A320834 - OEIS

%I #29 Oct 26 2018 21:40:48

%S 1,2,1,2,12,2,1,2,15,1,2,2,2,2,2,2,2,1,19,2,2,2,2,2,1,2,24,2,1,2,2,2,

%T 1,2,24,1,2,1,24,1,2,2,2,1,2,24,2,1,2,2,2,2,2,2,2,1,31,2,1,2,12,2,1,2,

%U 2,2,1,2,12,2,1,2,31,1,2,1,24,1,2,1,31,2,2,2,2,2,1,2,24,1,2,1,2,288,2,1,2,1,24,2,1,2,2,2,1,2,24,2,1,2,2,2,2,2,41,24,1,2,1,2,144,2,1,2,1,24,2,1,2,2,2,2,2,2,2,1,15,2,1,2,12,2,1,2,41,12,2,1,2,1,288

%N Continued fraction of a constant t with partial denominators {a(n), n>=0} such that the continued fraction of 3*t yields partial denominators {4*a(n), n>=0}.

%C Is this constant transcendental?

%C Compare to the continued fraction expansions of sqrt(3) and 3*sqrt(3), which are related by a factor of 5: sqrt(3) = [1; 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, ...] and 3*sqrt(3) = [5; 5, 10, 5, 10, 5, 10, 5, 10, 5, 10, ...].

%C Further, let CF(x) denote the simple continued fraction expansion of x, then we have the related identities which hold for n >= 1:

%C (C1) CF( (4*n+1) * sqrt((n+1)/n) ) = (4*n+3) * CF( sqrt((n+1)/n) ),

%C (C2) CF( (2*n+1) * sqrt((n+2)/n) ) = (2*n+3) * CF( sqrt((n+2)/n) ).

%H Paul D. Hanna, <a href="/A320834/b320834.txt">Table of n, a(n) for n = 0..4000</a>

%F Given t = [a(0); a(1), a(2), a(3), a(4), a(5), a(6), ...], some related simple continued fractions are:

%F (1) 3*t = [4*a(0); 4*a(1), 4*a(2), 4*a(3), 4*a(4), 4*a(5), ...],

%F (2) 3*t/4 = [a(0); 16*a(1), a(2), 16*a(3), a(4), 16*a(5), a(6), ...],

%F (3) 12*t = [16*a(0); a(1), 16*a(2), a(3), 16*a(4), a(5), 16*a(6), ...],

%F (4) 3*t/2 = [2*a(0); 8*a(1), 2*a(2), 8*a(3), 2*a(4), 8*a(5), 2*a(6), ...],

%F (5) 6*t = [8*a(0); 2*a(1), 8*a(2), 2*a(3), 8*a(4), 2*a(5), 8*a(6), ...].

%e The decimal expansion of this constant t begins:

%e t = 1.373774416411892779838549506639605383800713002607842189327990402...

%e The simple continued fraction expansion of t begins:

%e t = [1; 2, 1, 2, 12, 2, 1, 2, 15, 1, 2, 2, 2, 2, 2, 2, 2, 1, 19, 2, 2, 2, 2, 2, 1, 2, 24, 2, 1, 2, 2, 2, 1, 2, 24, ... , a(n), ...].

%e such that the simple continued fraction expansion of 3*t begins:

%e 3*t = [4; 8, 4, 8, 48, 8, 4, 8, 60, 4, 8, 8, 8, 8, 8, 8, 8, 4, 76, 8, 8, 8, 8, 8, 4, 8, 96, 8, 4, 8, 8, 8, 4, 8, 96, ... , 4*a(n), ...].

%e The initial 1000 terms in the continued fraction expansion of t are

%e t = [1;2,1,2,12,2,1,2,15,1,2,2,2,2,2,2,2,1,19,2,2,2,2,2,1,2,

%e 24,2,1,2,2,2,1,2,24,1,2,1,24,1,2,2,2,1,2,24,2,1,2,2,2,

%e 2,2,2,2,1,31,2,1,2,12,2,1,2,2,2,1,2,12,2,1,2,31,1,2,1,

%e 24,1,2,1,31,2,2,2,2,2,1,2,24,1,2,1,2,288,2,1,2,1,24,2,1,

%e 2,2,2,1,2,24,2,1,2,2,2,2,2,41,24,1,2,1,2,144,2,1,2,1,24,

%e 2,1,2,2,2,2,2,2,2,1,15,2,1,2,12,2,1,2,41,12,2,1,2,1,288,

%e 1,2,1,2,12,41,2,1,2,24,2,1,2,2,2,1,2,12,2,1,2,31,1,2,1,

%e 24,1,2,1,2,3456,2,1,2,1,24,1,2,1,31,2,1,2,12,2,1,2,2,2,1,

%e 2,12,2,1,2,31,1,2,2,2,2,2,2,2,1,2,24,2,1,2,2,2,1,54,288,

%e 1,2,1,2,12,2,1,2,191,1,2,2,2,2,2,2,2,2,2,31,1,2,2,2,2,

%e 2,2,2,1,2,24,2,1,2,2,2,1,2,24,1,2,1,19,2,1,2,12,2,1,2,

%e 15,1,2,2,2,2,2,2,2,1,54,144,2,1,2,1,24,1,2,1,383,2,2,2,2,

%e 2,2,2,2,2,1,15,2,1,54,24,1,2,1,2,288,2,1,2,1,24,2,1,2,2,

%e 2,2,2,2,2,1,15,2,1,2,12,2,1,2,41,12,2,1,2,1,288,1,2,1,2,

%e 12,2,1,2,4607,1,2,2,2,2,2,2,2,2,2,31,1,2,1,24,1,2,1,40,1,

%e 2,2,2,2,2,2,2,1,15,2,1,2,12,2,1,2,2,2,1,2,12,2,1,2,15,

%e 1,2,2,2,2,2,2,2,1,40,1,2,1,24,2,1,2,2,2,1,2,24,2,1,2,

%e 2,2,2,2,2,2,1,31,2,1,2,12,2,1,2,2,2,1,2,12,72,3456,1,2,1,

%e 2,12,2,1,2,15,1,2,2,2,2,2,2,2,1,254,12,2,1,2,2,2,1,2,24,

%e 2,1,2,2,2,1,2,24,2,1,2,41,12,2,1,2,2,2,1,2,24,2,1,2,2,

%e 2,1,2,12,2,1,2,31,1,2,2,2,2,2,2,2,1,2,24,1,2,1,2,288,1,

%e 2,1,2,12,25,2,1,2,12,2,1,2,15,1,2,2,2,2,2,2,2,1,19,2,2,

%e 2,2,2,1,2,24,2,1,2,2,2,1,2,24,1,2,1,71,2,1,191,2,1,2,12,

%e 2,1,2,1,288,1,2,1,2,12,510,1,2,2,2,1,2,24,2,1,2,2,2,1,2,

%e 24,2,1,2,2,2,2,2,19,1,2,2,2,2,2,71,1,2,31,1,2,1,24,1,2,

%e 1,2,3456,2,1,2,1,24,1,2,1,31,2,1,2,12,2,1,2,2,2,1,2,24,2,

%e 1,2,2,2,1,2,12,20,24,1,2,1,2,144,2,1,2,1,24,54,1,2,15,1,2,

%e 2,2,2,2,2,2,2,2,383,1,2,1,24,1,2,1,2,144,2,1,2,1,24,6142,1,

%e 2,1,24,2,1,2,2,2,1,2,24,2,1,2,2,2,1,2,24,41,2,2,2,2,2,

%e 2,2,31,1,2,1,24,1,2,1,52,1,2,1,24,2,1,2,2,2,1,2,24,2,1,

%e 2,2,2,2,2,19,1,2,2,2,2,2,2,2,1,15,2,1,2,12,2,1,2,2,2,

%e 1,2,12,2,1,2,15,1,2,2,2,2,2,2,2,1,19,2,2,2,2,2,1,2,24,

%e 2,1,2,2,2,1,2,24,1,2,1,52,1,2,1,24,1,2,1,31,2,1,2,12,2,

%e 1,2,2,2,1,2,12,2,1,2,31,1,2,2,2,2,2,2,2,1,2,24,2,1,2,

%e 2,2,1,2,24,1,2,1,40,1,2,2,2,2,2,2,2,1,15,2,1,2,12,2,1,

%e 2,2,2,1,2,12,2,1,2,15,1,2,95,1,2,4607,1,2,1,24,1,2,1,2,144,

%e 2,1,2,1,24,20,12,2,1,2,2,2,1,2,24,2,1,2,2,2,1,2,12,338,1,

%e 2,15,1,2,2,2,2,2,2,2,1,2,24,1,2,1,2,288,2,1,2,1,24,2,1,

%e 2,2,2,2,2,2,2,1,31,2,1,2,12,2,1,2,54,2,1,15,2,1,2,12,2,

%e 1,2,2,2,1,2,12,2,1,2,31,1,2,2,2,2,2,2,2,1,2,24,1,2, ...].

%e ...

%e The initial 1000 digits in the decimal expansion of t are

%e t = 1.37377441641189277983854950663960538380071300260784\

%e 21893279904024826733155241042545366869230907298089\

%e 13984845658100424411419330720447452186846496871309\

%e 93611537665995205176080835177406950895125756891119\

%e 80915134888807230285924622832070894390440105854297\

%e 09681249062135643627285425138701026540290422483564\

%e 88492730997578199229485083989821808079068883171984\

%e 07031141426574548209572058371283039520211641927528\

%e 87513532330143554990150517422840360714935764293437\

%e 57458872431528639369813728159265563110330718628799\

%e 96977704230617847144362021013013790221614910568231\

%e 27450894001787415768703790366780081874847582449346\

%e 46012195526919702308642260551245557225670489756926\

%e 10073356625047859682542328603265305024792731699661\

%e 07707346790590697402355533935869626197258295389474\

%e 34717509042975989479155252899700525467717059228960\

%e 29994752938551798255237161498633969001654212726415\

%e 44019114620209689305131806997427047626502096289167\

%e 37623579663973917818431418184110200421986541799594\

%e 29845246183090799528184841814574898135012164288045...

%e GENERATING METHOD.

%e Start with CF = [1] and repeat (PARI code):

%e t = (1/3)*contfracpnqn(4*CF)[1,1]/contfracpnqn(4*CF)[2,1]; CF = contfrac(t)

%e This method can be illustrated as follows.

%e t0 = [1] = 1 ;

%e t1 = (1/3)*[4] = [1; 3] = 4/3 ;

%e t2 = (1/3)*[4; 12] = [1; 2, 1, 3, 3] = 49/36 ;

%e t3 = (1/3)*[4; 8, 4, 12, 12] = [1; 2, 1, 2, 12, 4, 36] = 20116/14643 ;

%e t4 = (1/3)*[4; 8, 4, 8, 48, 16, 144] = [1; 2, 1, 2, 12, 2, 1, 2, 15, 1, 2, 5, 432] = 124502344/90627939;

%e t5 = (1/3)*[4; 8, 4, 8, 48, 8, 4, 8, 60, 4, 8, 20, 1728] = [1; 2, 1, 2, 12, 2, 1, 2, 15, 1, 2, 2, 2, 2, 2, 2, 2, 1, 19, 2, 2, 2, 2, 2, 1, 6, 5184] = 1018841077754176/741636374635107 ; ...

%e where this constant t equals the limit of the iterations of the above process.

%e The above method generates terms with exponential growth; the number of terms with each iteration begins:

%e [1, 2, 5, 7, 13, 27, 59, 119, 231, 463, 913, 1817, 3565, 7033, 13989, 27658, ...].

%o (PARI) /* Generate over 3500 terms in the continued fraction */

%o CF=[1];

%o {for(i=1, 12, t = (1/3)*contfracpnqn(4*CF)[1, 1]/contfracpnqn(4*CF)[2, 1];

%o CF = contfrac(t) ); CF}

%Y Cf. A320831, A320832, A320833, A320410, A320952.

%K nonn,cofr

%O 0,2

%A _Paul D. Hanna_, Oct 23 2018