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A321092
Decimal 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.
9
4, 6, 9, 6, 7, 4, 3, 2, 8, 5, 9, 7, 0, 0, 2, 7, 9, 0, 9, 0, 3, 7, 9, 7, 1, 3, 5, 0, 6, 1, 4, 8, 9, 9, 6, 9, 5, 9, 6, 7, 6, 8, 9, 0, 3, 4, 9, 8, 2, 6, 6, 3, 9, 7, 4, 4, 9, 7, 6, 0, 6, 5, 8, 7, 6, 8, 9, 4, 9, 3, 0, 1, 8, 0, 5, 9, 9, 4, 9, 8, 2, 2, 7, 4, 6, 7, 9, 3, 8, 7, 8, 8, 4, 5, 2, 6, 6, 8, 7, 6, 6, 7, 3, 5, 5, 4, 3, 3, 7, 3, 7, 9, 1, 7, 9, 7, 2, 8, 0, 8, 7, 7, 1, 0, 8, 6, 8, 2, 2, 0, 9, 5, 7, 9, 0, 8, 6, 6, 2, 6, 1, 3, 6, 1, 4, 0, 3, 2, 7, 9, 8, 3, 3, 3, 8, 8, 4, 0, 5, 7, 2, 7, 5, 2, 0, 0, 8, 9, 1, 3, 3, 8, 2, 5, 0, 6, 5, 1, 0, 4, 5, 5, 3, 0, 6, 3, 8, 3, 3, 4, 7, 1, 8, 5, 7, 2, 3, 6, 7, 8, 1
OFFSET
1,1
LINKS
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, ..., A321091(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, ..., A321091(n) + 10, ...].
EXTENDED TERMS.
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...
...
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, ...].
Is there a pattern to the above terms?
...
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 3600 digits */
{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=1, 200, print1(floor(10^(n-1)*z)%10, ", "))
KEYWORD
nonn,cons
AUTHOR
Paul D. Hanna, Oct 27 2018
STATUS
approved