%I #22 Oct 26 2018 21:39:52

%S 1,1,1,5,1,1,1,120,1,1,1,5,1,1,1,2880,1,1,1,5,1,1,1,120,1,1,1,5,1,1,1,

%T 69120,1,1,1,5,1,1,1,120,1,1,1,5,1,1,1,2880,1,1,1,5,1,1,1,120,1,1,1,5,

%U 1,1,1,1658880,1,1,1,5,1,1,1,120,1,1,1,5,1,1,1,2880,1,1,1,5,1,1,1,120,1,1,1,5,1,1,1,69120,1,1,1,5,1,1,1,120,1,1,1,5,1,1,1,2880,1,1,1,5,1,1,1,120,1,1,1,5,1,1,1,39813120,1,1,1,5,1,1,1,120,1,1,1,5,1,1,1,2880,1

%N Continued fraction of a constant t with partial denominators {a(n), n>=0} such that the continued fraction of 4*t yields partial denominators {6*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="/A320410/b320410.txt">Table of n, a(n) for n = 0..4000</a>

%F G.f.: 1/(1-x) + 4*x^3/(1-x^4) + Sum_{n>=3} 115*24^(n-3)*x^(2^n-1)/(1 - x^(2^n)).

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

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

%F (3) 8*t = [12*a(0); 3*a(1), 12*a(2), 3*a(3), 12*a(4), 3*a(5), 12*a(6), ...],

%F (4) 6*t = [9*a(0); 4*a(1), 9*a(2), 4*a(3), 9*a(4), 4*a(5), 9*a(6), ...],

%F (5) 8*t/3 = [4*a(0); 9*a(1), 4*a(2), 9*a(3), 4*a(4), 9*a(5), 4*a(6), ...],

%F (6) 12*t = [18*a(0); 2*a(1), 18*a(2), 2*a(3), 18*a(4), 2*a(5), 18*a(6), ...],

%F (7) 4*t/3 = [2*a(0); 18*a(1), 2*a(2), 18*a(3), 2*a(4), 18*a(5), 2*a(6), ...],

%F (8) 24*t = [36*a(0); a(1), 36*a(2), a(3), 36*a(4), a(5), 36*a(6), ...],

%F (9) 2*t/3 = [a(0); 36*a(1), a(2), 36*a(3), a(4), 36*a(5), a(6), ...].

%F Conjecture: If for a positive integer k, n = 2^(k+2) mod 2^(k+3) then a(n) = 5*24^k else a(n) = 1. - _Tristan Cam_, Oct 25 2018

%e The decimal expansion of this constant t begins:

%e t = 1.540546562581850919585092443277493630777443307008831680830472910...

%e The simple continued fraction expansion of t begins:

%e t = [1; 1, 1, 5, 1, 1, 1, 120, 1, 1, 1, 5, 1, 1, 1, 2880, 1, 1, 1, 5, 1, 1, 1, 120, 1, 1, 1, 5, 1, 1, 1, 69120, 1, 1, 1, 5, ... , a(n), ...]

%e such that the simple continued fraction expansion of 4*t begins:

%e 4*t = [6; 6, 6, 30, 6, 6, 6, 720, 6, 6, 6, 30, 6, 6, 6, 17280, 6, 6, 6, 30, 6, 6, 6, 720, 6, 6, 6, 30, 6, 6, 6, 414720, 6, 6, 6, 30, ... , 6*a(n), ...].

%e The initial 528 terms in the continued fraction expansion of t are

%e t = [1;1,1,5,1,1,1,120,1,1,1,5,1,1,1,2880,1,1,1,5,1,1,1,120,

%e 1,1,1,5,1,1,1,69120,1,1,1,5,1,1,1,120,1,1,1,5,1,1,1,2880,

%e 1,1,1,5,1,1,1,120,1,1,1,5,1,1,1,1658880,1,1,1,5,1,1,1,120,

%e 1,1,1,5,1,1,1,2880,1,1,1,5,1,1,1,120,1,1,1,5,1,1,1,69120,

%e 1,1,1,5,1,1,1,120,1,1,1,5,1,1,1,2880,1,1,1,5,1,1,1,120,

%e 1,1,1,5,1,1,1,39813120,1,1,1,5,1,1,1,120,1,1,1,5,1,1,1,2880,

%e 1,1,1,5,1,1,1,120,1,1,1,5,1,1,1,69120,1,1,1,5,1,1,1,120,

%e 1,1,1,5,1,1,1,2880,1,1,1,5,1,1,1,120,1,1,1,5,1,1,1,1658880,

%e 1,1,1,5,1,1,1,120,1,1,1,5,1,1,1,2880,1,1,1,5,1,1,1,120,

%e 1,1,1,5,1,1,1,69120,1,1,1,5,1,1,1,120,1,1,1,5,1,1,1,2880,

%e 1,1,1,5,1,1,1,120,1,1,1,5,1,1,1,955514880,1,1,1,5,1,1,1,120,

%e 1,1,1,5,1,1,1,2880,1,1,1,5,1,1,1,120,1,1,1,5,1,1,1,69120,

%e 1,1,1,5,1,1,1,120,1,1,1,5,1,1,1,2880,1,1,1,5,1,1,1,120,

%e 1,1,1,5,1,1,1,1658880,1,1,1,5,1,1,1,120,1,1,1,5,1,1,1,2880,

%e 1,1,1,5,1,1,1,120,1,1,1,5,1,1,1,69120,1,1,1,5,1,1,1,120,

%e 1,1,1,5,1,1,1,2880,1,1,1,5,1,1,1,120,1,1,1,5,1,1,1,39813120,

%e 1,1,1,5,1,1,1,120,1,1,1,5,1,1,1,2880,1,1,1,5,1,1,1,120,

%e 1,1,1,5,1,1,1,69120,1,1,1,5,1,1,1,120,1,1,1,5,1,1,1,2880,

%e 1,1,1,5,1,1,1,120,1,1,1,5,1,1,1,1658880,1,1,1,5,1,1,1,120,

%e 1,1,1,5,1,1,1,2880,1,1,1,5,1,1,1,120,1,1,1,5,1,1,1,69120,

%e 1,1,1,5,1,1,1,120,1,1,1,5,1,1,1,2880,1,1,1,5,1,1,1,120,

%e 1,1,1,5,1,1,1,22932357120,1,1,1,5,1,1,1,120,1,1,1,5,1,1,1,2880, ...].

%e ...

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

%e t = 1.54054656258185091958509244327749363077744330700883\

%e 16808304729109136071030553148439916006575514781913\

%e 07822865753878621983372629507214585549765601216855\

%e 82787499037533808802512182117029618023317430110304\

%e 68568548056881176292575460160397807605484819864109\

%e 97669631806987876538679522272871069799153383954670\

%e 27608251838807810531236786249017655750340240450431\

%e 39500820756134042709955133598762098341500330573137\

%e 99939807838257483585300963279557471021988938954240\

%e 72201092140208383999296983006564129234149599645137\

%e 97256177580945556498719774507283674158287467111752\

%e 77844466938413199720373428083674374904367749851098\

%e 21424896071080046214454673955084637450020810561194\

%e 71345165789766421139563276393063288950816919055445\

%e 03342967346596573587017045564281666931351950277058\

%e 62590013151270263215075119134200880720738116296046\

%e 68850174252499836432241162269740091282255326351721\

%e 57235179003932955261950046664886398112095558223840\

%e 82192509104134957717517226738372957214588632890700\

%e 21161549012492823004735717009953284171956638331978...

%e ...

%e GENERATING METHOD.

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

%e t = (1/4)*contfracpnqn(6*CF)[1,1]/contfracpnqn(6*CF)[2,1]; CF = contfrac(t)

%e This method can be illustrated as follows.

%e t0 = [1] = 1 ;

%e t1 = (1/4)*[6] = [1; 2] = 3/2 ;

%e t2 = (1/4)*[6; 12] = [1; 1, 1, 11, 2] = 73/48 ;

%e t3 = (1/4)*[6; 6, 6, 66, 12] = [1; 1, 1, 5, 1, 1, 1, 264, 3] = 45312/29413 ;

%e t4 = (1/4)*[6; 6, 6, 30, 6, 6, 6, 1584, 18] = [1; 1, 1, 5, 1, 1, 1, 120, 1, 1, 1, 5, 1, 1, 1, 6336, 4, 2] = 22476134901/14589714746 ;

%e t5 = (1/4)*[6; 6, 6, 30, 6, 6, 6, 720, 6, 6, 6, 30, 6, 6, 6, 38016, 24, 12] = [1; 1, 1, 5, 1, 1, 1, 120, 1, 1, 1, 5, 1, 1, 1, 2880, 1, 1, 1, 5, 1, 1, 1, 120, 1, 1, 1, 5, 1, 1, 1, 152064, 6, 48] = 19666327045008787868161/12765811513057655118144 ; ...

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

%e The number of terms generated in each iteration of the above method begins:

%e [1, 2, 5, 9, 18, 34, 66, 133, 265, 530, 1061, 2122, 4242, 8485, 16969, 33938, ...].

%e ...

%e GENERATING FUNCTION.

%e The terms of this continued fraction may be expressed by the generating function

%e A(x) = 1 + x + x^2 + 5*x^3 + x^4 + x^5 + x^6 + 120*x^7 + x^8 + x^9 + x^10 + 5*x^11 + x^12 + x^13 + x^14 + 2880*x^15 + ...

%e where A(x) = 1/(1-x) + 4*x^3/(1-x^4) + 115*x^7/(1-x^8) + 2760*x^15/(1-x^16) + 66240*x^31/(1-x^32) + 1589760*x^63/(1-x^64) + 38154240*x^127/(1-x^128) + ... + 115*24^(n-3)*x^(2^n-1)/(1 - x^(2^n)) + ...

%e ...

%o (PARI) /* Generate over 4200 terms of the continued fraction */

%o CF=[1];

%o {for(i=1, 12, t = (1/4)*contfracpnqn(6*CF)[1, 1]/contfracpnqn(6*CF)[2, 1];

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

%Y Cf. A320411, A320832, A320834, A320952.

%K nonn

%O 0,4

%A _Paul D. Hanna_, Oct 24 2018