A321091 Continued fraction expansion of the constant z that satisfies: CF(3*z, n) = CF(z, n) + 10, for n >= 0, where CF(z, n) denotes the n-th partial denominator in the continued fraction expansion of z.

4, 1, 2, 3, 2, 1, 3, 2, 1, 3, 1, 2, 3, 1, 2, 3, 2, 1, 3, 1, 2, 3, 1, 2, 3, 2, 1, 3, 1, 2, 3, 2, 1, 3, 2, 1, 3, 1, 2, 3, 2, 1, 3, 2, 1, 3, 1, 2, 3, 1, 2, 3, 2, 1, 3, 1, 2, 3, 2, 1, 3, 2, 1, 3, 1, 2, 3, 2, 1, 3, 2, 1, 3, 1, 2, 3, 1, 2, 3, 2, 1, 3, 1, 2, 3, 2, 1, 3, 2, 1, 3, 1, 2, 3, 1, 2, 3, 2, 1, 3, 1, 2, 3, 1, 2, 3, 2, 1, 3, 1, 2, 3, 2, 1, 3, 2, 1, 3, 1, 2, 3, 1, 2, 3, 2, 1, 3, 1, 2, 3, 1, 2, 3, 2, 1, 3, 1, 2, 3, 2, 1, 3, 2, 1, 3, 1, 2, 3, 2, 1, 3, 2, 1, 3, 1, 2, 3, 1, 2, 3, 2, 1, 3, 1, 2, 3, 2, 1, 3, 2, 1, 3, 1, 2, 3, 1, 2, 3, 2, 1, 3, 1, 2, 3, 1, 2, 3, 2, 1, 3, 1, 2, 3, 2, 1, 3, 2, 1, 3, 1, 2

Offset

0,1

Links

Paul D. Hanna, Table of n, a(n) for n = 0..10000

Formula

Formula for terms:

(1) a(0) = 4,

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

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

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

(5) a(9*n+7) = 2 for n >= 0,

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

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

a(n) = A321090(n) + 1 for n >= 1, with a(0) = 4.

a(n) = A321093(n) - A321090(n), for n >= 0.

a(n) = A321095(n) - 2*A321090(n), for n >= 0.

a(n) = A321097(n) - 3*A321090(n), for n >= 0.

Example

The decimal expansion of this constant z begins:
z = 4.69674328597002790903797135061489969596768903498266...
The simple continued fraction expansion of z begins:
z = [4; 1, 2, 3, 2, 1, 3, 2, 1, 3, 1, 2, 3, 1, 2, 3, ..., a(n), ...];
such that the simple continued fraction expansion of 3*z begins:
3*z = [14; 11, 12, 13, 12, 11, 13, 12, 11, 13, 11, 12, ..., a(n) + 10, ...].
RELATED CONTINUED FRACTIONS.
Continued fractions for z/2 and z/3 are also interesting:
z/2 = [2; 2, 1, 6, 1, 2, 1, 1, 2, 8, 2, 1, 1, 2, 1, 6, 1, 2, 1, 1, 5, 1, 1, 5, 1, 1, 2, 8, 2, 1, 1, 5, 1, 1, 5, 1, 1, 2, 1, 6, 1, 2, 1, 1, 2, 8, 2, ...];
z/3 = [1; 1, 1, 3, 3, 4, 1, 8, 2, 1, 3, 2, 1, 3, 3, 4, 1, 4, 3, 8, 1, 8, 2, 1, 3, 3, 4, 1, 8, 2, 1, 3, 3, 4, 1, 8, 2, 1, 3, 2, 1, 3, 3, 4, 1, 4, 3, 3, ...].
EXTENDED TERMS.
The initial 1020 terms of the continued fraction of z are
Z = [4;1,2,3,2,1,3,2,1,3,1,2,3,1,2,3,2,1,3,1,2,3,1,2,3,2,1,3,1,2,3,
2,1,3,2,1,3,1,2,3,2,1,3,2,1,3,1,2,3,1,2,3,2,1,3,1,2,3,2,1,3,
2,1,3,1,2,3,2,1,3,2,1,3,1,2,3,1,2,3,2,1,3,1,2,3,2,1,3,2,1,3,
1,2,3,1,2,3,2,1,3,1,2,3,1,2,3,2,1,3,1,2,3,2,1,3,2,1,3,1,2,3,
1,2,3,2,1,3,1,2,3,1,2,3,2,1,3,1,2,3,2,1,3,2,1,3,1,2,3,2,1,3,
2,1,3,1,2,3,1,2,3,2,1,3,1,2,3,2,1,3,2,1,3,1,2,3,1,2,3,2,1,3,
1,2,3,1,2,3,2,1,3,1,2,3,2,1,3,2,1,3,1,2,3,1,2,3,2,1,3,1,2,3,
1,2,3,2,1,3,1,2,3,2,1,3,2,1,3,1,2,3,2,1,3,2,1,3,1,2,3,1,2,3,
2,1,3,1,2,3,2,1,3,2,1,3,1,2,3,1,2,3,2,1,3,1,2,3,1,2,3,2,1,3,
1,2,3,2,1,3,2,1,3,1,2,3,2,1,3,2,1,3,1,2,3,1,2,3,2,1,3,1,2,3,
2,1,3,2,1,3,1,2,3,2,1,3,2,1,3,1,2,3,1,2,3,2,1,3,1,2,3,2,1,3,
2,1,3,1,2,3,1,2,3,2,1,3,1,2,3,1,2,3,2,1,3,1,2,3,2,1,3,2,1,3,
1,2,3,2,1,3,2,1,3,1,2,3,1,2,3,2,1,3,1,2,3,2,1,3,2,1,3,1,2,3,
2,1,3,2,1,3,1,2,3,1,2,3,2,1,3,1,2,3,2,1,3,2,1,3,1,2,3,1,2,3,
2,1,3,1,2,3,1,2,3,2,1,3,1,2,3,2,1,3,2,1,3,1,2,3,1,2,3,2,1,3,
1,2,3,1,2,3,2,1,3,1,2,3,2,1,3,2,1,3,1,2,3,2,1,3,2,1,3,1,2,3,
1,2,3,2,1,3,1,2,3,2,1,3,2,1,3,1,2,3,1,2,3,2,1,3,1,2,3,1,2,3,
2,1,3,1,2,3,2,1,3,2,1,3,1,2,3,2,1,3,2,1,3,1,2,3,1,2,3,2,1,3,
1,2,3,2,1,3,2,1,3,1,2,3,2,1,3,2,1,3,1,2,3,1,2,3,2,1,3,1,2,3,
2,1,3,2,1,3,1,2,3,1,2,3,2,1,3,1,2,3,1,2,3,2,1,3,1,2,3,2,1,3,
2,1,3,1,2,3,2,1,3,2,1,3,1,2,3,1,2,3,2,1,3,1,2,3,2,1,3,2,1,3,
1,2,3,2,1,3,2,1,3,1,2,3,1,2,3,2,1,3,1,2,3,2,1,3,2,1,3,1,2,3,
1,2,3,2,1,3,1,2,3,1,2,3,2,1,3,1,2,3,2,1,3,2,1,3,1,2,3,1,2,3,
2,1,3,1,2,3,1,2,3,2,1,3,1,2,3,2,1,3,2,1,3,1,2,3,2,1,3,2,1,3,
1,2,3,1,2,3,2,1,3,1,2,3,2,1,3,2,1,3,1,2,3,1,2,3,2,1,3,1,2,3,
1,2,3,2,1,3,1,2,3,2,1,3,2,1,3,1,2,3,2,1,3,2,1,3,1,2,3,1,2,3,
2,1,3,1,2,3,2,1,3,2,1,3,1,2,3,2,1,3,2,1,3,1,2,3,1,2,3,2,1,3,
1,2,3,2,1,3,2,1,3,1,2,3,1,2,3,2,1,3,1,2,3,1,2,3,2,1,3,1,2,3,
2,1,3,2,1,3,1,2,3,1,2,3,2,1,3,1,2,3,1,2,3,2,1,3,1,2,3,2,1,3,
2,1,3,1,2,3,2,1,3,2,1,3,1,2,3,1,2,3,2,1,3,1,2,3,2,1,3,2,1,3,
1,2,3,1,2,3,2,1,3,1,2,3,1,2,3,2,1,3,1,2,3,2,1,3,2,1,3,1,2,3,
1,2,3,2,1,3,1,2,3,1,2,3,2,1,3,1,2,3,2,1,3,2,1,3,1,2,3,2,1,3,
2,1,3,1,2,3,1,2,3,2,1,3,1,2,3,2,1,3,2,1,3,1,2,3,1,2,3,2,1,3,
1,2,3,1,2,3,2,1,3,1,2,3,2,1,3,2,1,3,1,2,3,2,1,3,2,1,3,1,2,3, ...].
...
The initial 1000 digits in the decimal expansion of z are
z = 4.69674328597002790903797135061489969596768903498266\
39744976065876894930180599498227467938788452668766\
73554337379179728087710868220957908662613614032798\
33388405727520089133825065104553063833471857236781\
30005222692602670820049366351838443544785254486817\
36713432024147916330132466297271088195927148747751\
61081134504127224293197514421215072180791056995531\
25282254039576642227973626045065085745006810232418\
31151864601858865871496634133860069172907530299184\
80599130262819282015136761682297696200354791299160\
72099154048193374507277756176907149849150248588944\
75702959853076891940422757945105365738782828309624\
89182729519410181321396021772327752921026193693551\
52235778358918181495624102484144418903334090227672\
68450214362152729231740655406007216125545132538964\
63321922981643984915752295515263408732183582572996\
74985150593020685391286450747231245540741605404683\
64053241113305130300450809189365250050022411738651\
75283281124343007394094179023437577924273245108697\
96469643376214173186511094645179990191104328112759...
...
GENERATING METHOD.
Start with CF = [4] and repeat (PARI code):
{M = contfracpnqn(CF + vector(#CF,i, 10));
z = (1/3)*M[1,1]/M[2,1]; CF = contfrac(z)}
This method can be illustrated as follows.
z0 = [4] = 4 ;
z1 = (1/3)*[14] = [4; 1, 2] = 14/3 ;
z2 = (1/3)*[14; 11, 12] = [4, 1, 2, 3, 2, 1, 3, 3] = 1874/399 ;
z3 = (1/3)*[14; 11, 12, 13, 12, 11, 13, 13] = [4; 1, 2, 3, 2, 1, 3, 2, 1, 3, 1, 2, 3, 1, 2, 3, 2, 1, 3, 1, 2, 4] = 187227588/39863279 ;
z4 = (1/3)*[14; 11, 12, 13, 12, 11, 13, 12, 11, 13, 11, 12, 13, 11, 12, 13, 12, 11, 13, 11, 12, 14] = [4; 1, 2, 3, 2, 1, 3, 2, 1, 3, 1, 2, 3, 1, 2, 3, 2, 1, 3, 1, 2, 3, 1, 2, 3, 2, 1, 3, 1, 2, 3, 2, 1, 3, 2, 1, 3, 1, 2, 3, 2, 1, 3, 2, 1, 3, 1, 2, 3, 1, 2, 3, 2, 1, 3, 1, 2, 3, 2, 1, 3, 2, 1, 4] = 254413274852180460063336/54168017999228580539039 ; ...
where this constant z equals the limit of the iterations of the above process.

Prog

(PARI) /* Generate over 5000 terms */
{CF=[4]; for(i=1, 8, M = contfracpnqn( CF + vector(#CF, i, 10) ); z = (1/3)*M[1, 1]/M[2, 1]; CF = contfrac(z) )}
for(n=0, 200, print1(CF[n+1], ", "))

Crossrefs

Cf. A321090, A321092, A321093, A321094, A321095, A321096, A321097, A321098.

Sequence in context: A010473 A176041 A131100 * A375576 A064124 A106654

Adjacent sequences: A321088 A321089 A321090 * A321092 A321093 A321094

Keyword

nonn,cofr

Author

Paul D. Hanna, Oct 27 2018

Status

approved