A320952 Continued fraction of a constant t with partial denominators {a(n), n>=0} such that the continued fraction of 4*t yields partial denominators {5*a(n), n>=0}.

1, 3, 1, 3, 20, 3, 1, 3, 24, 1, 3, 3, 1, 1, 4, 1, 1, 3, 3, 1, 29, 3, 2, 1, 14, 1, 1, 3, 3, 2, 1, 4, 1, 1, 4, 1, 1, 4, 1, 2, 3, 3, 1, 1, 14, 1, 2, 3, 36, 60, 2, 1, 1, 4, 1, 1, 17, 20, 1, 3, 1, 3, 60, 2, 1, 1, 4, 1, 1, 4, 1, 1, 4, 1, 2, 3, 4, 1, 3, 1, 20, 5, 20, 2, 1, 1, 14, 1, 1, 3, 3, 2, 1, 4, 1, 1, 17, 20, 2, 1, 1, 14, 1, 1, 44, 1, 1, 299, 1, 1, 2, 20, 1, 3, 1, 4, 3, 2, 1, 4, 1, 1, 20, 1, 3, 24, 1, 3, 1, 60, 1, 3, 1, 3, 1200, 2, 1, 1, 4, 1, 2, 3, 4, 1, 3, 1, 20, 5, 20, 1, 3, 1, 4, 3, 2, 1, 9, 1, 1, 3, 3, 1, 4, 3, 2, 1, 14, 1, 2, 3, 24, 1, 3, 6, 400, 2, 1, 1, 4, 1, 2, 3, 17, 3, 2, 1, 4, 1, 1, 3, 3, 1, 3, 40, 1, 3, 1, 4, 3, 2, 1, 4, 1, 1, 20

Offset

0,2

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

Paul D. Hanna, Table of n, a(n) for n = 0..5000

Formula

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

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

(2) 4*t/5 = [a(0); 25*a(1), a(2), 25*a(3), a(4), 25*a(5), a(6), ...],

(3) 20*t = [25*a(0); a(1), 25*a(2), a(3), 25*a(4), a(5), 25*a(6), ...].

Example

The decimal expansion of this constant t begins:
t = 1.266450214578651613778772929010259237236966545054753300558099305...
The simple continued fraction expansion of t begins:
t = [1; 3, 1, 3, 20, 3, 1, 3, 24, 1, 3, 3, 1, 1, 4, 1, 1, 3, 3, 1, 29, 3, 2, 1, 14, 1, 1, 3, 3, 2, 1, 4, 1, 1, 4, 1, 1, 4, 1, 2, ... , a(n), ...]
such that the simple continued fraction expansion of 4*t begins:
4*t = [5; 15, 5, 15, 100, 15, 5, 15, 120, 5, 15, 15, 5, 5, 20, 5, 5, 15, 15, 5, 145, 15, 10, 5, 70, 5, 5, 15, 15, 10, 5, 20, 5, 5, ... , 5*a(n), ...].
...
The initial 528 terms in the continued fraction expansion of t are
t = [1;3,1,3,20,3,1,3,24,1,3,3,1,1,4,1,1,3,3,1,29,3,2,1,14,
1,1,3,3,2,1,4,1,1,4,1,1,4,1,2,3,3,1,1,14,1,2,3,36,
60,2,1,1,4,1,1,17,20,1,3,1,3,60,2,1,1,4,1,1,4,1,1,4,
1,2,3,4,1,3,1,20,5,20,2,1,1,14,1,1,3,3,2,1,4,1,1,17,
20,2,1,1,14,1,1,44,1,1,299,1,1,2,20,1,3,1,4,3,2,1,4,1,
1,20,1,3,24,1,3,1,60,1,3,1,3,1200,2,1,1,4,1,2,3,4,1,3,
1,20,5,20,1,3,1,4,3,2,1,9,1,1,3,3,1,4,3,2,1,14,1,2,
3,24,1,3,6,400,2,1,1,4,1,2,3,17,3,2,1,4,1,1,3,3,1,3,
40,1,3,1,4,3,2,1,4,1,1,20,1,3,24,1,3,2,3,2,1,4,1,1,
17,20,1,3,1,54,3,2,1,4,1,1,373,3,2,1,4,1,1,2,400,1,3,1,
3,20,5,60,2,1,1,4,1,1,4,1,1,4,1,2,3,24,1,3,1,60,30,20,
3,1,3,1,1200,1,3,1,3,20,3,1,3,1499,1,3,2,3,2,1,4,1,1,4,
1,1,4,1,1,2,60,5,20,3,1,3,1,400,6,3,1,24,3,2,1,14,1,2,
3,4,1,3,3,1,1,9,1,2,3,11,20,1,3,1,3,60,1,3,1,4,3,1,
3,40,1,3,1,16,1,3,1,40,3,1,3,29,1,3,1,60,7,1,1,1999,1,1,
2,20,1,3,1,4,3,2,1,9,1,1,3,3,1,20,1,1,14,1,1,2,20,5,
20,1,3,1,3,60,1,3,1,3,800,1,3,1,3,20,5,60,2,1,1,4,1,1,
4,1,1,4,1,2,3,24,1,3,1,60,30,20,3,1,3,2,3,1,3,40,1,3,
1,4,3,2,1,4,1,1,20,1,3,24,1,3,1,60,1,3,1,66,1,3,3,1,
1,9,1,2,3,4,1,3,1,20,466,3,1,3,40,1,3,1,4,3,2,1,4,1,
1,2,8000,1,3,1,3,20,3,1,3,24,1,3,6,1200,2,1,1,4,1,2,3,4,
1,3,1,20,5,20,1,3,1,4,3,2,1,9,1,1,3,3,1,29,3,2,1, ...].
...
The initial 1000 digits in the decimal expansion of t are
t = 1.26645021457865161377877292901025923723696654505475\
33005580993053624272184801354973578017721161093902\
83186837335260733121712163054501018742646583790643\
54875833008007285407818429195692996453113827495305\
54926835731865875321451747328930568555229448988608\
05786435870905452173007164597651669613272647900654\
83894404364902408305059276949480386935767775514765\
94968299951493100831611976267451605948933885299869\
07863756473354702971788204675428850345468192538035\
85810263279458657769543507292924716821458977316162\
73791864200695026130548141231704085623717435331603\
39200767062558929533702255394800104584764699667049\
25260260287699076614416015170423275489461158402478\
17149290630968365611810185428241284526255760519651\
94435407858521668152963500082382089516858099107232\
93046639826593759422087336891654489562762696517528\
97543506563610723909866277201274334313308836147313\
53053566946629955949015183419229625850457388237270\
77912709245178910220379095072759755794410149465362\
47442129159574142863549600623749608271595134828851...
...
GENERATING METHOD.
Start with CF = [1] and repeat (PARI code):
t = (1/4)*contfracpnqn(5*CF)[1,1]/contfracpnqn(5*CF)[2,1]; CF = contfrac(t)
This method can be illustrated as follows.
t0 = [1] = 1 ;
t1 = (1/4)*[5] = [1; 4] = 5/4 ;
t2 = (1/4)*[5; 20] = [1; 3, 1, 4, 4] = 101/80 ;
t3 = (1/4)*[5; 15, 5, 20, 20] = [1; 3, 1, 3, 20, 5, 80] = 155905/123104 ;
t4 = (1/4)*[5; 15, 5, 15, 100, 25, 400] = [1; 3, 1, 3, 20, 3, 1, 3, 24, 1, 3, 6, 1600] = 5857775885/4625350304 ;
t5 = (1/4)*[5; 15, 5, 15, 100, 15, 5, 15, 120, 5, 15, 30, 8000] = [1; 3, 1, 3, 20, 3, 1, 3, 24, 1, 3, 3, 1, 1, 4, 1, 1, 3, 3, 1, 29, 3, 2, 1, 14, 1, 1, 7, 32000] = 1487626504292631625/1174642703809375904 ; ...
where this constant t equals the limit of the iterations of the above process.
The number of terms generated in each iteration of the above method begins:
[1, 2, 5, 7, 13, 29, 63, 138, 286, 582, 1223, 2534, 5249, 10932, 22944, 48264, ...].

Prog

(PARI) /* Generate over 5200 terms of the continued fraction */
CF=[1];
{for(i=1, 12, t = (1/4)*contfracpnqn(5*CF)[1, 1]/contfracpnqn(5*CF)[2, 1];
CF = contfrac(t) ); CF}

Crossrefs

Cf. A320953, A320412, A320410.

Sequence in context: A361839 A160708 A040173 * A128777 A286892 A215828

Adjacent sequences: A320949 A320950 A320951 * A320953 A320954 A320955

Keyword

nonn,cofr

Author

Paul D. Hanna, Oct 26 2018

Status

approved