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5th binomial transform of (0,0,0,1,0,0,0,0,......).
8

%I #29 Sep 08 2022 08:45:09

%S 0,0,0,1,20,250,2500,21875,175000,1312500,9375000,64453125,429687500,

%T 2792968750,17773437500,111083984375,683593750000,4150390625000,

%U 24902343750000,147857666015625,869750976562500,5073547363281250

%N 5th binomial transform of (0,0,0,1,0,0,0,0,......).

%C Starting at 1, four-fold convolution of A000351 (powers of 5).

%C With a different offset, number of n-permutations (n=4)of 6 objects u, v, w, z, x, y with repetition allowed, containing exactly three u's. Example: a(4)=20 because we have uuuv, uuvu, uvuu, vuuu, uuuw, uuwu, uwuu, wuuu, uuuz, uuzu, uzuu, zuuu, uuux, uuxu, uxuu, xuuu, uuuy, uuyu, uyuu and yuuu. - _Zerinvary Lajos_, Jun 03 2008

%H Vincenzo Librandi, <a href="/A081143/b081143.txt">Table of n, a(n) for n = 0..300</a>

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (20,-150,500,-625).

%F a(n) = 20*a(n-1) - 150*a(n-2) + 500*a(n-3) - 625*a(n-4), with a(0)=a(1)=a(2)=0, a(3)=1.

%F a(n) = 5^(n-3)*binomial(n,3).

%F G.f.: x^3/(1-5*x)^4.

%F E.g.f.: x^3*exp(x)/6. - _G. C. Greubel_, Mar 05 2020

%F From _Amiram Eldar_, Jan 04 2022: (Start)

%F Sum_{n>=3} 1/a(n) = 240*log(5/4) - 105/2.

%F Sum_{n>=3} (-1)^(n+1)/a(n) = 540*log(6/4) - 195/2. (End)

%p seq(binomial(n,3)*5^(n-3), n=0..25); # _Zerinvary Lajos_, Jun 03 2008

%t CoefficientList[Series[x^3/(1-5x)^4, {x, 0, 30}], x] (* _Vincenzo Librandi_, Aug 06 2013 *)

%t LinearRecurrence[{20,-150,500,-625}, {0,0,0,1}, 30] (* _Harvey P. Dale_, Dec 24 2015 *)

%o (Sage) [lucas_number2(n, 5, 0)*binomial(n,3)/5^3 for n in range(0, 22)] # _Zerinvary Lajos_, Mar 12 2009

%o (Magma) [5^(n-3) * Binomial(n, 3): n in [0..25]]; // _Vincenzo Librandi_, Aug 06 2013

%o (PARI) vector(31, n, my(m=n-1); 5^(m-3)*binomial(m,3)) \\ _G. C. Greubel_, Mar 05 2020

%Y Cf. A038846, A036216, A001789, A081144, A081135.

%K easy,nonn

%O 0,5

%A _Paul Barry_, Mar 08 2003