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A321093
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.
9
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, 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, 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
OFFSET
0,1
FORMULA
Formula for terms:
(1) a(0) = 6,
(2) a(3*n) = 5 for n >= 1,
(3) a(3*n+2) = 4 - a(3*n+1) for n >= 0,
(4) a(9*n+1) = 1 for n >= 0,
(5) a(9*n+7) = 3 for n >= 0,
(6) a(9*n+4) = 4 - a(3*n+1) for n >= 0.
a(3*n+1) = 2*A189706(n+1) + 1 for n >= 0.
a(n) = 2*A321090(n) + 1 for n >= 1, with a(0) = 6.
a(n) = A321091(n) + A321090(n), for n >= 0.
a(n) = A321095(n) - A321090(n), for n >= 0.
a(n) = A321097(n) - 2*A321090(n), for n >= 0.
EXAMPLE
The decimal expansion of this constant z begins:
z = 6.76134218926005257700977872218164992635251510297623...
The simple continued fraction expansion of z begins:
z = [6; 1, 3, 5, 3, 1, 5, 3, 1, 5, 1, 3, 5, 1, 3, 5, 3, ..., a(n), ...];
such that the simple continued fraction expansion of 4*z begins:
4*z = [27; 22, 24, 26, 24, 22, 26, 24, 22, 26, 22, 24, ..., a(n) + 21, ...].
EXTENDED TERMS.
The initial 1020 terms of the continued fraction of z are
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,
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,
3,1,5,1,3,5,3,1,5,3,1,5,1,3,5,1,3,5,3,1,5,1,3,5,3,1,5,3,1,5,
1,3,5,1,3,5,3,1,5,1,3,5,1,3,5,3,1,5,1,3,5,3,1,5,3,1,5,1,3,5,
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,
1,3,5,1,3,5,3,1,5,1,3,5,3,1,5,3,1,5,1,3,5,1,3,5,3,1,5,1,3,5,
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,
1,3,5,3,1,5,3,1,5,1,3,5,3,1,5,3,1,5,1,3,5,1,3,5,3,1,5,1,3,5,
3,1,5,3,1,5,1,3,5,3,1,5,3,1,5,1,3,5,1,3,5,3,1,5,1,3,5,3,1,5,
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,
1,3,5,3,1,5,3,1,5,1,3,5,1,3,5,3,1,5,1,3,5,3,1,5,3,1,5,1,3,5,
3,1,5,3,1,5,1,3,5,1,3,5,3,1,5,1,3,5,3,1,5,3,1,5,1,3,5,1,3,5,
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,
3,1,5,1,3,5,3,1,5,3,1,5,1,3,5,3,1,5,3,1,5,1,3,5,1,3,5,3,1,5,
1,3,5,3,1,5,3,1,5,1,3,5,3,1,5,3,1,5,1,3,5,1,3,5,3,1,5,1,3,5,
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,
3,1,5,1,3,5,3,1,5,3,1,5,1,3,5,1,3,5,3,1,5,1,3,5,3,1,5,3,1,5,
1,3,5,3,1,5,3,1,5,1,3,5,1,3,5,3,1,5,1,3,5,3,1,5,3,1,5,1,3,5,
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,
1,3,5,3,1,5,1,3,5,3,1,5,3,1,5,1,3,5,3,1,5,3,1,5,1,3,5,1,3,5,
3,1,5,1,3,5,3,1,5,3,1,5,1,3,5,3,1,5,3,1,5,1,3,5,1,3,5,3,1,5,
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,
1,3,5,1,3,5,3,1,5,1,3,5,1,3,5,3,1,5,1,3,5,3,1,5,3,1,5,1,3,5,
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,
1,3,5,1,3,5,3,1,5,1,3,5,3,1,5,3,1,5,1,3,5,3,1,5,3,1,5,1,3,5, ...].
...
The initial 1000 digits in the decimal expansion of z are
z = 6.76134218926005257700977872218164992635251510297623\
82400543083129376225709036432953076227076440718046\
31453442451274796099781430687357859993593539615424\
85162251478631501261337528918395764663363544002143\
30566359813785491386584191187803105986019925125817\
32384559737235970178404167070340277684175375836022\
44216229551462644458639369128033581124838287016422\
22465666402121115108802903360423241018391788636516\
34399014131667135507041306777557247262858680853566\
90766431107666031620853273656348374218315881851585\
44479630341871788962324209888156077547855028352382\
55925436674100345760249501114992870759647092383954\
96007641595484979939448371742959430380908428799273\
24364811976473376482587750783735175570880836352529\
00570146062437832931408193409264288195544895081069\
61581071442858044513706835327745695285899243308879\
49562004677663100500730098181693102336025369092373\
55940511090331284958638975881509935173849203422329\
94160267792889652591286734230036989339097946760406\
04381955878548467565031395792149142079025146844422...
...
GENERATING METHOD.
Start with CF = [6] and repeat (PARI code):
{M = contfracpnqn(CF + vector(#CF,i, 21));
z = (1/4)*M[1,1]/M[2,1]; CF = contfrac(z)}
This method can be illustrated as follows.
z0 = [6] = 6 ;
z1 = (1/4)*[27] = [6; 1, 3] = 27/4 ;
z2 = (1/4)*[27; 22, 24] = [6; 1, 3, 5, 3, 1, 5, 4] = 14307/2116 ;
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 ;
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 ; ...
where this constant z equals the limit of the iterations of the above process.
PROG
(PARI) /* Generate over 5000 terms */
{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) )}
for(n=0, 200, print1(CF[n+1], ", "))
KEYWORD
nonn,cofr
AUTHOR
Paul D. Hanna, Oct 28 2018
STATUS
approved