%I #31 Sep 08 2022 08:45:09
%S 0,0,1,15,150,1250,9375,65625,437500,2812500,17578125,107421875,
%T 644531250,3808593750,22216796875,128173828125,732421875000,
%U 4150390625000,23345947265625,130462646484375,724792480468750
%N 5th binomial transform of (0,0,1,0,0,0, ...).
%C Starting at 1, three-fold convolution of A000351 (powers of 5).
%H Vincenzo Librandi, <a href="/A081135/b081135.txt">Table of n, a(n) for n = 0..300</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (15,-75,125).
%F a(n) = 15*a(n-1) - 75*a(n-2) + 125*a(n-3), a(0)=a(1)=0, a(2)=1.
%F a(n) = 5^(n-2)*binomial(n, 2).
%F G.f.: x^2/(1-5*x)^3.
%F E.g.f.: (x^2/2)*exp(5*x). - _G. C. Greubel_, May 14 2021
%F From _Amiram Eldar_, Jan 05 2022: (Start)
%F Sum_{n>=2} 1/a(n) = 10 - 40*log(5/4).
%F Sum_{n>=2} (-1)^n/a(n) = 60*log(6/5) - 10. (End)
%p seq(n*(n-1)*5^(n-2)/2, n=0..30); # _Zerinvary Lajos_, May 03 2007
%t CoefficientList[Series[x^2/(1-5x)^3, {x, 0, 30}], x] (* _Vincenzo Librandi_, Aug 06 2013 *)
%t LinearRecurrence[{15,-75,125},{0,0,1},30] (* _Harvey P. Dale_, Sep 13 2017 *)
%o (Sage) [5^(n-2)*binomial(n,2) for n in range(0, 30)] # _Zerinvary Lajos_, Mar 12 2009
%o (Magma) [5^(n-2)*Binomial(n, 2): n in [0..30]]; // _Vincenzo Librandi_, Aug 06 2013
%Y Sequences similar to the form q^(n-2)*binomial(n, 2): A000217 (q=1), A001788 (q=2), A027472 (q=3), A038845 (q=4), this sequence (q=5), A081136 (q=6), A027474 (q=7), A081138 (q=8), A081139 (q=9), A081140 (q=10), A081141 (q=11), A081142 (q=12), A027476 (q=15).
%K easy,nonn
%O 0,4
%A _Paul Barry_, Mar 08 2003