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A023365
a(n) = 2^(3^(n-1)).
7
2, 8, 512, 134217728, 2417851639229258349412352, 14134776518227074636666380005943348126619871175004951664972849610340958208
OFFSET
1,1
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
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
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
MATHEMATICA
NestList[#^3&, 2, 6] (* Harvey P. Dale, Aug 10 2014 *)
PROG
(Magma) [Floor(2^(3^(n-1))): n in [1..10]]; // Vincenzo Librandi, Aug 11 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
KEYWORD
nonn
STATUS
approved