A007400 -id:A007400 - cp's OEIS frontend

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

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

A232734 Decimal expansion of Integral {x=0..infinity} 1/2^(2^x) dx.

Original entry on oeis.org

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

Offset: 0

Jean-François Alcover, Nov 29 2013

			0.546306835952482741736098769624101388937635539081659135416783399176163689841...

I. S. Gradsteyn, I. M. Ryzhik, Table of integrals, series and products, (1980) 8.212.

Cf. A007400, A007404 (sum instead of integral).

RealDigits[-ExpIntegralEi[-Log[2]]/Log[2], 10, 100] // First

eint1(log(2))/log(2) \\ Charles R Greathouse IV, Dec 02 2013

-Ei(-log(2))/log(2), where Ei is the exponential integral function.
Also equals (2*Integral_{x = 0..1/2} log(log(1/x)) dx - log(log(2)))/(2*log(2)).
From Peter Bala, Feb 05 2024: (Start)
Equals 1/log(2) * Integral_{x >= 1} 1/(x * 2^x) dx.
Equals 1/log(4) * Integral_{x = 0..1} 1/(log(2) - log(x)) dx.
Equals Integral_{x >= 1} log(x)/2^x dx = (log(2))^2 * Integral_{x >= 0} x*(2^x) /(2^(2^x)). See Gradsteyn and Ryzhik, Section 8.212, formulas (4) and (16). (End)

A073096 Maximal element in continued fraction for s(n) = sum( k>=n,1/2^(2^k) ).

Original entry on oeis.org

6, 6, 18, 258, 65538, 4294967298, 18446744073709551618, 340282366920938463463374607431768211458, 115792089237316195423570985008687907853269984665640564039457584007913129639938

Offset: 0

Benoit Cloitre, Aug 18 2002

nonn

			Continued fraction expansion of 1/2^(2^5)+1/2^(2^6)+1/2^(2^7)+...is [0, 4294967295, 4294967298, 4294967296, 4294967296,... ] where the maximum element is 4294967298, hence a(5)=4294967298

Cf. A007400 for case n=0.

a(0)=6; for n>0 a(n)=2^(2^n)+2.

a(n) = A063486(n)-3 for n>0. - R. J. Mathar, Apr 22 2007

A081778 Maximum element is the continued fraction for n*sum(k>=0,1/2^(2^k)).

Original entry on oeis.org

6, 2, 19, 4, 33, 9, 46, 7, 60, 16, 72, 12, 85, 23, 99, 16, 79, 29, 126, 22, 139, 35, 153, 23, 166, 43, 180, 30, 192, 49, 205, 31, 219, 56, 232, 40, 246, 62, 259, 39, 273, 69, 286, 48, 300, 76, 303, 48, 229, 83, 223, 58, 352, 89, 366, 55, 367, 95, 380, 67, 406, 102, 296, 63

Offset: 1

Benoit Cloitre, Apr 10 2003

nonn

Cf. A007400.

a(n)=n for n=2, 4, 12, 16, 48, 80, 108, 112, 123, 144.....

A173995 Continued fraction expansion of sum of reciprocals of Fermat primes.

Original entry on oeis.org

0, 1, 1, 2, 9, 1, 3, 5, 1, 2, 1, 1, 1, 1, 3, 1, 7, 1, 31, 1, 2, 4, 5

Offset: 1

Jonathan Vos Post, Mar 04 2010

If there are only five Fermat primes, a(24) = 2 is the last term of this sequence. Otherwise, a(24) = a(25) = 1 and a(26) is large (billions of digits).

This sequence is finite if and only if A019434 is finite.

			(1/3) + (1/5) + (1/17) + (1/257) + (1/65537) = 2560071829/4294967295 = 0 + 1/1+ 1/1+ 1/2+ 1/9+ 1/1+ 1/3+ 1/5+ 1/1+ 1/2+ 1/1+ 1/1+ 1/1+ 1/1+ 1/3+ 1/1+ 1/7+ 1/1+ 1/31+ 1/1+ 1/2+ 1/4+ 1/5+ 1/2.

S. W. Golomb, Irrationality of the sum of reciprocals of fermat numbers and other functions, NASA Technical Report 19630013175, Accession ID 63N23055, Contract/grant NAS7-100, 4 pp., Jet Propulsion Laboratory, Jan 01 1962.

Cf. A019434, A000215, A159611, A173898 (sum of reciprocals of Mersenne primes), A007400.

(* Assuming 65537 is the largest Fermat prime *) ContinuedFraction[Sum[1/(2^(2^n) + 1), {n, 0, 4}]] (* Alonso del Arte, Apr 21 2013 *)

Continued fraction of Sum_{i >= 1} 1/A019434(i).

Sequence corrected and comments added by Charles R Greathouse IV, Feb 04 2011

A235063 Continued fraction expansion of Sum(i=1..inf, 1/2^(2^i+1) ).

Original entry on oeis.org

2, 2, 4, 2, 8, 3, 8, 1, 8, 3, 4, 2, 12, 2, 8, 1, 8, 3, 4, 2, 8, 3, 8, 1, 12, 2, 4, 2, 12, 2, 8, 1, 8, 3, 4, 2, 8, 3, 8, 1, 8, 3, 4, 2, 12, 2, 8, 1, 12, 2, 4, 2, 8, 3, 8, 1, 12, 2, 4, 2, 12, 2, 8, 1, 8, 3, 4, 2, 8, 3, 8, 1, 8, 3, 4, 2, 12, 2, 8, 1, 8, 3, 4, 2, 8, 3, 8, 1, 12, 2, 4, 2, 12, 2, 8, 1, 12, 2, 4

Offset: 0

Ralf Stephan, Jan 03 2014

			0.40821075451094657... = 2/(2+1/(2+1/(4+1/(2+1/(8+1/(3+1/8...

H. Cohn, Symmetry and specializability in continued fractions. arXiv preprint math/0008221 (2000), published in Acta Arithmetica 75 (1996): 4.
Jeffrey Shallit, Simple continued fractions for some irrational numbers, J. Number Theory 11 (1979), no. 2, 209-217.
Index entries for continued fractions for constants

Cf. A007400.

a(n)=contfrac(suminf(i=0, 1/2^(2^i+1)))[n+1]

0.40821075451094657... = (1/2) A007400.

Recurrence: a(8n)=1, a(8n+4)=a(16n+14)=a(32n+26)=2, a(16n+6)=a(32n+10)=3, a(8n+3)=4, a(8n+7)=a(16n+5)=a(32n+9)=8, a(16n+13)=a(32n+25)=12, a(8n+1)=a(4n+1), a(8n+2)=a(4n+2), starting 2,2,4 (conjectured).

A347589 Continued fraction for Sum_{k>=0} 1/2^(3^k).

Original entry on oeis.org

0, 1, 1, 1, 2, 7, 1, 1, 1, 1, 1, 511, 2, 1, 1, 1, 7, 2, 1, 1, 1, 134217727, 2, 1, 2, 7, 1, 1, 1, 2, 511, 1, 1, 1, 1, 1, 7, 2, 1, 1, 1, 2417851639229258349412351, 2, 1, 2, 7, 1, 1, 1, 1, 1, 511, 2, 1, 1, 1, 7, 2, 1, 2, 134217727, 1, 1, 1, 2, 7, 1, 1, 1, 2, 511, 1, 1, 1, 1, 1, 7, 2, 1, 1, 1

Offset: 0

Benoit Cloitre, Sep 11 2021

Cf. A007400.

ContinuedFraction[N[Sum[1/2^(3^k),{k,0,Infinity}],250]] (* Stefano Spezia, Sep 11 2021 *)

my(v=contfrac(suminf(k=0, 1/2^(3^k)))); Vec(v, #v-1) \\ Michel Marcus, Sep 11 2021 and Sep 30 2024

a(5*2^k+2) = 2^(3^(k+1)) - 1. Other terms are obtained by symmetry around (5*2^k,5*2^k+1,5*2^k+2,5*2^k+3). For instance 1, 1, 1, 2, 7, 1, 1, 1, (1, 1, 511, 2), 1, 1, 1, 7, 2, 1, 1, 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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A232734 Decimal expansion of Integral {x=0..infinity} 1/2^(2^x) dx.

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A073096 Maximal element in continued fraction for s(n) = sum( k>=n,1/2^(2^k) ).

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A081778 Maximum element is the continued fraction for n*sum(k>=0,1/2^(2^k)).

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A173995 Continued fraction expansion of sum of reciprocals of Fermat primes.

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A235063 Continued fraction expansion of Sum(i=1..inf, 1/2^(2^i+1) ).

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A347589 Continued fraction for Sum_{k>=0} 1/2^(3^k).

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