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

2, 8, 512, 134217728, 2417851639229258349412352, 14134776518227074636666380005943348126619871175004951664972849610340958208

Offset: 1

Author

David W. Wilson

Keywords

• nonn

Comments

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

Links

• 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

Crossrefs

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

Programs

Magma

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

Mathematica

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

Python

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

Sage

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

Formula

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