A321093 - OEIS

%I #19 Oct 29 2018 03:58:42

%S 6,1,3,5,3,1,5,3,1,5,1,3,5,1,3,5,3,1,5,1,3,5,1,3,5,3,1,5,1,3,5,3,1,5,

%T 3,1,5,1,3,5,3,1,5,3,1,5,1,3,5,1,3,5,3,1,5,1,3,5,3,1,5,3,1,5,1,3,5,3,

%U 1,5,3,1,5,1,3,5,1,3,5,3,1,5,1,3,5,3,1,5,3,1,5,1,3,5,1,3,5,3,1,5,1,3,5,1,3,5,3,1,5,1,3,5,3,1,5,3,1,5,1,3,5,1,3,5,3,1,5,1,3,5,1,3,5,3,1,5,1,3,5,3,1,5,3,1,5,1,3,5,3,1,5,3,1,5,1,3,5,1,3,5,3,1,5,1,3,5,3,1,5,3,1,5,1,3,5,1,3,5,3,1,5,1,3,5,1,3,5,3,1,5,1,3,5,3,1,5,3,1,5,1,3

%N Continued fraction expansion of the constant z that satisfies: CF(4*z, n) = CF(z, n) + 21, for n >= 0, where CF(z, n) denotes the n-th partial denominator in the continued fraction expansion of z.

%F Formula for terms:

%F (1) a(0) = 6,

%F (2) a(3*n) = 5 for n >= 1,

%F (3) a(3*n+2) = 4 - a(3*n+1) for n >= 0,

%F (4) a(9*n+1) = 1 for n >= 0,

%F (5) a(9*n+7) = 3 for n >= 0,

%F (6) a(9*n+4) = 4 - a(3*n+1) for n >= 0.

%F a(3*n+1) = 2*A189706(n+1) + 1 for n >= 0.

%F a(n) = 2*A321090(n) + 1 for n >= 1, with a(0) = 6.

%F a(n) = A321091(n) + A321090(n), for n >= 0.

%F a(n) = A321095(n) - A321090(n), for n >= 0.

%F a(n) = A321097(n) - 2*A321090(n), for n >= 0.

%e The decimal expansion of this constant z begins:

%e z = 6.76134218926005257700977872218164992635251510297623...

%e The simple continued fraction expansion of z begins:

%e z = [6; 1, 3, 5, 3, 1, 5, 3, 1, 5, 1, 3, 5, 1, 3, 5, 3, ..., a(n), ...];

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

%e 4*z = [27; 22, 24, 26, 24, 22, 26, 24, 22, 26, 22, 24, ..., a(n) + 21, ...].

%e EXTENDED TERMS.

%e The initial 1020 terms of the continued fraction of z are

%e z = [6;1,3,5,3,1,5,3,1,5,1,3,5,1,3,5,3,1,5,1,3,5,1,3,5,3,1,5,1,3,5,

%e 3,1,5,3,1,5,1,3,5,3,1,5,3,1,5,1,3,5,1,3,5,3,1,5,1,3,5,3,1,5,

%e 3,1,5,1,3,5,3,1,5,3,1,5,1,3,5,1,3,5,3,1,5,1,3,5,3,1,5,3,1,5,

%e 1,3,5,1,3,5,3,1,5,1,3,5,1,3,5,3,1,5,1,3,5,3,1,5,3,1,5,1,3,5,

%e 1,3,5,3,1,5,1,3,5,1,3,5,3,1,5,1,3,5,3,1,5,3,1,5,1,3,5,3,1,5,

%e 3,1,5,1,3,5,1,3,5,3,1,5,1,3,5,3,1,5,3,1,5,1,3,5,1,3,5,3,1,5,

%e 1,3,5,1,3,5,3,1,5,1,3,5,3,1,5,3,1,5,1,3,5,1,3,5,3,1,5,1,3,5,

%e 1,3,5,3,1,5,1,3,5,3,1,5,3,1,5,1,3,5,3,1,5,3,1,5,1,3,5,1,3,5,

%e 3,1,5,1,3,5,3,1,5,3,1,5,1,3,5,1,3,5,3,1,5,1,3,5,1,3,5,3,1,5,

%e 1,3,5,3,1,5,3,1,5,1,3,5,3,1,5,3,1,5,1,3,5,1,3,5,3,1,5,1,3,5,

%e 3,1,5,3,1,5,1,3,5,3,1,5,3,1,5,1,3,5,1,3,5,3,1,5,1,3,5,3,1,5,

%e 3,1,5,1,3,5,1,3,5,3,1,5,1,3,5,1,3,5,3,1,5,1,3,5,3,1,5,3,1,5,

%e 1,3,5,3,1,5,3,1,5,1,3,5,1,3,5,3,1,5,1,3,5,3,1,5,3,1,5,1,3,5,

%e 3,1,5,3,1,5,1,3,5,1,3,5,3,1,5,1,3,5,3,1,5,3,1,5,1,3,5,1,3,5,

%e 3,1,5,1,3,5,1,3,5,3,1,5,1,3,5,3,1,5,3,1,5,1,3,5,1,3,5,3,1,5,

%e 1,3,5,1,3,5,3,1,5,1,3,5,3,1,5,3,1,5,1,3,5,3,1,5,3,1,5,1,3,5,

%e 1,3,5,3,1,5,1,3,5,3,1,5,3,1,5,1,3,5,1,3,5,3,1,5,1,3,5,1,3,5,

%e 3,1,5,1,3,5,3,1,5,3,1,5,1,3,5,3,1,5,3,1,5,1,3,5,1,3,5,3,1,5,

%e 1,3,5,3,1,5,3,1,5,1,3,5,3,1,5,3,1,5,1,3,5,1,3,5,3,1,5,1,3,5,

%e 3,1,5,3,1,5,1,3,5,1,3,5,3,1,5,1,3,5,1,3,5,3,1,5,1,3,5,3,1,5,

%e 3,1,5,1,3,5,3,1,5,3,1,5,1,3,5,1,3,5,3,1,5,1,3,5,3,1,5,3,1,5,

%e 1,3,5,3,1,5,3,1,5,1,3,5,1,3,5,3,1,5,1,3,5,3,1,5,3,1,5,1,3,5,

%e 1,3,5,3,1,5,1,3,5,1,3,5,3,1,5,1,3,5,3,1,5,3,1,5,1,3,5,1,3,5,

%e 3,1,5,1,3,5,1,3,5,3,1,5,1,3,5,3,1,5,3,1,5,1,3,5,3,1,5,3,1,5,

%e 1,3,5,1,3,5,3,1,5,1,3,5,3,1,5,3,1,5,1,3,5,1,3,5,3,1,5,1,3,5,

%e 1,3,5,3,1,5,1,3,5,3,1,5,3,1,5,1,3,5,3,1,5,3,1,5,1,3,5,1,3,5,

%e 3,1,5,1,3,5,3,1,5,3,1,5,1,3,5,3,1,5,3,1,5,1,3,5,1,3,5,3,1,5,

%e 1,3,5,3,1,5,3,1,5,1,3,5,1,3,5,3,1,5,1,3,5,1,3,5,3,1,5,1,3,5,

%e 3,1,5,3,1,5,1,3,5,1,3,5,3,1,5,1,3,5,1,3,5,3,1,5,1,3,5,3,1,5,

%e 3,1,5,1,3,5,3,1,5,3,1,5,1,3,5,1,3,5,3,1,5,1,3,5,3,1,5,3,1,5,

%e 1,3,5,1,3,5,3,1,5,1,3,5,1,3,5,3,1,5,1,3,5,3,1,5,3,1,5,1,3,5,

%e 1,3,5,3,1,5,1,3,5,1,3,5,3,1,5,1,3,5,3,1,5,3,1,5,1,3,5,3,1,5,

%e 3,1,5,1,3,5,1,3,5,3,1,5,1,3,5,3,1,5,3,1,5,1,3,5,1,3,5,3,1,5,

%e 1,3,5,1,3,5,3,1,5,1,3,5,3,1,5,3,1,5,1,3,5,3,1,5,3,1,5,1,3,5, ...].

%e ...

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

%e z = 6.76134218926005257700977872218164992635251510297623\

%e 82400543083129376225709036432953076227076440718046\

%e 31453442451274796099781430687357859993593539615424\

%e 85162251478631501261337528918395764663363544002143\

%e 30566359813785491386584191187803105986019925125817\

%e 32384559737235970178404167070340277684175375836022\

%e 44216229551462644458639369128033581124838287016422\

%e 22465666402121115108802903360423241018391788636516\

%e 34399014131667135507041306777557247262858680853566\

%e 90766431107666031620853273656348374218315881851585\

%e 44479630341871788962324209888156077547855028352382\

%e 55925436674100345760249501114992870759647092383954\

%e 96007641595484979939448371742959430380908428799273\

%e 24364811976473376482587750783735175570880836352529\

%e 00570146062437832931408193409264288195544895081069\

%e 61581071442858044513706835327745695285899243308879\

%e 49562004677663100500730098181693102336025369092373\

%e 55940511090331284958638975881509935173849203422329\

%e 94160267792889652591286734230036989339097946760406\

%e 04381955878548467565031395792149142079025146844422...

%e ...

%e GENERATING METHOD.

%e Start with CF = [6] and repeat (PARI code):

%e {M = contfracpnqn(CF + vector(#CF,i, 21));

%e z = (1/4)*M[1,1]/M[2,1]; CF = contfrac(z)}

%e This method can be illustrated as follows.

%e z0 = [6] = 6 ;

%e z1 = (1/4)*[27] = [6; 1, 3] = 27/4 ;

%e z2 = (1/4)*[27; 22, 24] = [6; 1, 3, 5, 3, 1, 5, 4] = 14307/2116 ;

%e z3 = (1/4)*[27; 22, 24, 26, 24, 22, 26, 25] = [6; 1, 3, 5, 3, 1, 5, 3, 1, 5, 1, 3, 5, 1, 3, 5, 3, 1, 5, 1, 3, 6] = 32184125853/4760020267 ;

%e z4 = (1/4)*[27; 22, 24, 26, 24, 22, 26, 24, 22, 26, 22, 24, 26, 22, 24, 26, 24, 22, 26, 22, 24, 27] = [6; 1, 3, 5, 3, 1, 5, 3, 1, 5, 1, 3, 5, 1, 3, 5, 3, 1, 5, 1, 3, 5, 1, 3, 5, 3, 1, 5, 1, 3, 5, 3, 1, 5, 3, 1, 5, 1, 3, 5, 3, 1, 5, 3, 1, 5, 1, 3, 5, 1, 3, 5, 3, 1, 5, 1, 3, 5, 3, 1, 5, 3, 1, 6] = 668028962666848193442388706037/98801235607919425997683216483 ; ...

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

%o (PARI) /* Generate over 5000 terms */

%o {CF=[6]; for(i=1,8, M = contfracpnqn( CF + vector(#CF,i,21) ); z = (1/4)*M[1,1]/M[2,1]; CF = contfrac(z) )}

%o for(n=0,200,print1(CF[n+1],", "))

%Y Cf. A321090, A321091, A321092, A321094, A321095, A321096, A321097, A321098.

%K nonn,cofr

%O 0,1

%A _Paul D. Hanna_, Oct 28 2018