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Decimal expansion of the constant z that satisfies: CF(6*z, n) = CF(z, n) + 55, for n >= 0, where CF(z, n) denotes the n-th partial denominator in the continued fraction expansion of z.
8

%I #11 Nov 01 2018 06:47:19

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

%T 7,8,7,6,5,9,7,6,8,5,9,5,3,4,9,3,1,1,7,3,8,4,0,5,0,9,7,5,6,9,3,1,9,6,

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

%N Decimal expansion of the constant z that satisfies: CF(6*z, n) = CF(z, n) + 55, for n >= 0, where CF(z, n) denotes the n-th partial denominator in the continued fraction expansion of z.

%e The decimal expansion of this constant z begins:

%e z = 10.8363086385325049431870358764271278765976859534931...

%e The simple continued fraction expansion of z begins:

%e z = [10; 1, 5, 9, 5, 1, 9, 5, 1, 9, 1, 5, 9, 1, 5, 9, ..., A321097(n), ...];

%e such that the simple continued fraction expansion of 6*z begins:

%e 6*z = [65; 56, 60, 64, 60, 56, 64, 60, 56, 64, 56, ..., A321097(n) + 55, ...].

%e EXTENDED TERMS.

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

%e z = 10.83630863853250494318703587642712787659768595349311\

%e 73840509756931960001734110072237735215798152265637\

%e 46033548966251275029451235255917154869550414547346\

%e 45601898326659069509029896766448634581870902999261\

%e 78291099037993947368425232025742840508201019811150\

%e 20208289541868116590985746685817208034834182741861\

%e 61586263073936595659616093596727391439370392218179\

%e 08547782927594504604528661115974783060857978290729\

%e 53554586787471663938331763610007750862560295292956\

%e 22583160832720034539915220107654291931753328805663\

%e 44405451280922502018454665640681719991329902449206\

%e 06333718948414803434770198192597675071144159105469\

%e 40129387536502210902718153383173369508615022733071\

%e 21561771111264471719424048701509094587624798702003\

%e 98051339274318126056502629341701820569809346581703\

%e 17900710636878987844980734936343020769115171474588\

%e 61969660741202379814909712010859009313616125172084\

%e 48790047790048120552938902316397984428482656427316\

%e 35549261064509815824229787948467548551187722067240\

%e 97844304015886508690953002055508602378218606168521...

%e ...

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

%e z = [10;1,5,9,5,1,9,5,1,9,1,5,9,1,5,9,5,1,9,1,5,9,1,5,9,5,1,9,1,5,9,

%e 5,1,9,5,1,9,1,5,9,5,1,9,5,1,9,1,5,9,1,5,9,5,1,9,1,5,9,5,1,9,

%e 5,1,9,1,5,9,5,1,9,5,1,9,1,5,9,1,5,9,5,1,9,1,5,9,5,1,9,5,1,9,

%e 1,5,9,1,5,9,5,1,9,1,5,9,1,5,9,5,1,9,1,5,9,5,1,9,5,1,9,1,5,9,

%e 1,5,9,5,1,9,1,5,9,1,5,9,5,1,9,1,5,9,5,1,9,5,1,9,1,5,9,5,1,9,

%e 5,1,9,1,5,9,1,5,9,5,1,9,1,5,9,5,1,9,5,1,9,1,5,9,1,5,9,5,1,9,

%e 1,5,9,1,5,9,5,1,9,1,5,9,5,1,9,5,1,9,1,5,9,1,5,9,5,1,9,1,5,9,

%e 1,5,9,5,1,9,1,5,9,5,1,9,5,1,9,1,5,9,5,1,9,5,1,9,1,5,9,1,5,9,

%e 5,1,9,1,5,9,5,1,9,5,1,9,1,5,9,1,5,9,5,1,9,1,5,9,1,5,9,5,1,9,

%e 1,5,9,5,1,9,5,1,9,1,5,9,5,1,9,5,1,9,1,5,9,1,5,9,5,1,9,1,5,9,

%e 5,1,9,5,1,9,1,5,9,5,1,9,5,1,9,1,5,9,1,5,9,5,1,9,1,5,9,5,1,9,

%e 5,1,9,1,5,9,1,5,9,5,1,9,1,5,9,1,5,9,5,1,9,1,5,9,5,1,9,5,1,9,

%e 1,5,9,5,1,9,5,1,9,1,5,9,1,5,9,5,1,9,1,5,9,5,1,9,5,1,9,1,5,9,

%e 5,1,9,5,1,9,1,5,9,1,5,9,5,1,9,1,5,9,5,1,9,5,1,9,1,5,9,1,5,9,

%e 5,1,9,1,5,9,1,5,9,5,1,9,1,5,9,5,1,9,5,1,9,1,5,9,1,5,9,5,1,9,

%e 1,5,9,1,5,9,5,1,9,1,5,9,5,1,9,5,1,9,1,5,9,5,1,9,5,1,9,1,5,9,

%e 1,5,9,5,1,9,1,5,9,5,1,9,5,1,9,1,5,9,1,5,9,5,1,9,1,5,9,1,5,9,

%e 5,1,9,1,5,9,5,1,9,5,1,9,1,5,9,5,1,9,5,1,9,1,5,9,1,5,9,5,1,9,

%e 1,5,9,5,1,9,5,1,9,1,5,9,5,1,9,5,1,9,1,5,9,1,5,9,5,1,9,1,5,9,

%e 5,1,9,5,1,9,1,5,9,1,5,9,5,1,9,1,5,9,1,5,9,5,1,9,1,5,9,5,1,9,

%e 5,1,9,1,5,9,5,1,9,5,1,9,1,5,9,1,5,9,5,1,9,1,5,9,5,1,9,5,1,9,

%e 1,5,9,5,1,9,5,1,9,1,5,9,1,5,9,5,1,9,1,5,9,5,1,9,5,1,9,1,5,9,

%e 1,5,9,5,1,9,1,5,9,1,5,9,5,1,9,1,5,9,5,1,9,5,1,9,1,5,9,1,5,9,

%e 5,1,9,1,5,9,1,5,9,5,1,9,1,5,9,5,1,9,5,1,9,1,5,9,5,1,9,5,1,9,

%e 1,5,9,1,5,9,5,1,9,1,5,9,5,1,9,5,1,9,1,5,9,1,5,9,5,1,9,1,5,9,

%e 1,5,9,5,1,9,1,5,9,5,1,9,5,1,9,1,5,9,5,1,9,5,1,9,1,5,9,1,5,9,

%e 5,1,9,1,5,9,5,1,9,5,1,9,1,5,9,5,1,9,5,1,9,1,5,9,1,5,9,5,1,9,

%e 1,5,9,5,1,9,5,1,9,1,5,9,1,5,9,5,1,9,1,5,9,1,5,9,5,1,9,1,5,9,

%e 5,1,9,5,1,9,1,5,9,1,5,9,5,1,9,1,5,9,1,5,9,5,1,9,1,5,9,5,1,9,

%e 5,1,9,1,5,9,5,1,9,5,1,9,1,5,9,1,5,9,5,1,9,1,5,9,5,1,9,5,1,9,

%e 1,5,9,1,5,9,5,1,9,1,5,9,1,5,9,5,1,9,1,5,9,5,1,9,5,1,9,1,5,9,

%e 1,5,9,5,1,9,1,5,9,1,5,9,5,1,9,1,5,9,5,1,9,5,1,9,1,5,9,5,1,9,

%e 5,1,9,1,5,9,1,5,9,5,1,9,1,5,9,5,1,9,5,1,9,1,5,9,1,5,9,5,1,9,

%e 1,5,9,1,5,9,5,1,9,1,5,9,5,1,9,5,1,9,1,5,9,5,1,9,5,1,9,1,5,9, ...].

%e ...

%e GENERATING METHOD.

%e Start with CF = [10] and repeat (PARI code):

%e {M = contfracpnqn(CF + vector(#CF,i, 55));

%e z = (1/6)*M[1,1]/M[2,1]; CF = contfrac(z)}

%e This method can be illustrated as follows.

%e z0 = [10] = 10;

%e z1 = (1/6)*[65] = [10; 1, 5] = 65/6;

%e z2 = (1/6)*[65; 56, 60] = [10; 1, 5, 9, 5, 1, 9, 6] = 218525/20166;

%e z3 = (1/6)*[65; 56, 60, 64, 60, 56, 64, 61] = [10; 1, 5, 9, 5, 1, 9, 5, 1, 9, 1, 5, 9, 1, 5, 9, 5, 1, 9, 1, 5, 10] = 30617277049665/2825434201901;

%e z4 = (1/6)*[65; 56, 60, 64, 60, 56, 64, 60, 56, 64, 56, 60, 64, 56, 60, 64, 60, 56, 64, 56, 60, 65] = [10; 1, 5, 9, 5, 1, 9, 5, 1, 9, 1, 5, 9, 1, 5, 9, 5, 1, 9, 1, 5, 9, 1, 5, 9, 5, 1, 9, 1, 5, 9, 5, 1, 9, 5, 1, 9, 1, 5, 9, 5, 1, 9, 5, 1, 9, 1, 5, 9, 1, 5, 9, 5, 1, 9, 1, 5, 9, 5, 1, 9, 5, 1, 10] = 235326213809918755668077578309692661245/21716455451732827969266335806481498321;

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

%o (PARI) /* Generate over 6000 digits */

%o {CF=[10]; for(i=1,8, M = contfracpnqn( CF + vector(#CF,i,55) ); z = (1/6)*M[1,1]/M[2,1]; CF = contfrac(z) )}

%o for(n=0,200,print1(floor(10^(n-1)*z)%10,", "))

%Y Cf. A321090, A321091, A321092, A321093, A321094, A321095, A321096, A321097.

%K nonn,cons

%O 0,3

%A _Paul D. Hanna_, Oct 28 2018

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Last modified September 22 20:46 EDT 2024. Contains 376140 sequences. (Running on oeis4.)