%I #18 Nov 04 2018 20:34:45
%S 1,5,4,0,5,4,6,5,6,2,5,8,1,8,5,0,9,1,9,5,8,5,0,9,2,4,4,3,2,7,7,4,9,3,
%T 6,3,0,7,7,7,4,4,3,3,0,7,0,0,8,8,3,1,6,8,0,8,3,0,4,7,2,9,1,0,9,1,3,6,
%U 0,7,1,0,3,0,5,5,3,1,4,8,4,3,9,9,1,6,0,0,6,5,7,5,5,1,4,7,8,1,9,1,3,0,7,8,2,2,8,6,5,7,5,3,8,7,8,6,2,1,9,8,3,3,7,2,6,2,9,5,0,7,2,1,4,5,8
%N Decimal expansion of the constant t having the continued fraction expansion {d(n), n>=0} such that the continued fraction expansion of 4*t yields partial denominators {6*d(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="/A320411/b320411.txt">Table of n, a(n) for n = 1..4001</a>
%F Given t = [d(0); d(1), d(2), d(3), d(4), d(5), d(6), ... , A320410(n), ...], some related simple continued fractions are:
%F (1) 4*t = [6*d(0); 6*d(1), 6*d(2), 6*d(3), 6*d(4), 6*d(5), ...],
%F (2) 2*t = [3*d(0); 12*d(1), 3*d(2), 12*d(3), 3*d(4), 12*d(5), 3*d(6), ...],
%F (3) 8*t = [12*d(0); 3*d(1), 12*d(2), 3*d(3), 12*d(4), 3*d(5), 12*d(6), ...],
%F (4) 6*t = [9*d(0); 4*d(1), 9*d(2), 4*d(3), 9*d(4), 4*d(5), 9*d(6), ...],
%F (5) 8*t/3 = [4*d(0); 9*d(1), 4*d(2), 9*d(3), 4*d(4), 9*d(5), 4*d(6), ...],
%F (6) 12*t = [18*d(0); 2*d(1), 18*d(2), 2*d(3), 18*d(4), 2*d(5), 18*d(6), ...],
%F (7) 4*t/3 = [2*d(0); 18*d(1), 2*d(2), 18*d(3), 2*d(4), 18*d(5), 2*d(6), ...],
%F (8) 24*t = [36*d(0); d(1), 36*d(2), d(3), 36*d(4), d(5), 36*d(6), ...],
%F (9) 2*t/3 = [d(0); 36*d(1), d(2), 36*d(3), d(4), 36*d(5), d(6), ...].
%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, ... , A320410(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*A320410(n), ...].
%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 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 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.
%o (PARI) /* Generate over 5000 digits in the decimal expansion */
%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) ); }
%o for(n=1, 150, print1(floor(10^(n-1)*t)%10, ", "))
%Y Cf. A320410, A320833, A320831, A320953.
%K nonn,cons
%O 1,2
%A _Paul D. Hanna_, Oct 24 2018