A376867 -id:A376867 - cp's OEIS frontend

A023365 a(n) = 2^(3^(n-1)).

2, 8, 512, 134217728, 2417851639229258349412352, 14134776518227074636666380005943348126619871175004951664972849610340958208

Offset: 1

David W. Wilson

nonn

a(n+1) = a(n) converted to base 8 from base 2 (written in base 10).
Number of disjunctive-normal forms of n-1 variables (either with x, or x-negated or without x). - Labos Elemer, Jul 24 2003
a(n)*Psi(3^n,x), with the (monic) minimal polynomial Psi(3^n,x) of cos(2*Pi/3^n), becomes an integer polynomial with coefficient 1 of x^0.
  E.g., 8*Psi(9,x)=8*(x^3 - (3/4)*x + 1/8) = 8*x^3 - 6*x + 1.
  See A181875/A181876, A181877 and the W. Lang link under A181875. - Wolfdieter Lang, Feb 24 2011
The next term (a(7)) has 220 digits. - Harvey P. Dale, Aug 10 2014
These seem to be the reduced denominators of Newton's iteration for 1/sqrt(2), starting with 1/2. - Steven Finch, Oct 08 2024

Vincenzo Librandi, Table of n, a(n) for n = 1..8
W. van der Aalst, J. Buijs and B. van Dongen, Towards Improving the Representational Bias of Process Mining, 2012. - From _N. J. A. Sloane_, Feb 03 2013
X. Gourdon and P. Sebah, Pythagoras' Constant. - From _Steven Finch_, Oct 07 2024

Cf. A000079, A000244, A010503, A181875, A181876, A181877, A376867.

[Floor(2^(3^(n-1))): n in [1..10]]; // Vincenzo Librandi, Aug 11 2014

NestList[#^3&,2,6] (* Harvey P. Dale, Aug 10 2014 *)

def A023365(n): return 1<<3**(n-1) # Chai Wah Wu, Oct 11 2024

[2^(3^(n-1)) for n in range(1,8)] # Stefano Spezia, Oct 15 2024

a(n) = a(n-1)^3.
a(n) = A000079(A000244(n-1)).
a(n+1) is conjectured to be the reduced denominator of b(n) = b(n-1)*(3/2 - b(n-1)^2); b(0) = 1/2. - Steven Finch, Oct 08 2024
Limit_{n -> oo} A376867(n-1)/a(n) = 1/sqrt(2) = A010503. - Steven Finch, Oct 08 2024

A376870 Reduced numerators of Newton's iteration for 1/sqrt(3), starting with 1/3.

Original entry on oeis.org

1, 4, 130, 2739685, 21055737501685791580, 9337539302589041654242365815942422114384262970589593842110

Offset: 0

Steven Finch, Oct 07 2024

An explicit formula for a(n) is not known, although it arises from a recurrence and the corresponding denominators are simply 3^((3^n + 1)/2) = 3*A134799(n).

Next term is too large to include.

			a(1) = 4 because b(1) = (3/2)*(1/3)*(1 - 1/9) = 4/9.
1/3, 4/9, 130/243, 2739685/4782969, ... = A376870(n)/(3*A134799(n)).

Paolo Xausa, Table of n, a(n) for n = 0..7
X. Gourdon and P. Sebah, Pythagoras' Constant.

Cf. A020760, A134799, A376867.

Module[{n = 0}, NestList[#*(3^(3^n++ + 1) - #^2)/2 &, 1, 6]] (* Paolo Xausa, Oct 17 2024 *)

from itertools import count, islice
def A376870_gen(): # generator of terms
    p = 1
    for k in count(0):
        yield p
        p = p*(3**(3**k+1)-p**2)>>1
A376870_list = list(islice(A376870_gen(),6)) # Chai Wah Wu, Oct 11 2024

a(n) is the reduced numerator of b(n) = (3/2)*b(n-1)*(1 - b(n-1)^2); b(0) = 1/3.
Limit_{n -> oo} a(n)/(3*A134799(n)) = 1/sqrt(3) = A020760.
a(n+1) = a(n)*(3^(3^n+1)-a(n)^2)/2. - Chai Wah Wu, Oct 11 2024

cp's OEIS Frontend

A023365 a(n) = 2^(3^(n-1)).

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