A081771 -id:A081771 - cp's OEIS frontend

A061678 Continued fraction for Sum_{n>=0} 1/3^(3^n).

0, 2, 1, 2, 3, 26, 1, 2, 2, 1, 2, 19682, 1, 1, 1, 2, 2, 1, 26, 3, 2, 1, 2, 7625597484986, 1, 1, 1, 2, 3, 26, 1, 2, 2, 1, 1, 1, 19682, 2, 1, 2, 2, 1, 26, 3, 2, 1, 2, 443426488243037769948249630619149892802, 1, 1, 1, 2, 3, 26, 1, 2, 2, 1, 2, 19682

Offset: 0

Jason Earls, Jun 23 2001

The continued fraction has a "folded" overall structure. Apart from a(0) and from the record values of the form 3^(3^k)-1 (k >= 0), the only terms are 1 and 3. This follows from the theorem in Shallit's paper. - Georg Fischer, Aug 29 2022

			0.370421175633926798495743189411...

Harry J. Smith, Table of n, a(n) for n = 0..382
Jeffrey O. Shallit, Simple Continued Fractions for Some Irrational Numbers II, J. Number Theory 14 (1982), 228-231.

Cf. A004200, A006464, A006465, A006466, A006467, A007400, A081771.

ContinuedFraction[Sum[1/3^(3^i), {i, 0, 5}]] (* Michael De Vlieger, Jul 01 2018 *)

{ allocatemem(932245000); default(realprecision, 8000); x=contfrac(suminf(n=0, 1/3^(3^n))); for (n=0, 382, write("b061678.txt", n, " ", x[n+1])) } \\ Harry J. Smith, Jul 26 2009

A283874 The Pierce expansion of the number Sum_{k>=1} 1/3^((2^k) - 1).

Original entry on oeis.org

2, 3, 4, 9, 10, 81, 82, 6561, 6562, 43046721, 43046722, 1853020188851841, 1853020188851842, 3433683820292512484657849089281, 3433683820292512484657849089282, 11790184577738583171520872861412518665678211592275841109096961, 11790184577738583171520872861412518665678211592275841109096962

Offset: 0

Kutlwano Loeto, Mar 24 2017

nonn

This sequence is the Pierce expansion of the number 3*s(3) - 1 = 0.370827687432918983346475478500709113969827799141493576... where s(u) = Sum_{k>=0} 1/u^(2^k) for u=3 has been considered by N. J. A. Sloane in A004200.

The continued fraction expansion of the number 3*s(3)-1 is essentially A081771.

			The Pierce expansion of 0.3708276874329189833 starts as 1/2 - 1/(2*3) + 1/(2*3*4) - 1/(2*3*4*9) + 1/(2*3*4*9*10) - 1/(2*3*4*9*10*81) + ...

Jeffrey Shallit, Simple continued fractions for some irrational numbers. J. Number Theory 11 (1979), no. 2, 209-217.

L:=[2]: for k from 0 to 6 do: L:=[op(L),3^(2^k),3^(2^k)+1]: od: print(L);

a(n) = if (n==0, 2, if (n%2, 3^(2^((n-1)/2)), 3^(2^((n-2)/2))+1)); \\ Michel Marcus, Mar 31 2017

a(0) = 2, a(2k+1) = 3^(2^k), a(2k+2) = 3^(2^k) + 1, k >= 0.

cp's OEIS Frontend

A061678 Continued fraction for Sum_{n>=0} 1/3^(3^n).

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