A006466 -id:A006466 - cp's OEIS frontend

A076214 Decimal expansion of C = Sum_{k>=0} 1/2^(2^k-1).

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

Offset: 1

Benoit Cloitre, Nov 03 2002

This constant has a nice continued fraction expansion (i.e. only 1 and 2 occur). C arises when looking for a sequence b(n) such that : b(1) = 0, b(n+1) is the smallest integer > b(n) such that the continued fraction for 1/2^b(1) + 1/2^b(2) + ... + 1/2^b(n+1) contains only 1's or 2's. Because b(n) = 2^n-1 and C = Sum_{k>=0} 1/2^b(k).

			1.632843018043786287416159475061050443406622751841105608682421807686111...

Harry J. Smith, Table of n, a(n) for n = 1..20000
Boris Adamczewski, The Many Faces of the Kempner Number, Journal of Integer Sequences, Vol. 16 (2013), Article 13.2.15.
Michael Ian Shamos, Property Enumerators and a Partial Sum Theorem, 2011; alternative link.

Cf. A006466 (continued fraction), A007404, A078585.

Take[ RealDigits[ 2*NSum[1/2^2^k, {k, 0, Infinity}, WorkingPrecision -> 120]][[1]], 105] (* Jean-François Alcover, Nov 15 2011 *)

default(realprecision, 20080); x=suminf(k=0, 1/2^(2^k)); x*=2; for (n=1, 20000, d=floor(x); x=(x-d)*10; write("b076214.txt", n, " ", d)); \\ Harry J. Smith, May 09 2009

Equals 2 * Sum_{k>=0} 1/2^(2^k) = 2 * A007404. - Harry J. Smith, May 09 2009
From Amiram Eldar, Mar 12 2024: (Start)
Equals 1 + 2 * A078585.
Equals 1 + Sum_{k>=1} floor(log_2(k))/2^k (Shamos, 2011, p. 8). (End)

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.

A283526 Pierce expansion of the number Sum_{k >= 1} 1/(2^(2^k - 1)).

Original entry on oeis.org

1, 2, 3, 4, 5, 16, 17, 256, 257, 65536, 65537, 4294967296, 4294967297, 18446744073709551616, 18446744073709551617, 340282366920938463463374607431768211456, 340282366920938463463374607431768211457

Offset: 0

Kutlwano Loeto, Mar 10 2017

nonn

This sequence is the Pierce expansion of the number 2*s(2) - 1 = 0.632843018043786287416159475061... where s(u) = Sum_{k>=0} 1/u^(2^k) that has been considered by J. Shallit in A007400. The continued fraction expansion of this number is essentially A006466.

			The Pierce expansion of 0.6328430180437862 starts as 1 - 1/2 + 1/(2*3) - 1/(2*3*4) + 1/(2*3*4*5) - 1/(2*3*4*5*16) + ...

Michael De Vlieger, Table of n, a(n) for n = 0..24
Jeffrey Shallit, Simple continued fractions for some irrational numbers. J. Number Theory 11 (1979), no. 2, 209-217.

Cf. A006466, A007400, A076214.

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

{1}~Join~Map[{#, # + 1} &, 2^2^Range[0, 8]] // Flatten (* Michael De Vlieger, Mar 18 2017 *)

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

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

A076216 Partial sum of coefficients in the continued fraction expansion of sum(k>=0,1/2^(2^k-1)).

Original entry on oeis.org

1, 2, 3, 4, 6, 7, 8, 9, 10, 11, 12, 13, 15, 16, 17, 18, 19, 21, 22, 23, 24, 25, 27, 28, 30, 31, 32, 33, 34, 36, 37, 38, 39, 40, 41, 42, 43, 45, 46, 47, 48, 49, 51, 52, 54, 55, 56, 57, 58, 59, 60, 61, 63, 64, 65, 66, 67, 69, 70, 72, 73, 74, 75, 76, 78, 79, 80, 81, 82, 84, 85, 86

Offset: 1

Benoit Cloitre, Nov 03 2002

nonn

a(n)=sum(k=1, n, A006466(k)). lim n ->infinity a(n)/n = 6/5 and 0<=6*n-5*a(n)<=7

Corrected by T. D. Noe, Nov 02 2006

A081769 a(n)-th term of the continued fraction for sum(k>=0,1/2^(2^k)) is 2.

Original entry on oeis.org

5, 13, 18, 23, 25, 30, 38, 43, 45, 53, 58, 60, 65, 70, 78, 83, 85, 93, 98, 103, 105, 110, 118, 120, 125, 133, 138, 140, 145, 150, 158, 163, 165, 173, 178, 183, 185, 190, 198, 203, 205, 213, 218, 220, 225, 230, 238, 240, 245, 253, 258, 263, 265, 270, 278, 280

Offset: 1

Benoit Cloitre, Apr 09 2003

			The continued fraction for sum(k>=0,1/2^(2^k)) is : 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1... hence a(1)=5

Cf. A006466, A073089.

a(n)=5*n + 0 or + 3 and it appears that a(n)=5*n+3*A073089(n+2)

cp's OEIS Frontend

A076214 Decimal expansion of C = Sum_{k>=0} 1/2^(2^k-1).

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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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A283526 Pierce expansion of the number Sum_{k >= 1} 1/(2^(2^k - 1)).

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

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A076216 Partial sum of coefficients in the continued fraction expansion of sum(k>=0,1/2^(2^k-1)).

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A081769 a(n)-th term of the continued fraction for sum(k>=0,1/2^(2^k)) is 2.

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