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A036071 Expansion of 1/(1-5*x)^5.

%I A036071 #16 Dec 07 2019 12:18:21
%S A036071 1,25,375,4375,43750,393750,3281250,25781250,193359375,1396484375,
%T A036071 9775390625,66650390625,444335937500,2905273437500,18676757812500,
%U A036071 118286132812500,739288330078125,4566192626953125,27904510498046875
%N A036071 Expansion of 1/(1-5*x)^5.
%C A036071 With a different offset, number of n-permutations (n=5) of 6 objects u, v, w, z, x, y with repetition allowed, containing exactly four (4)u's. Example: a(1)=25 because we have uuuuv, uuuvu, uuvuu, uvuuu, vuuuu, uuuuw, uuuwu, uuwuu, uwuuu, wuuuu, uuuuz, uuuzu, uuzuu, uzuuu, zuuuu, uuuux, uuuxu, uuxuu, uxuuu, xuuuu uuuuy, uuuyu, uuyuu, uyuuu, yuuuu. - _Zerinvary Lajos_, Jun 12 2008
%H A036071 <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (25, -250, 1250, -3125, 3125).
%F A036071 a(n) = binomial(n+4, 4)*5^n;
%F A036071 g.f. 1/(1-5*x)^5.
%F A036071 a(n) = 25*a(n-1) - 250*a(n-2) + 1250*a(n-3) - 3125*a(n-4) + 3125*a(n-5), a(0)=1, a(1)=25, a(2)=375, a(3)=4375, a(4)=43750. - _Harvey P. Dale_, Mar 20 2013
%p A036071 seq(binomial(n+4,4)*5^n,n=0..18); # _Zerinvary Lajos_, Jun 12 2008
%t A036071 CoefficientList[Series[1/(1-5x)^5,{x,0,30}],x] (* or *) LinearRecurrence[ {25,-250,1250,-3125,3125},{1,25,375,4375,43750},30] (* _Harvey P. Dale_, Mar 20 2013 *)
%o A036071 (Sage) [lucas_number2(n, 5, 0)*binomial(n,4)/5^4 for n in range(4, 23)] # _Zerinvary Lajos_, Mar 12 2009
%Y A036071 Cf. A001787, A038846.
%K A036071 easy,nonn
%O A036071 0,2
%A A036071 _Wolfdieter Lang_