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A320411 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}. 5
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, 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, 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 (list; constant; graph; refs; listen; history; text; internal format)
OFFSET
1,2
COMMENTS
Is this constant transcendental?
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, ...].
Further, let CF(x) denote the simple continued fraction expansion of x, then we have the related identities which hold for n >= 1:
(C1) CF( (4*n+1) * sqrt((n+1)/n) ) = (4*n+3) * CF( sqrt((n+1)/n) ),
(C2) CF( (2*n+1) * sqrt((n+2)/n) ) = (2*n+3) * CF( sqrt((n+2)/n) )
LINKS
FORMULA
Given t = [d(0); d(1), d(2), d(3), d(4), d(5), d(6), ... , A320410(n), ...], some related simple continued fractions are:
(1) 4*t = [6*d(0); 6*d(1), 6*d(2), 6*d(3), 6*d(4), 6*d(5), ...],
(2) 2*t = [3*d(0); 12*d(1), 3*d(2), 12*d(3), 3*d(4), 12*d(5), 3*d(6), ...],
(3) 8*t = [12*d(0); 3*d(1), 12*d(2), 3*d(3), 12*d(4), 3*d(5), 12*d(6), ...],
(4) 6*t = [9*d(0); 4*d(1), 9*d(2), 4*d(3), 9*d(4), 4*d(5), 9*d(6), ...],
(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), ...],
(6) 12*t = [18*d(0); 2*d(1), 18*d(2), 2*d(3), 18*d(4), 2*d(5), 18*d(6), ...],
(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), ...],
(8) 24*t = [36*d(0); d(1), 36*d(2), d(3), 36*d(4), d(5), 36*d(6), ...],
(9) 2*t/3 = [d(0); 36*d(1), d(2), 36*d(3), d(4), 36*d(5), d(6), ...].
EXAMPLE
The decimal expansion of this constant t begins:
t = 1.540546562581850919585092443277493630777443307008831680830472910...
The simple continued fraction expansion of t begins:
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), ...]
such that the simple continued fraction expansion of 4*t begins:
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), ...].
The initial 1000 digits in the decimal expansion of t are
t = 1.54054656258185091958509244327749363077744330700883\
16808304729109136071030553148439916006575514781913\
07822865753878621983372629507214585549765601216855\
82787499037533808802512182117029618023317430110304\
68568548056881176292575460160397807605484819864109\
97669631806987876538679522272871069799153383954670\
27608251838807810531236786249017655750340240450431\
39500820756134042709955133598762098341500330573137\
99939807838257483585300963279557471021988938954240\
72201092140208383999296983006564129234149599645137\
97256177580945556498719774507283674158287467111752\
77844466938413199720373428083674374904367749851098\
21424896071080046214454673955084637450020810561194\
71345165789766421139563276393063288950816919055445\
03342967346596573587017045564281666931351950277058\
62590013151270263215075119134200880720738116296046\
68850174252499836432241162269740091282255326351721\
57235179003932955261950046664886398112095558223840\
82192509104134957717517226738372957214588632890700\
21161549012492823004735717009953284171956638331978...
...
The initial 528 terms in the continued fraction expansion of t are
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,
1,1,1,5,1,1,1,69120,1,1,1,5,1,1,1,120,1,1,1,5,1,1,1,2880,
1,1,1,5,1,1,1,120,1,1,1,5,1,1,1,1658880,1,1,1,5,1,1,1,120,
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,
1,1,1,5,1,1,1,120,1,1,1,5,1,1,1,69120,1,1,1,5,1,1,1,120,
1,1,1,5,1,1,1,2880,1,1,1,5,1,1,1,120,1,1,1,5,1,1,1,1658880,
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,
1,1,1,5,1,1,1,120,1,1,1,5,1,1,1,955514880,1,1,1,5,1,1,1,120,
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,
1,1,1,5,1,1,1,1658880,1,1,1,5,1,1,1,120,1,1,1,5,1,1,1,2880,
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,
1,1,1,5,1,1,1,120,1,1,1,5,1,1,1,2880,1,1,1,5,1,1,1,120,
1,1,1,5,1,1,1,69120,1,1,1,5,1,1,1,120,1,1,1,5,1,1,1,2880,
1,1,1,5,1,1,1,120,1,1,1,5,1,1,1,1658880,1,1,1,5,1,1,1,120,
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,
1,1,1,5,1,1,1,22932357120,1,1,1,5,1,1,1,120,1,1,1,5,1,1,1,2880, ...].
...
GENERATING METHOD.
Start with CF = [1] and repeat (PARI code):
t = (1/4)*contfracpnqn(6*CF)[1,1]/contfracpnqn(6*CF)[2,1]; CF = contfrac(t)
This method can be illustrated as follows.
t0 = [1] = 1 ;
t1 = (1/4)*[6] = [1; 2] = 3/2 ;
t2 = (1/4)*[6; 12] = [1; 1, 1, 11, 2] = 73/48 ;
t3 = (1/4)*[6; 6, 6, 66, 12] = [1; 1, 1, 5, 1, 1, 1, 264, 3] = 45312/29413 ;
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 ;
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 ; ...
where this constant t equals the limit of the iterations of the above process.
PROG
(PARI) /* Generate over 5000 digits in the decimal expansion */
CF=[1];
{for(i=1, 12, t = (1/4)*contfracpnqn(6*CF)[1, 1]/contfracpnqn(6*CF)[2, 1];
CF = contfrac(t) ); }
for(n=1, 150, print1(floor(10^(n-1)*t)%10, ", "))
CROSSREFS
Sequence in context: A309699 A153106 A021189 * A197012 A200137 A197384
KEYWORD
nonn,cons
AUTHOR
Paul D. Hanna, Oct 24 2018
STATUS
approved

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Last modified April 22 07:37 EDT 2024. Contains 371892 sequences. (Running on oeis4.)