%I #56 Sep 08 2022 08:45:00
%S 0,1,10,75,500,3125,18750,109375,625000,3515625,19531250,107421875,
%T 585937500,3173828125,17089843750,91552734375,488281250000,
%U 2593994140625,13732910156250,72479248046875,381469726562500
%N a(n) = n*5^(n-1).
%C With a different offset, number of n-permutations of 6 objects u, v, w, z, x, y with repetition allowed, containing exactly one u. - _Zerinvary Lajos_, Dec 28 2007
%D A. H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, pp. 194-196.
%H Vincenzo Librandi, <a href="/A053464/b053464.txt">Table of n, a(n) for n = 0..500</a>
%H Frank Ellermann, <a href="/A001792/a001792.txt">Illustration of binomial transforms</a>
%H INRIA Algorithms Project, <a href="http://ecs.inria.fr/services/structure?nbr=756">Encyclopedia of Combinatorial Structures 756</a>.
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (10,-25).
%F a(n) = Sum_{k=0..n} 5^(n-k)*binomial(n-k+1, k)*binomial(1, (k+1)/2)*(1-(-1)^k)/2. - _Paul Barry_, Oct 15 2004
%F a(n) = 10*a(n-1) - 25*a(n-2); n>1; a(0)=0, a(1)=1.
%F Fourth binomial transform of n (starting 0, 1, 10...) Convolution of powers of 5.
%F G.f.: x/(1-5*x)^2; E.g.f.: x*exp(5*x). - _Paul Barry_, Jul 22 2003
%F a(n) = - 25^n * a(-n) for all n in Z. - _Michael Somos_, Jun 26 2017
%F From _Amiram Eldar_, Oct 28 2020: (Start)
%F Sum_{n>=1} 1/a(n) = 5*log(5/4).
%F Sum_{n>=1} (-1)^(n+1)/a(n) = 5*log(6/5). (End)
%t Join[{a=0,b=1},Table[c=10*b-25*a;a=b;b=c,{n,60}]] (* _Vladimir Joseph Stephan Orlovsky_, Jan 27 2011 *)
%t Table[n*5^(n-1),{n,0,20}] (* or *) LinearRecurrence[{10,-25},{0,1},30] (* _Harvey P. Dale_, Jul 22 2014 *)
%o (PARI) {a(n) = n*5^(n-1)}; /* _Michael Somos_, Sep 12 2005 */
%o (Sage) [lucas_number1(n,10,25) for n in range(0, 21)] # _Zerinvary Lajos_, Apr 26 2009
%o (Magma) [n*(5^(n-1)): n in [0..30]]; // _Vincenzo Librandi_, Jun 09 2011
%Y Cf. A002697, A027471, A001787.
%K easy,nonn
%O 0,3
%A _Barry E. Williams_, Jan 13 2000
%E More terms from _James A. Sellers_, Feb 02 2000