A321094 Decimal 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.

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

Offset

1,1

Links

Table of n, a(n) for n=1..154.

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, ..., A321093(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, ..., A321093(n) + 21, ...].
EXTENDED TERMS.
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...
...
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, ...].
...
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 4600 digits */
{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=1, 200, print1(floor(10^(n-1)*z)%10, ", "))

Crossrefs

Cf. A321090, A321091, A321092, A321093, A321095, A321096, A321097, A321098.

Sequence in context: A035414 A020505 A171538 * A330114 A011423 A239134

Adjacent sequences: A321091 A321092 A321093 * A321095 A321096 A321097

Keyword

nonn,cons

Author

Paul D. Hanna, Oct 28 2018

Status

approved