A320832 Continued fraction of a constant t with partial denominators {a(n), n>=0} such that the continued fraction of 2*t yields partial denominators {3*a(n), n>=0}.

1, 1, 1, 1, 6, 1, 1, 1, 8, 1, 1, 1, 6, 1, 1, 1, 11, 1, 1, 1, 6, 1, 1, 1, 8, 1, 1, 1, 6, 1, 1, 1, 16, 6, 1, 1, 1, 1, 36, 1, 1, 1, 1, 6, 12, 6, 1, 1, 1, 1, 36, 1, 1, 1, 1, 6, 24, 36, 1, 1, 1, 1, 6, 1, 1, 1, 53, 1, 1, 1, 6, 1, 1, 1, 1, 36, 18, 36, 1, 1, 1, 1, 6, 1, 1, 1, 53, 1, 1, 1, 6, 1, 1, 1, 1, 36, 36, 216, 1, 1, 1, 1, 6, 1, 1, 1, 8, 1, 1, 1, 6, 1, 1, 1, 79, 6, 1, 1, 1, 1, 36, 1, 1, 1, 1, 6, 1, 1, 1, 53, 1, 1, 26, 1, 1, 53, 1, 1, 1, 6, 1, 1, 1, 1, 36, 1, 1, 1, 1, 6

Offset

0,5

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

Paul D. Hanna, Table of n, a(n) for n = 0..3000

Formula

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

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

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

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

Example

The decimal expansion of this constant t begins:
t = 1.6514904747308816380867968444430065786661319067679...
The simple continued fraction expansion of t begins:
t = [1; 1, 1, 1, 6, 1, 1, 1, 8, 1, 1, 1, 6, 1, 1, 1, 11, 1, 1, 1, 6, 1, 1, 1, 8, 1, 1, 1, 6, 1, 1, 1, 16, 6, ..., a(n), ...]
such that the simple continued fraction expansion of 2*t begins:
2*t = [3; 3, 3, 3, 18, 3, 3, 3, 24, 3, 3, 3, 18, 3, 3, 3, 33, 3, 3, 3, 18, 3, 3, 3, 24, 3, 3, 3, 18, 3, 3, 3, 48, 18, ..., 3*a(n), ...].
...
The initial 1000 terms of the simple continued fraction expansion of t are
t = [1;1,1,1,6,1,1,1,8,1,1,1,6,1,1,1,11,1,1,1,6,1,1,1,8,1,
1,1,6,1,1,1,16,6,1,1,1,1,36,1,1,1,1,6,12,6,1,1,1,1,36,
1,1,1,1,6,24,36,1,1,1,1,6,1,1,1,53,1,1,1,6,1,1,1,1,36,
18,36,1,1,1,1,6,1,1,1,53,1,1,1,6,1,1,1,1,36,36,216,1,1,1,
1,6,1,1,1,8,1,1,1,6,1,1,1,79,6,1,1,1,1,36,1,1,1,1,6,
1,1,1,53,1,1,26,1,1,53,1,1,1,6,1,1,1,1,36,1,1,1,1,6,79,
1,1,1,6,1,1,1,8,1,1,1,6,1,1,1,1,216,54,1296,1,1,1,1,6,1,
1,1,8,1,1,1,6,1,1,1,11,1,1,1,6,1,1,1,8,1,1,1,6,1,1,
1,118,36,1,1,1,1,6,1,1,1,53,1,1,1,6,1,1,1,1,36,1,1,1,1,
6,79,1,1,1,6,39,6,1,1,1,79,6,1,1,1,1,36,1,1,1,1,6,1,1,
1,53,1,1,1,6,1,1,1,1,36,118,1,1,1,6,1,1,1,8,1,1,1,6,1,
1,1,11,1,1,1,6,1,1,1,8,1,1,1,6,1,1,1,1,1296,81,7776,1,1,1,
1,6,1,1,1,8,1,1,1,6,1,1,1,11,1,1,1,6,1,1,1,8,1,1,1,
6,1,1,1,16,6,1,1,1,1,36,1,1,1,1,6,12,6,1,1,1,1,36,1,1,
1,1,6,177,216,1,1,1,1,6,1,1,1,8,1,1,1,6,1,1,1,79,6,1,1,
1,1,36,1,1,1,1,6,1,1,1,53,1,1,1,6,1,1,1,1,36,118,1,1,1,
6,1,1,1,8,1,1,58,36,1,1,1,1,6,118,1,1,8,1,1,1,6,1,1,1,
1,216,1,1,1,1,6,1,1,1,8,1,1,1,6,1,1,1,79,6,1,1,1,1,36,
1,1,1,1,6,1,1,1,53,1,1,176,1,1,1,6,1,1,1,8,1,1,1,6,1,
1,1,11,1,1,1,6,1,1,1,8,1,1,1,6,1,1,1,16,6,1,1,1,1,36,
1,1,1,1,6,12,6,1,1,1,1,36,1,1,1,1,6,1,1,1,1943,1,1,121,46656,
1,1,1,1,6,1,1,1,8,1,1,1,6,1,1,1,11,1,1,1,6,1,1,1,8,
1,1,1,6,1,1,1,16,6,1,1,1,1,36,1,1,1,1,6,12,6,1,1,1,1,
36,1,1,1,1,6,24,36,1,1,1,1,6,1,1,1,53,1,1,1,6,1,1,1,1,
36,18,36,1,1,1,1,6,1,1,1,53,1,1,1,6,1,1,1,1,36,265,1,1,323,
1,1,1,6,1,1,1,1,36,1,1,1,1,6,12,6,1,1,1,1,36,1,1,1,1,
6,118,1,1,8,1,1,1,6,1,1,1,1,216,1,1,1,1,6,1,1,1,8,1,1,
1,6,1,1,1,79,6,1,1,1,1,36,1,1,1,1,6,1,1,1,53,1,1,176,1,
1,1,6,1,1,1,8,1,1,1,6,1,1,1,11,1,1,1,6,87,216,1,1,1,1,
6,1,1,1,8,1,1,176,1,1,1,6,12,6,1,1,1,1,36,1,1,1,1,6,1,
1,1,323,1,1,1,6,1,1,1,1,36,1,1,1,1,6,12,6,1,1,1,1,36,1,
1,1,1,6,118,1,1,8,1,1,1,6,1,1,1,1,216,1,1,1,1,6,1,1,1,
8,1,1,1,6,1,1,1,79,6,1,1,1,263,1,1,1,6,1,1,1,8,1,1,1,
6,1,1,1,11,1,1,1,6,1,1,1,8,1,1,1,6,1,1,1,16,6,1,1,1,
1,36,1,1,1,1,6,12,6,1,1,1,1,36,1,1,1,1,6,24,36,1,1,1,1,
6,1,1,1,53,1,1,1,6,1,1,1,1,36,18,36,1,1,1,1,6,1,1,1,53,
1,1,1,6,1,1,1,1,36,1,1,1,1,6,2914,1,1,1,6,181,1,1,69983,1,1,
1,6,1,1,1,1,36,1,1,1,1,6,12,6,1,1,1,1,36,1,1,1,1,6,16,
1,1,1,6,1,1,1,8,1,1,1,6,1,1,1,11,1,1,1,6,1,1,1,8,1,
1,1,6,1,1,1,23,1,1,8,1,1,1,6,1,1,1,1,216,1,1,1,1,6, ...].
...
The initial 1000 digits of constant t are
t = 1.65149047473088163808679684444300657866613190676791\
28737714830270861698007448681310643103951379451671\
54114507647362346731655343736679236451104324525424\
37390179560582497835710083063843943037795949756748\
97440995664382615322263468614614477737059496505552\
08331434371348884420941679386949556166027674330855\
36885812669651061905701771944991773909674239753715\
13170920756343463276796919694602222128457275152421\
55344241016460665248526708333389356493782455903465\
51804053948550486808342254050578114807877804220843\
95652198152342146601232467945702955643471515688899\
60645600818470508698165499450873578310068410378749\
13752238554732608875401892185360771843689733221676\
17468480327947937646720546426241804321256694683699\
70360475351525492330828064189641281507963387604040\
39551897256028156761617895973905334533561413827981\
22417201150520589772826034062665207814244141469100\
63982517271814133206125657700492407641669772124988\
49178505556132383776229702780090744171002180938101\
95839491920468860210551810985143159776520025263372...
GENERATING METHOD.
Start with CF = [1] and repeat (PARI code):
t = (1/2)*contfracpnqn(3*CF)[1,1]/contfracpnqn(3*CF)[2,1]; CF = contfrac(t)
This method can be illustrated as follows.
t0 = [1] = 1 ;
t1 = (1/2)*[3] = [1; 2] = 3/2 ;
t2 = (1/2)*[3; 6] = [1; 1, 1, 2, 2] = 19/12 ;
t3 = (1/2)*[3; 3, 3, 6, 6] = [1, 1, 1, 1, 6, 3, 12] = 1281/776 ;
t4 = (1/2)*[3; 3, 3, 3, 18, 9, 36] = [1; 1, 1, 1, 6, 1, 1, 1, 8, 1, 1, 4, 72] = 652299/394976 ;
t5 = (1/2)*[3; 3, 3, 3, 18, 3, 3, 3, 24, 3, 3, 12, 216] = [1, 1, 1, 1, 6, 1, 1, 1, 8, 1, 1, 1, 6, 1, 1, 1, 11, 1, 1, 1, 6, 6, 432] = 43763571081/26499438992 ; ...
The above method generates terms with exponential growth; the number of terms with each iteration begins:
[1, 2, 5, 7, 13, 23, 41, 71, 121, 214, 377, 662, 1170, 2082, 3681, 6535, 11632, 20700, 36799, 65472,...].

Prog

(PARI) /* Generates over 3600 terms of the continued fraction */
CF=[1];
{for(i=1, 14, t = (contfracpnqn(3*CF)[1, 1]/contfracpnqn(3*CF)[2, 1])/2;
CF = contfrac(t) ); CF }

Crossrefs

Cf. A320833, A320834, A320410, A320952.

Sequence in context: A344697 A364917 A364944 * A034460 A063919 A308135

Adjacent sequences: A320829 A320830 A320831 * A320833 A320834 A320835

Keyword

nonn,cofr

Author

Paul D. Hanna, Oct 22 2018

Status

approved