A320412 - OEIS

%I #12 Oct 26 2018 17:59:49

%S 1,1,3,1,1,1,20,1,1,1,3,1,1,28,35,28,1,1,3,1,1,1,20,1,1,1,3,1,1,784,

%T 61,3,1,48,3,1,1,28,5,3,1,1,28,1,1,3,34,1,3,1,1,1,6,1,1,1,3,1,4,1,1,6,

%U 1,1,1,3,1,1371,3,1,106,84,1,1,3,83,1,3,5,28,1,1,3,48,1,3,8,1,1,20,1,1,1,3,1,1,784,1,1,3,1,1,1,20,1,1,59,28,5,3,1,1,28,1,1,3,10,3,1,1,28,1,1,3,5,28,7,28,1,1,3,10,3,1,1,28,1,1,3,5,28,2399,3,1,4,1,1,6,1,1,185,2352,1

%N Continued fraction of a constant t with partial denominators {a(n), n>=0} such that the continued fraction of 4*t yields partial denominators {7*a(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="/A320412/b320412.txt">Table of n, a(n) for n = 0..5000</a>

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

%F (1) 4*t = [7*a(0); 7*a(1), 7*a(2), 7*a(3), 7*a(4), 7*a(5), ...],

%F (2) 4*t/7 = [a(0); 49*a(1), a(2), 49*a(3), a(4), 49*a(5), a(6), ...],

%F (3) 28*t = [49*a(0); a(1), 49*a(2), a(3), 49*a(4), a(5), 49*a(6), ...].

%e The decimal expansion of this constant t begins:

%e t = 1.785474560974062575656044807894470237170401346194715783864279820...

%e The simple continued fraction expansion of t begins:

%e t = [1; 1, 3, 1, 1, 1, 20, 1, 1, 1, 3, 1, 1, 28, 35, 28, 1, 1, 3, 1, 1, 1, 20, 1, 1, 1, 3, 1, 1, 784, 61, 3, 1, 48, 3, 1, 1, 28, ... , a(n), ...]

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

%e 4*t = [7; 7, 21, 7, 7, 7, 140, 7, 7, 7, 21, 7, 7, 196, 245, 196, 7, 7, 21, 7, 7, 7, 140, 7, 7, 7, 21, 7, 7, 5488, 427, 21, 7, 679, ... , 7*a(n), ...].

%e ...

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

%e t = [1;1,3,1,1,1,20,1,1,1,3,1,1,28,35,28,1,1,3,1,1,1,20,1,1,

%e 1,3,1,1,784,61,3,1,48,3,1,1,28,5,3,1,1,28,1,1,3,34,1,3,

%e 1,1,1,6,1,1,1,3,1,4,1,1,6,1,1,1,3,1,1371,3,1,106,84,1,

%e 1,3,83,1,3,5,28,1,1,3,48,1,3,8,1,1,20,1,1,1,3,1,1,784,

%e 1,1,3,1,1,1,20,1,1,59,28,5,3,1,1,28,1,1,3,10,3,1,1,28,

%e 1,1,3,5,28,7,28,1,1,3,10,3,1,1,28,1,1,3,5,28,2399,3,1,4,

%e 1,1,6,1,1,185,2352,1,1,3,1,1,1,20,1,1,144,1,3,1,1,1,20,1,

%e 1,8,3,1,48,3,1,1,28,5,3,1,83,3,1,1,84,14,28,1,1,3,34,1,

%e 3,1,1,1,6,1,1,1,3,1,4,1,1,6,1,1,1,3,1,1371,3,1,1,28,

%e 5,3,1,1,28,1,1,3,34,1,3,1,1,1,6,1,1,102,1,3,48,1,3,8,

%e 1,1,20,1,1,1,3,1,1,784,1,1,3,1,1,1,20,1,1,17,84,1,1,3,

%e 1,1,1,195,1,1,1,3,1,1,84,8,1,3,48,1,3,12,784,1,1,3,1,1,

%e 1,20,1,1,17,84,1,1,3,1,1,1,195,1,1,1,3,1,1,84,8,1,3,48,

%e 1,3,4198,84,1,1,3,6,1,3,1,1,1,6,1,1,10,28,1,1,3,323,1,1,

%e 16463,1,1,1,3,1,1,84,1,1,3,1,1,1,6,1,1,34,1,1,6,1,1,1,

%e 3,1,251,3,1,1,84,1,1,3,1,1,1,6,1,1,34,1,1,6,1,1,1,3,

%e 1,13,3,1,4,1,1,6,1,1,83,1,1,20,1,1,1,3,1,1,784,8,1,3,

%e 5,28,145,3,1,4,1,1,6,1,1,1,3,1,146,3,1,23,1,3,48,1,3,1,

%e 1,1,6,1,1,4,1,3,59,3,1,1,84,1,1,3,1,1,1,6,1,1,10,28,

%e 1,1,3,1,1,1,20,1,1,1,3,1,6,3,1,1,28,10,1,1,6,1,1,1,

%e 3,1,1,84,1,1,3,2399,84,1,1,3,1,1,1,195,1,1,8,3,1,4,1,1,

%e 6,1,1,1,3,1,48,3,1,1,28,5,3,1,58,1,3,1,1,1,20,1,1, ...].

%e ...

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

%e t = 1.78547456097406257565604480789447023717040134619471\

%e 57838642798207580399578548165115122069732837050490\

%e 61286404635705077102774547249610751367094378190709\

%e 85401921611227678484369568036324073754957652390489\

%e 39943379031303125762694739475382462408109374813120\

%e 00422151267123870355135847303104193193077189589829\

%e 99475276214685582004074592162502265622639066962207\

%e 90264315701034911225864026241008924068247829601272\

%e 51264292894203497339763733333382142446328347420764\

%e 99374490081465556514449152015450022286517069550755\

%e 66197498206618148152838524399347921835127003551875\

%e 27576125361547533763998734682666496571601524379160\

%e 71274633004761476783978729230894485114372110439235\

%e 32434589015973787010030486968307426525808424138019\

%e 30747406263726641419798279952039943508589013753829\

%e 08707915177794914067867811104635718728142844757658\

%e 92545040853582478172771767319724386460332134939668\

%e 53646394220175623167681377901091602489813663749106\

%e 20051189089949357240864658051397802563633845923264\

%e 45649305634826816712996792415779406659039071052772...

%e ...

%e GENERATING METHOD.

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

%e t = (1/4)*contfracpnqn(7*CF)[1,1]/contfracpnqn(7*CF)[2,1]; CF = contfrac(t)

%e This method can be illustrated as follows.

%e t0 = [1] = 1 ;

%e t1 = (1/4)*[7] = [1; 1, 3] = 7/4 ;

%e t2 = (1/4)*[7; 7, 21] = [1; 1, 3, 1, 1, 1, 20, 2] = 1057/592

%e t3 = (1/4)*[7; 7, 21, 7, 7, 7, 140, 14] = [1; 1, 3, 1, 1, 1, 20, 1, 1, 1, 3, 1, 1, 28, 35, 56] = 745628689/417608128 ;

%e t4 = (1/4)*[7; 7, 21, 7, 7, 7, 140, 7, 7, 7, 21, 7, 7, 196, 245, 392] = [1; 1, 3, 1, 1, 1, 20, 1, 1, 1, 3, 1, 1, 28, 35, 28, 1, 1, 3, 1, 1, 1, 20, 1, 1, 1, 3, 1, 1, 784, 61, 3, 1, 97, 4] = 381285245640002405521/213548405546586390784 ; ...

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

%e The number of terms generated in each iteration of the above method begins:

%e [1, 3, 8, 16, 35, 80, 172, 387, 908, 2164, 5120, 12111, 28492, 67075, ...].

%e ...

%o (PARI) /* Generate over 5100 terms of the continued fraction */

%o CF=[1];

%o {for(i=1, 10, t = (1/4)*contfracpnqn(7*CF)[1, 1]/contfracpnqn(7*CF)[2, 1];

%o CF = contfrac(t) ); CF}

%Y Cf. A320413, A320410.

%K nonn,cofr

%O 0,3

%A _Paul D. Hanna_, Oct 26 2018