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Number of n X n matrices over GF(5) with rank 1.
2

%I #17 Jan 12 2017 01:56:52

%S 4,144,3844,97344,2439844,61027344,1525839844,38146777344,

%T 953673339844,23841853027344,596046423339844,14901161071777344,

%U 372529029235839844,9313225743103027344,232830643638610839844,5820766091270446777344

%N Number of n X n matrices over GF(5) with rank 1.

%H Harry J. Smith, <a href="/A060870/b060870.txt">Table of n, a(n) for n = 1..200</a>

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (31,-155,125).

%F a(n) = 1/4 * (5^n - 1)^2.

%F G.f.: -4*x*(5*x+1) / ((x-1)*(5*x-1)*(25*x-1)). [_Colin Barker_, Dec 23 2012]

%e a(2) = 144 because there are 145 (the second element in sequence A060720) singular 2 X 2 matrices over GF(5), that have rank <= 1 of which only the zero matrix has rank zero so a(2) = 145 - 1 = 144.

%t Table[(5^n-1)^2/4,{n,20}] (* or *) LinearRecurrence[{31,-155,125},{4,144,3844},20] (* _Harvey P. Dale_, Dec 06 2014 *)

%o (PARI) { for (n=1, 200, write("b060870.txt", n, " ", (5^n - 1)^2 / 4); ) } \\ _Harry J. Smith_, Jul 13 2009

%Y Cf. A060720.

%K nonn,easy

%O 1,1

%A Ahmed Fares (ahmedfares(AT)my-deja.com), May 04 2001

%E More terms from Larry Reeves (larryr(AT)acm.org), May 07 2001