A004200 -id:A004200 - cp's OEIS frontend

A078885 Decimal expansion of Sum {n>=0} 1/3^(2^n).

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

Offset: 0

Robert G. Wilson v, Dec 11 2002

			0.456942562477639661115491826166903037989942599713831192091056874309982...

Harry J. Smith, Table of n, a(n) for n = 0..20000
Aubrey J. Kempner, On Transcendental Numbers, Transactions of the American Mathematical Society, volume 17, number 4, October 1916, pages 476-482.
Index entries for transcendental numbers

Cf. A004200 (continued fraction), A011764.

Similar sums: A007404, A078585, A078886, A078887, A078888, A078889, A078890, A036987.

RealDigits[ N[ Sum[1/3^(2^n), {n, 0, Infinity}], 110]] [[1]]

default(realprecision, 20080); x=suminf(n=0, 1/3^(2^n)); x*=10; for (n=0, 20000, d=floor(x); x=(x-d)*10; write("b078885.txt", n, " ", d)); \\ Harry J. Smith, May 10 2009

Equals -Sum_{k>=1} mu(2*k)/(3^k - 1), where mu is the Möbius function (A008683). - Amiram Eldar, Jul 12 2020

A006464 Continued fraction for Sum_{n>=0} 1/4^(2^n).

Original entry on oeis.org

0, 3, 6, 4, 4, 2, 4, 6, 4, 2, 6, 4, 2, 4, 4, 6, 4, 2, 6, 4, 4, 2, 4, 6, 2, 4, 6, 4, 2, 4, 4, 6, 4, 2, 6, 4, 4, 2, 4, 6, 4, 2, 6, 4, 2, 4, 4, 6, 2, 4, 6, 4, 4, 2, 4, 6, 2, 4, 6, 4, 2, 4, 4, 6, 4, 2, 6, 4, 4, 2, 4, 6, 4, 2, 6, 4, 2, 4, 4, 6, 4, 2, 6, 4, 4, 2, 4, 6, 2, 4, 6, 4, 2, 4, 4, 6, 2, 4, 6, 4, 4, 2, 4, 6, 4

Offset: 0

N. J. A. Sloane

a(n)=A004200(n) if n=0; A004200(n)+1 if n>0 (according to case u=3, b=1 of Theorem 5 (of the reference) which states that: if B(u,infinity) = Sum_{n>=0} 1/u^(2^n) = [a0, a1, a2, ...] then B(u + b,infinity) = [a0, a1+b, a2+b, a3+b,... ] (u >= 3, b >= 0)).
The sum is equal to 0.316421509021893143708079...= A078585.
After computing the first 10^5 terms and dropping the first two (0 & 3), only the numbers 2, 4 & 6 occur. Further I found no two 0's in a row and no three 2's or three 1's in a row. - Robert G. Wilson v, Dec 01 2002

			0.316421509021893143708079737... = 0 + 1/(3 + 1/(6 + 1/(4 + 1/(4 + ...)))). - _Harry J. Smith_, May 11 2009

N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Harry J. Smith, Table of n, a(n) for n = 0..20000
J. Shallit, Letter to N. J. A. Sloane with attachment, Aug. 1979
Jeffrey Shallit, Simple continued fractions for some irrational numbers. J. Number Theory 11 (1979), no. 2, 209-217 [DOI]

u := 4: v := 7: Buv := [u,1,[0,u-1,u+1]]: for k from 2 to v do n := nops(Buv[3]): Buv := [u,Buv[2]+1,[seq(Buv[3][i],i=1..n-1),Buv[3][n]+1,Buv[3][n]-1,seq(Buv[3][n-i],i=1..n-2)]] od:seq(Buv[3][i],i=1..2^v);# first 2^v terms of A006464, Antonio G. Astudillo (aft_astudillo(AT)hotmail.com), Dec 02 2002

ContinuedFraction[ N[ Sum[1/4^(2^n), {n, 0, Infinity}], 1000]]

{ allocatemem(932245000); default(realprecision, 25000); x=suminf(n=0, 1/4^(2^n)); x=contfrac(x); for (n=1, 20001, write("b006464.txt", n-1, " ", x[n])); } \\ Harry J. Smith, May 11 2009

Better description and more terms from Antonio G. Astudillo (afg_astudillo(AT)hotmail.com), Jun 19 2001

A076157 Continued fraction expansion for c=sum_{k>=0} 1/2^(k!).

Original entry on oeis.org

1, 3, 1, 3, 4, 4095, 1, 3, 3, 1, 3, 4722366482869645213695, 1, 2, 1, 3, 3, 1, 4095, 4, 3, 1, 3, 3121748550315992231381597229793166305748598142664971150859156959625371738819765620120306103063491971159826931121406622895447975679288285306290175

Offset: 1

Benoit Cloitre, Nov 02 2002

Observation: if b(k) denotes the sequence of all elements of the continued fraction for c, b(k) = 4095 if k==6 or 19 (mod 24); b(k) = 4722366482869645213695 if k==12 or 37 (mod 48); .... If b(k) is not congruent to 5 (mod 10), it seems that b(k) = 1,2,3 or 4 only.
Conjecture: a(3*2^n) = -1 + 2^[(n+1)((n+2)!) ]. - Ralf Stephan, May 17 2005
The conjecture follows from the theorem in Shallit's paper. The continued fraction has a "folded" overall structure. - Georg Fischer, Aug 29 2022

Rick L. Shepherd, Table of n, a(n) for n = 1..47
Jeffrey O. Shallit, Simple Continued Fractions for Some Irrational Numbers II, J. Number Theory 14 (1982), 228-231.
Rick L. Shepherd, Table of n, a(n) for n = 1..383 (has larger terms than b-files support)

Cf. A076152, A076154, A076187.

Cf. A007400, A004200, A006466.

{allocatemem(220000000);
default(realprecision, 1000000);
contfrac(suminf(k=0, 1/(2^(k!))))}

c=1.2656250596046447753906250000000000007... = A076187.

More terms from Ralf Stephan, May 17 2005

b-file, a-file, PARI program, and corrected conjecture by Rick L. Shepherd, Jun 07 2013

A081771 Continued fraction for Sum_{k>=0} 1/3^(2^k-1).

Original entry on oeis.org

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

Offset: 1

Benoit Cloitre, Apr 10 2003

Contains only elements 1 <= a(n) <= 3. See Judnick, example 12.

S. W. Judnick, Patterns in continued fraction expansions, Thesis submitted to the faculty of the graduate school of the University of Minnesota, May 2013.

Cf. A004200, A006466.

A088435 1/2 + half of the (n+1)-st component of the continued fraction expansion of sum(k>=1,1/3^(2^k)).

Original entry on oeis.org

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

Offset: 1

Benoit Cloitre, Nov 08 2003

nonn

To construct the sequence use the rule : a(1)=3, then a(a(1)+a(2)+...+a(n)+1)=2+(-1)^n and fill in any undefined place with 2.

			Example to illustrate the comment : a(a(1)+1) = a(4) = 2+(-1)^1 = 1 and a(2), a(3) are undefined. The rule forces a(2) = a(3) = 2.

Antti Karttunen, Table of n, a(n) for n = 1..8192 (computed from the b-file of A004200 provided by _Harry J. Smith_)

Cf. A088431.

a(n) = (1/2) * (1+A004200(n+1)).

a(a(1)+a(2)+...+a(n)+1) = 2+(-1)^n.

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

Original entry on oeis.org

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

A384939 Maximum element in the continued fraction for (1/n) * Sum_{k>=0} 1/3^(2^k).

Original entry on oeis.org

5, 11, 17, 24, 29, 35, 42, 48, 53, 60, 66, 72, 77, 84, 90, 96, 102, 108, 114, 121, 126, 132, 139, 145, 151, 156, 163, 169, 175, 181, 187, 193, 200, 205, 211, 218, 224, 229, 235, 242, 248, 253, 260, 266, 272, 279, 284, 290, 297, 303, 308, 314, 321, 327, 332, 339, 345, 351, 357, 363, 369, 375, 381, 387, 393, 400, 406, 411, 418, 424, 430, 436, 442, 448, 454, 460, 466, 472, 479, 484, 326, 497, 503, 508, 515, 521, 527, 532, 539, 545, 551, 558, 563, 569, 576, 582, 587, 594, 600, 606

Offset: 1

Vaclav Kotesovec, Jul 22 2025

nonn

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

Cf. A004200, A078885, A081846.

m = 10; terms = 100; t1 = ConstantArray[0, 2*terms]; t2 = ConstantArray[1, 2*terms]; Until[t1 == t2, m++; PrintTemporary["m=",m]; s = Sum[1/3^(2^k), {k, 0, m}]; t1 = t2; t2 = Table[Max[ContinuedFraction[s/n]], {n, 1, 2*terms}]]; Take[t2, terms]

A081772 Continued fraction for sum(k>=0,1/4^(2^k-1)).

Original entry on oeis.org

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

Offset: 1

Benoit Cloitre, Apr 10 2003

Contains only elements 1<=a(n)<=4

Cf. A004200, A006464.

A081773 Continued fraction for sum(k>=0,1/5^(2^k-1)).

Original entry on oeis.org

1, 4, 1, 4, 5, 4, 1, 4, 4, 1, 4, 4, 1, 3, 1, 4, 4, 1, 4, 5, 4, 1, 4, 4, 1, 3, 1, 4, 5, 4, 1, 4, 4, 1, 3, 1, 4, 4, 1, 4, 4, 1, 4, 5, 4, 1, 4, 4, 1, 3, 1, 4, 5, 4, 1, 4, 4, 1, 4, 4, 1, 3, 1, 4, 4, 1, 4, 5, 4, 1, 3, 1, 4, 4, 1, 4, 5, 4, 1, 4, 4, 1, 3, 1, 4, 4, 1, 4, 4, 1, 4, 5, 4, 1, 4, 4, 1, 3, 1, 4, 5, 4, 1, 4, 4

Offset: 1

Benoit Cloitre, Apr 10 2003

Contains only elements 1<=a(n)<=5 ( but 2 never occurs)

Cf. A004200.

A081774 Continued fraction for sum(k>=0,1/6^(2^k-1)).

Original entry on oeis.org

1, 5, 1, 5, 6, 5, 1, 5, 5, 1, 5, 5, 1, 4, 1, 5, 5, 1, 5, 6, 5, 1, 5, 5, 1, 4, 1, 5, 6, 5, 1, 5, 5, 1, 4, 1, 5, 5, 1, 5, 5, 1, 5, 6, 5, 1, 5, 5, 1, 4, 1, 5, 6, 5, 1, 5, 5, 1, 5, 5, 1, 4, 1, 5, 5, 1, 5, 6, 5, 1, 4, 1, 5, 5, 1, 5, 6, 5, 1, 5, 5, 1, 4, 1, 5, 5, 1, 5, 5, 1, 5, 6, 5, 1, 5, 5, 1, 4, 1, 5, 6, 5, 1, 5, 5

Offset: 1

Benoit Cloitre, Apr 10 2003

Contains only elements 1<=a(n)<=6 ( but 2 and 3 never occur)

Cf. A004200.

cp's OEIS Frontend

A078885 Decimal expansion of Sum {n>=0} 1/3^(2^n).

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A006464 Continued fraction for Sum_{n>=0} 1/4^(2^n).

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A076157 Continued fraction expansion for c=sum_{k>=0} 1/2^(k!).

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A081771 Continued fraction for Sum_{k>=0} 1/3^(2^k-1).

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A088435 1/2 + half of the (n+1)-st component of the continued fraction expansion of sum(k>=1,1/3^(2^k)).

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

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A384939 Maximum element in the continued fraction for (1/n) * Sum_{k>=0} 1/3^(2^k).

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Mathematica

A081772 Continued fraction for sum(k>=0,1/4^(2^k-1)).

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A081773 Continued fraction for sum(k>=0,1/5^(2^k-1)).

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