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%I #42 Nov 22 2022 02:35:12
%S 0,0,16,162,768,2500,6480,14406,28672,52488,90000,146410,228096,
%T 342732,499408,708750,983040,1336336,1784592,2345778,3040000,3889620,
%U 4919376,6156502,7630848,9375000,11424400,13817466,16595712,19803868,23490000,27705630,32505856
%N a(n) = n^5 - n^4.
%C For n >= 1, a(n) is equal to the number of functions f:{1,2,3,4,5}->{1,2,...,n} such that for a fixed x in {1,2,3,4,5} and a fixed y in {1,2,...,n} we have f(x) <> y. - Aleksandar M. Janjic and _Milan Janjic_, Mar 13 2007
%H Vincenzo Librandi, <a href="/A085538/b085538.txt">Table of n, a(n) for n = 0..10000</a>
%H Milan Janjic, <a href="https://pmf.unibl.org/wp-content/uploads/2017/10/enumfun.pdf">Enumerative Formulas for Some Functions on Finite Sets</a>.
%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (6,-15,20,-15,6,-1).
%F G.f.: 2*x^2*(x^3 + 18*x^2 + 33*x + 8)/(x-1)^6. - _Colin Barker_, Nov 06 2012
%F Sum_{n>=2} 1/a(n) = 4 - zeta(2) - zeta(3) - zeta(4). - _Amiram Eldar_, Jul 05 2020
%F Product_{n>=2} (1 - 1/a(n)) = A146492. - _Amiram Eldar_, Nov 22 2022
%p a:=n->sum(sum(n^3, j=1..n),k=2..n): seq(a(n), n=0..31); # _Zerinvary Lajos_, May 09 2007
%t Table[n^5 - n^4, {n, 0, 40}] (* or *) LinearRecurrence[{6, -15, 20, -15, 6, -1}, {0, 0, 16, 162, 768, 2500}, 40] (* _Vladimir Joseph Stephan Orlovsky_, Feb 20 2012 *)
%o (Magma) [n^5-n^4: n in [0..50]]; // _Vincenzo Librandi_, Feb 12 2012
%o (PARI) a(n)=n^5-n^4 \\ _Charles R Greathouse IV_, Oct 07 2015
%Y A diagonal of A228273.
%Y Cf. A000583, A000584, A146492.
%K nonn,easy
%O 0,3
%A _N. J. A. Sloane_, Jul 05 2003
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