A320953 - OEIS

%I #14 Nov 01 2018 06:47:02

%S 1,2,6,6,4,5,0,2,1,4,5,7,8,6,5,1,6,1,3,7,7,8,7,7,2,9,2,9,0,1,0,2,5,9,

%T 2,3,7,2,3,6,9,6,6,5,4,5,0,5,4,7,5,3,3,0,0,5,5,8,0,9,9,3,0,5,3,6,2,4,

%U 2,7,2,1,8,4,8,0,1,3,5,4,9,7,3,5,7,8,0,1,7,7,2,1,1,6,1,0,9,3,9,0,2,8,3,1,8,6,8,3,7,3,3,5,2,6,0,7,3,3,1,2,1,7,1,2,1,6,3,0,5,4,5,0,1,0,1,8,7,4,2,6,4,6,5,8,3,7,9,0,6,4

%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 {5*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="/A320953/b320953.txt">Table of n, a(n) for n = 1..5000</a>

%F Given t = [d(0); d(1), d(2), d(3), d(4), d(5), d(6), ..., A320952(n), ...], some related simple continued fractions are:

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

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

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

%e The decimal expansion of this constant t begins:

%e t = 1.266450214578651613778772929010259237236966545054753300558099305...

%e The simple continued fraction expansion of t begins:

%e 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, ..., A320952(n), ...]

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

%e 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*A320952(n), ...].

%e ...

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

%e t = 1.26645021457865161377877292901025923723696654505475\

%e 33005580993053624272184801354973578017721161093902\

%e 83186837335260733121712163054501018742646583790643\

%e 54875833008007285407818429195692996453113827495305\

%e 54926835731865875321451747328930568555229448988608\

%e 05786435870905452173007164597651669613272647900654\

%e 83894404364902408305059276949480386935767775514765\

%e 94968299951493100831611976267451605948933885299869\

%e 07863756473354702971788204675428850345468192538035\

%e 85810263279458657769543507292924716821458977316162\

%e 73791864200695026130548141231704085623717435331603\

%e 39200767062558929533702255394800104584764699667049\

%e 25260260287699076614416015170423275489461158402478\

%e 17149290630968365611810185428241284526255760519651\

%e 94435407858521668152963500082382089516858099107232\

%e 93046639826593759422087336891654489562762696517528\

%e 97543506563610723909866277201274334313308836147313\

%e 53053566946629955949015183419229625850457388237270\

%e 77912709245178910220379095072759755794410149465362\

%e 47442129159574142863549600623749608271595134828851...

%e ...

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

%e 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,

%e 1,1,3,3,2,1,4,1,1,4,1,1,4,1,2,3,3,1,1,14,1,2,3,36,

%e 60,2,1,1,4,1,1,17,20,1,3,1,3,60,2,1,1,4,1,1,4,1,1,4,

%e 1,2,3,4,1,3,1,20,5,20,2,1,1,14,1,1,3,3,2,1,4,1,1,17,

%e 20,2,1,1,14,1,1,44,1,1,299,1,1,2,20,1,3,1,4,3,2,1,4,1,

%e 1,20,1,3,24,1,3,1,60,1,3,1,3,1200,2,1,1,4,1,2,3,4,1,3,

%e 1,20,5,20,1,3,1,4,3,2,1,9,1,1,3,3,1,4,3,2,1,14,1,2,

%e 3,24,1,3,6,400,2,1,1,4,1,2,3,17,3,2,1,4,1,1,3,3,1,3,

%e 40,1,3,1,4,3,2,1,4,1,1,20,1,3,24,1,3,2,3,2,1,4,1,1,

%e 17,20,1,3,1,54,3,2,1,4,1,1,373,3,2,1,4,1,1,2,400,1,3,1,

%e 3,20,5,60,2,1,1,4,1,1,4,1,1,4,1,2,3,24,1,3,1,60,30,20,

%e 3,1,3,1,1200,1,3,1,3,20,3,1,3,1499,1,3,2,3,2,1,4,1,1,4,

%e 1,1,4,1,1,2,60,5,20,3,1,3,1,400,6,3,1,24,3,2,1,14,1,2,

%e 3,4,1,3,3,1,1,9,1,2,3,11,20,1,3,1,3,60,1,3,1,4,3,1,

%e 3,40,1,3,1,16,1,3,1,40,3,1,3,29,1,3,1,60,7,1,1,1999,1,1,

%e 2,20,1,3,1,4,3,2,1,9,1,1,3,3,1,20,1,1,14,1,1,2,20,5,

%e 20,1,3,1,3,60,1,3,1,3,800,1,3,1,3,20,5,60,2,1,1,4,1,1,

%e 4,1,1,4,1,2,3,24,1,3,1,60,30,20,3,1,3,2,3,1,3,40,1,3,

%e 1,4,3,2,1,4,1,1,20,1,3,24,1,3,1,60,1,3,1,66,1,3,3,1,

%e 1,9,1,2,3,4,1,3,1,20,466,3,1,3,40,1,3,1,4,3,2,1,4,1,

%e 1,2,8000,1,3,1,3,20,3,1,3,24,1,3,6,1200,2,1,1,4,1,2,3,4,

%e 1,3,1,20,5,20,1,3,1,4,3,2,1,9,1,1,3,3,1,29,3,2,1, ...].

%e ...

%e GENERATING METHOD.

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

%e t = (1/4)*contfracpnqn(5*CF)[1,1]/contfracpnqn(5*CF)[2,1]; CF = contfrac(t)

%e This method can be illustrated as follows.

%e t0 = [1] = 1;

%e t1 = (1/4)*[5] = [1; 4] = 5/4;

%e t2 = (1/4)*[5; 20] = [1; 3, 1, 4, 4] = 101/80;

%e t3 = (1/4)*[5; 15, 5, 20, 20] = [1; 3, 1, 3, 20, 5, 80] = 155905/123104;

%e 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;

%e 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; ...

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

%o (PARI) /* Generate over 5700 digits in the decimal expansion */

%o CF=[1];

%o {for(i=1, 12, t = (1/4)*contfracpnqn(5*CF)[1, 1]/contfracpnqn(5*CF)[2, 1];

%o CF = contfrac(t) ); }

%o for(n=1, 150, print1(floor(10^(n-1)*t)%10, ", "))

%Y Cf. A320952, A320413, A320411.

%K nonn,cons

%O 1,2

%A _Paul D. Hanna_, Oct 26 2018