0,1,2,0,1,2,1,0,2,0,1,2,1,0,
%e 2,1,0,2,0,1,2,1,0,2,1,0,2,0,1,2,0,1,2,1,0,2,0,1,2,1,0,2,1,0,
%e 2,0,1,2,0,1,2,1,0,2,0,1,2,0,1,2,1,0,2,0,1,2,1,0,2,1,0,2,0,1,
%e 2,0,1,2,1,0,2,0,1,2,0,1,2,1,0,2,0,1,2,1,0,2,1,0,2,0,1,2,1,0,
%e 2,1,0,2,0,1,2,0,1,2,1,0,2,0,1,2,1,0,2,1,0,2,0,1,2,0,1,2,1,0,
%e 2,0,1,2,0,1,2,1,0,2,0,1,2,1,0,2,1,0,2,0,1,2,0,1,2,1,0,2,0,1,
%e 2,0,1,2,1,0,2,0,1,2,1,0,2,1,0,2,0,1,2,1,0,2,1,0,2,0,1,2,0,1,
%e 2,1,0,2,0,1,2,1,0,2,1,0,2,0,1,2,0,1,2,1,0,2,0,1,2,0,1,2,1,0,
%e 2,0,1,2,1,0,2,1,0,2,0,1,2,1,0,2,1,0,2,0,1,2,0,1,2,1,0,2,0,1,
%e 2,1,0,2,1,0,2,0,1,2,1,0,2,1,0,2,0,1,2,0,1,2,1,0,2,0,1,2,1,0,
%e 2,1,0,2,0,1,2,0,1,2,1,0,2,0,1,2,0,1,2,1,0,2,0,1,2,1,0,2,1,0,
%e 2,0,1,2,1,0,2,1,0,2,0,1,2,0,1,2,1,0,2,0,1,2,1,0,2,1,0,2,0,1,
%e 2,1,0,2,1,0,2,0,1,2,0,1,2,1,0,2,0,1,2,1,0,2,1,0,2,0,1,2,0,1,
%e 2,1,0,2,0,1,2,0,1,2,1,0,2,0,1,2,1,0,2,1,0,2,0,1,2,0,1,2,1,0,
%e 2,0,1,2,0,1,2,1,0,2,0,1,2,1,0,2,1,0,2,0,1,2,1,0,2,1,0,2,0,1,
%e 2,0,1,2,1,0,2,0,1,2,1,0,2,1,0,2,0,1,2,0,1,2,1,0,2,0,1,2,0,1,
%e 2,1,0,2,0,1,2,1,0,2,1,0,2,0,1,2,1,0,2,1,0,2,0,1,2,0,1,2,1,0,
%e 2,0,1,2,1,0,2,1,0,2,0,1,2,1,0,2,1,0,2,0,1,2,0,1,2,1,0,2,0,1,
%e 2,1,0,2,1,0,2,0,1,2,0,1,2,1,0,2,0,1,2,0,1,2,1,0,2,0,1,2,1,0,
%e 2,1,0,2,0,1,2,1,0,2,1,0,2,0,1,2,0,1,2,1,0,2,0,1,2,1,0,2,1,0,
%e 2,0,1,2,1,0,2,1,0,2,0,1,2,0,1,2,1,0,2,0,1,2,1,0,2,1,0,2,0,1,
%e 2,0,1,2,1,0,2,0,1,2,0,1,2,1,0,2,0,1,2,1,0,2,1,0,2,0,1,2,0,1,
%e 2,1,0,2,0,1,2,0,1,2,1,0,2,0,1,2,1,0,2,1,0,2,0,1,2,1,0,2,1,0,
%e 2,0,1,2,0,1,2,1,0,2,0,1,2,1,0,2,1,0,2,0,1,2,0,1,2,1,0,2,0,1,
%e 2,0,1,2,1,0,2,0,1,2,1,0,2,1,0,2,0,1,2,1,0,2,1,0,2,0,1,2,0,1,
%e 2,1,0,2,0,1,2,1,0,2,1,0,2,0,1,2,1,0,2,1,0,2,0,1,2,0,1,2,1,0,
%e 2,0,1,2,1,0,2,1,0,2,0,1,2,0,1,2,1,0,2,0,1,2,0,1,2,1,0,2,0,1,
%e 2,1,0,2,1,0,2,0,1,2,0,1,2,1,0,2,0,1,2,0,1,2,1,0,2,0,1,2,1,0,
%e 2,1,0,2,0,1,2,1,0,2,1,0,2,0,1,2,0,1,2,1,0,2,0,1,2,1,0,2,1,0,
%e 2,0,1,2,0,1,2,1,0,2,0,1,2,0,1,2,1,0,2,0,1,2,1,0,2,1,0,2,0,1,
%e 2,0,1,2,1,0,2,0,1,2,0,1,2,1,0,2,0,1,2,1,0,2,1,0,2,0,1,2,1,0,
%e 2,1,0,2,0,1,2,0,1,2,1,0,2,0,1,2,1,0,2,1,0,2,0,1,2,0,1,2,1,0,
%e 2,0,1,2,0,1,2,1,0,2,0,1,2,1,0,2,1,0,2,0,1,2,1,0,2,1,0,2,0,1, ...].
%e ...
%o (PARI) /* Generate over 5000 terms */
%o {CF=[4]; for(i=1,8, M = contfracpnqn( CF + vector(#CF,i,10) ); z = (1/3)*M[1,1]/M[2,1]; CF = contfrac(z) )}
%o for(n=0,200,print1(CF[n+1]-1-0^n,", "))
%o (PARI) /* Using formula for individual terms */
%o {a(n) = if(n%3==0, 2,
%o if(n%3==2, 1 - a(n-1),
%o if(n%9==1, 0,
%o if(n%9==7, 1,
%o if(n%9==4, 1 - a((n-1)/3) )))))}
%o for(n=0,200,print1(a(n),","))
%Y Cf. A321100, A321091, A321092, A321093, A321094, A321095, A321096, A321097, A321098.
%K nonn
%O 0,1
%A _Paul D. Hanna_, Oct 27 2018
%E Revised example and comments sections. - _Paul D. Hanna_, Nov 03 2018
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