Kutlwano Loeto - cp's OEIS frontend

User: Kutlwano Loeto

Kutlwano Loeto has authored 3 sequences.

A286016 Signed continued fraction expansion with all signs negative of tanh(1).

1, 5, 2, 2, 2, 2, 9, 2, 2, 2, 2, 2, 2, 2, 2, 13, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 17, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 21, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 25, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2

Offset: 1

Kutlwano Loeto, Apr 30 2017

For any given sequence of signs (e_1, e_2, ..., e_n, ...) one may define the signed continued fraction expansion of a real number x by using floor or ceiling in the step i according to e_i = +1 or e_i = -1. In the present case for the sequence (-1, -1, -1, -1, ...) consisting of only negative signs the ceiling is taken at each step, and the formulas with x_0 = x are a_n = ceiling(x_n) and x_{n+1} = 1/(a_n - x_n). See chapter 1 and 2 of the book by Perron.

			a(2) = 5, a(3) = a(4) = a(5) = a(6) = 2, a(7) = 9, etc. These numbers are obtained from the partial quotients xj as follows:
x2 =  (1 +  e^2)/( 2 + 0e^2) ~4.17 so that a(2)=ceiling(x2)=5;
x3 =  (2 + 0e^2)/( 9 - e^2)  ~1.21 so that a(3)=ceiling(x3)=2;
x4 =  (9 -  e^2)/(16 - 2e^2) ~1.31 so that a(4)=ceiling(x4)=2;
x5 = (16 - 2e^2)/(23 - 3e^2) ~1.46 so that a(5)=ceiling(x5)=2;
x6 = (23 - 3e^2)/(30 - 4e^2) ~1.87 so that a(6)=ceiling(x6)=2;
x7 = (30 - 4e^2)/(37 - 5e^2) ~8.11 so that a(7)=ceiling(x7)=9.
The pairs of integers appearing in the xj's are obtained as the principal or as every other of the non-principal approximating fractions of e^2 in the sense of the A. Hurwitz reference.

Adolf Hurwitz, Über die Kettenbrüche, deren Teilnenner arithmetische Reihen bilden, Vierteljahrsschrift der Naturforschenden Gesellschaft in Zürich, Jahrg. 41 (1896), Jubelband 2, 34-64. Reprinted in Hurwitz's Mathematische Werke.
Oskar Perron, Die Lehre von den Kettenbrüchen, Teubner, Leipzig, 1930.

Cf. A004273 (continued fraction of tanh(1)), A280135, A280136.

x:=(exp(1)-exp(-1))/(exp(1)+exp(-1)):b:=ceil(x): x1:=1/(b-x):L:=[b]:
for k from 0 to 40 do:
b1:=ceil(x1): x1:=1/(b1-x1): L:=[op(L),b1]: od: print(L);

Using an obvious condensed notation we get for the sequence 1, 5, 2^(4), 9, 2^(8), 13, 2^(12), 17, 2^(16), 21, 2^(20), ... where 2^(m) means m copies of 2.

More terms from Jinyuan Wang, Jul 02 2022

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.

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.

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User: Kutlwano Loeto

A286016 Signed continued fraction expansion with all signs negative of tanh(1).

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A283874 The Pierce expansion of the number Sum_{k>=1} 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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