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a(n) = 3^floor(n^2/4).
0

%I #20 Sep 08 2022 08:45:08

%S 1,1,3,9,81,729,19683,531441,43046721,3486784401,847288609443,

%T 205891132094649,150094635296999121,109418989131512359209,

%U 239299329230617529590083,523347633027360537213511521,3433683820292512484657849089281,22528399544939174411840147874772641

%N a(n) = 3^floor(n^2/4).

%C Number of groves of order n.

%C Tropical version is sequence A002620. - _Michael Somos_, Mar 14 2020

%H Gabriel D. Carroll and David Speyer, <a href="https://doi.org/10.37236/1826">The cube recurrence</a>, The Electronic Journal of Combinatorics, Volume 11, Issue 1 (2004), R73.

%F a(n) = 9*a(n-2)^2/a(n-4). - _Michael Somos_, Sep 16 2005

%F 0 = a(n)*a(n+3) - 3*a(n+1)*a(n+2) for all n in Z. - _Michael Somos_, Jan 25 2014

%F a(n) = a(-n) for all n in Z. a(n) = 3^A002620(n). - _Michael Somos_, Mar 14 2020

%t Table[3^Floor[n^2/4],{n,0,20}] (* _Harvey P. Dale_, May 08 2011 *)

%t a[ n_] := 3^Quotient[n^2, 4]; (* _Michael Somos_, Mar 14 2020 *)

%o (PARI) {a(n) = 3^(n^2\4)} /* _Michael Somos_, Sep 16 2005 */

%o (Magma) [3^Floor(n^2/4): n in [0..20]]; // _Vincenzo Librandi_, Mar 17 2020

%Y Cf. A002620.

%K nonn

%O 0,3

%A _N. J. A. Sloane_, Dec 18 2002