A320953 - OEIS

A320953

Decimal expansion of the constant t having the continued fraction expansion {d(n), n>=0} such that the continued fraction expansion of 4*t yields partial denominators {5*d(n), n>=0}.

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

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OFFSET

1,2

COMMENTS

Is this constant transcendental?

Compare to the continued fraction expansions of sqrt(3) and 3*sqrt(3), which are related by a factor of 5: sqrt(3) = [1; 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, ...] and 3*sqrt(3) = [5; 5, 10, 5, 10, 5, 10, 5, 10, 5, 10, ...].

Further, let CF(x) denote the simple continued fraction expansion of x, then we have the related identities which hold for n >= 1:

(C1) CF( (4*n+1) * sqrt((n+1)/n) ) = (4*n+3) * CF( sqrt((n+1)/n) ),

(C2) CF( (2*n+1) * sqrt((n+2)/n) ) = (2*n+3) * CF( sqrt((n+2)/n) ).

LINKS

Paul D. Hanna, Table of n, a(n) for n = 1..5000

FORMULA

Given t = [d(0); d(1), d(2), d(3), d(4), d(5), d(6), ..., A320952(n), ...], some related simple continued fractions are:

(1) 4*t = [5*d(0); 5*d(1), 5*d(2), 5*d(3), 5*d(4), 5*d(5), ...],

(2) 4*t/5 = [d(0); 25*d(1), d(2), 25*d(3), d(4), 25*d(5), d(6), ...],

(3) 20*t = [25*d(0); d(1), 25*d(2), d(3), 25*d(4), d(5), 25*d(6), ...].

EXAMPLE

The decimal expansion of this constant t begins:

t = 1.266450214578651613778772929010259237236966545054753300558099305...

The simple continued fraction expansion of t begins:

t = [1; 3, 1, 3, 20, 3, 1, 3, 24, 1, 3, 3, 1, 1, 4, 1, 1, 3, 3, 1, 29, 3, 2, 1, 14, 1, 1, 3, 3, 2, 1, 4, 1, 1, 4, 1, 1, 4, 1, 2, ..., A320952(n), ...]

such that the simple continued fraction expansion of 4*t begins:

4*t = [5; 15, 5, 15, 100, 15, 5, 15, 120, 5, 15, 15, 5, 5, 20, 5, 5, 15, 15, 5, 145, 15, 10, 5, 70, 5, 5, 15, 15, 10, 5, 20, 5, 5, ..., 5*A320952(n), ...].

...

The initial 1000 digits in the decimal expansion of t are

t = 1.26645021457865161377877292901025923723696654505475\

33005580993053624272184801354973578017721161093902\

83186837335260733121712163054501018742646583790643\

54875833008007285407818429195692996453113827495305\

54926835731865875321451747328930568555229448988608\

05786435870905452173007164597651669613272647900654\

83894404364902408305059276949480386935767775514765\

94968299951493100831611976267451605948933885299869\

07863756473354702971788204675428850345468192538035\

85810263279458657769543507292924716821458977316162\

73791864200695026130548141231704085623717435331603\

39200767062558929533702255394800104584764699667049\

25260260287699076614416015170423275489461158402478\

17149290630968365611810185428241284526255760519651\

94435407858521668152963500082382089516858099107232\

93046639826593759422087336891654489562762696517528\

97543506563610723909866277201274334313308836147313\

53053566946629955949015183419229625850457388237270\

77912709245178910220379095072759755794410149465362\

47442129159574142863549600623749608271595134828851...

...

The initial 528 terms in the continued fraction expansion of t are

t = [1;3,1,3,20,3,1,3,24,1,3,3,1,1,4,1,1,3,3,1,29,3,2,1,14,

1,1,3,3,2,1,4,1,1,4,1,1,4,1,2,3,3,1,1,14,1,2,3,36,

60,2,1,1,4,1,1,17,20,1,3,1,3,60,2,1,1,4,1,1,4,1,1,4,

1,2,3,4,1,3,1,20,5,20,2,1,1,14,1,1,3,3,2,1,4,1,1,17,

20,2,1,1,14,1,1,44,1,1,299,1,1,2,20,1,3,1,4,3,2,1,4,1,

1,20,1,3,24,1,3,1,60,1,3,1,3,1200,2,1,1,4,1,2,3,4,1,3,

1,20,5,20,1,3,1,4,3,2,1,9,1,1,3,3,1,4,3,2,1,14,1,2,

3,24,1,3,6,400,2,1,1,4,1,2,3,17,3,2,1,4,1,1,3,3,1,3,

40,1,3,1,4,3,2,1,4,1,1,20,1,3,24,1,3,2,3,2,1,4,1,1,

17,20,1,3,1,54,3,2,1,4,1,1,373,3,2,1,4,1,1,2,400,1,3,1,

3,20,5,60,2,1,1,4,1,1,4,1,1,4,1,2,3,24,1,3,1,60,30,20,

3,1,3,1,1200,1,3,1,3,20,3,1,3,1499,1,3,2,3,2,1,4,1,1,4,

1,1,4,1,1,2,60,5,20,3,1,3,1,400,6,3,1,24,3,2,1,14,1,2,

3,4,1,3,3,1,1,9,1,2,3,11,20,1,3,1,3,60,1,3,1,4,3,1,

3,40,1,3,1,16,1,3,1,40,3,1,3,29,1,3,1,60,7,1,1,1999,1,1,

2,20,1,3,1,4,3,2,1,9,1,1,3,3,1,20,1,1,14,1,1,2,20,5,

20,1,3,1,3,60,1,3,1,3,800,1,3,1,3,20,5,60,2,1,1,4,1,1,

4,1,1,4,1,2,3,24,1,3,1,60,30,20,3,1,3,2,3,1,3,40,1,3,

1,4,3,2,1,4,1,1,20,1,3,24,1,3,1,60,1,3,1,66,1,3,3,1,

1,9,1,2,3,4,1,3,1,20,466,3,1,3,40,1,3,1,4,3,2,1,4,1,

1,2,8000,1,3,1,3,20,3,1,3,24,1,3,6,1200,2,1,1,4,1,2,3,4,

1,3,1,20,5,20,1,3,1,4,3,2,1,9,1,1,3,3,1,29,3,2,1, ...].

...

GENERATING METHOD.

Start with CF = [1] and repeat (PARI code):

t = (1/4)*contfracpnqn(5*CF)[1,1]/contfracpnqn(5*CF)[2,1]; CF = contfrac(t)

This method can be illustrated as follows.

t0 = [1] = 1;

t1 = (1/4)*[5] = [1; 4] = 5/4;

t2 = (1/4)*[5; 20] = [1; 3, 1, 4, 4] = 101/80;

t3 = (1/4)*[5; 15, 5, 20, 20] = [1; 3, 1, 3, 20, 5, 80] = 155905/123104;

t4 = (1/4)*[5; 15, 5, 15, 100, 25, 400] = [1; 3, 1, 3, 20, 3, 1, 3, 24, 1, 3, 6, 1600] = 5857775885/4625350304;

t5 = (1/4)*[5; 15, 5, 15, 100, 15, 5, 15, 120, 5, 15, 30, 8000] = [1; 3, 1, 3, 20, 3, 1, 3, 24, 1, 3, 3, 1, 1, 4, 1, 1, 3, 3, 1, 29, 3, 2, 1, 14, 1, 1, 7, 32000] = 1487626504292631625/1174642703809375904; ...

where this constant t equals the limit of the iterations of the above process.

PROG

(PARI) /* Generate over 5700 digits in the decimal expansion */

CF=[1];

{for(i=1, 12, t = (1/4)*contfracpnqn(5*CF)[1, 1]/contfracpnqn(5*CF)[2, 1];

CF = contfrac(t) ); }

for(n=1, 150, print1(floor(10^(n-1)*t)%10, ", "))

CROSSREFS

Cf. A320952, A320413, A320411.

Sequence in context: A316164 A319276 A141329 * A110388 A007507 A350031

Adjacent sequences: A320950 A320951 A320952 * A320954 A320955 A320956

KEYWORD

nonn,cons

AUTHOR

Paul D. Hanna, Oct 26 2018

STATUS

approved