A002489 a(n) = n^(n^2), or (n^n)^n.

1, 1, 16, 19683, 4294967296, 298023223876953125, 10314424798490535546171949056, 256923577521058878088611477224235621321607, 6277101735386680763835789423207666416102355444464034512896, 196627050475552913618075908526912116283103450944214766927315415537966391196809

Offset: 0

Author

N. J. A. Sloane

Keywords

• nonn
• easy
• nice

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

Examples

			a(3) = 19683 because (3^3)^3 = 3^(3^2) = 19683.
		

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]
• Eric Postpischil, Posting to sci.math newsgroup, May 21 1990.
• P. Rossier, Grands nombres, Elemente der Mathematik, Vol. 3 (1948), p. 20; alternative link.
• Index entries for sequences related to groupoids

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).

Cf. A002488, A001329, A002488, A023813, A076113, A090588, A000312, A258102.

Programs

Magma

[n^(n^2): n in [0..10]]; // Vincenzo Librandi, May 13 2011

Mathematica

Join[{1},Table[n^n^2,{n,10}]] (* Harvey P. Dale, Sep 06 2011 *)

PARI

a(n)=n^(n^2) \\ Charles R Greathouse IV, Nov 20 2012

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