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A002489
a(n) = n^(n^2), or (n^n)^n.
(Formerly M5030 N2170)
55
1, 1, 16, 19683, 4294967296, 298023223876953125, 10314424798490535546171949056, 256923577521058878088611477224235621321607, 6277101735386680763835789423207666416102355444464034512896, 196627050475552913618075908526912116283103450944214766927315415537966391196809
OFFSET
0,3
COMMENTS
The number of closed binary operations on a set of order n. Labeled groupoids.
The values of "googol" in base N: "10^100" in base 2 is 2^4=16; "10^100" in base 3 is 3^9=19683, etc. This is N^^3 by the "lower-valued" (left-associative) definition of the hyper4 or tetration operator (see Munafo webpage). - Robert Munafo, Jan 25 2010
n^(n^k) = (((n^n)^n)^...)^n, with k+1 n's, k >= 0. - Daniel Forgues, May 18 2013
REFERENCES
John S. Rose, A Course on Group Theory, Camb. Univ. Press, 1978, see p. 6.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
LINKS
Michael Lee, Table of n, a(n) for n = 0..26 (first 16 terms from Vincenzo Librandi)
Robert Munafo, Hyper4 Iterated Exponential Function [From Robert Munafo, Jan 25 2010]
P. Rossier, Grands nombres, Elemente der Mathematik, Vol. 3 (1948), p. 20; alternative link.
FORMULA
a(n) = [x^(n^2)] 1/(1 - n*x). - Ilya Gutkovskiy, Oct 10 2017
Sum_{n>=1} 1/a(n) = A258102. - Amiram Eldar, Nov 11 2020
EXAMPLE
a(3) = 19683 because (3^3)^3 = 3^(3^2) = 19683.
MATHEMATICA
Join[{1}, Table[n^n^2, {n, 10}]] (* Harvey P. Dale, Sep 06 2011 *)
PROG
(Magma) [n^(n^2): n in [0..10]]; // Vincenzo Librandi, May 13 2011
(PARI) a(n)=n^(n^2) \\ Charles R Greathouse IV, Nov 20 2012
CROSSREFS
a(n) = A079172(n) + A023814(n) = A079176(n) + A079179(n);
a(n) = A079182(n) + A023813(n) = A079186(n) + A079189(n);
a(n) = A079192(n) + A079195(n) + A079198(n) + A023815(n).
Sequence in context: A098175 A089232 A300615 * A060205 A140597 A017296
KEYWORD
nonn,easy,nice
STATUS
approved