A081143 5th binomial transform of (0,0,0,1,0,0,0,0,......).

0, 0, 0, 1, 20, 250, 2500, 21875, 175000, 1312500, 9375000, 64453125, 429687500, 2792968750, 17773437500, 111083984375, 683593750000, 4150390625000, 24902343750000, 147857666015625, 869750976562500, 5073547363281250

Offset

0,5

Comments

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

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

Links

Vincenzo Librandi, Table of n, a(n) for n = 0..300

Index entries for linear recurrences with constant coefficients, signature (20,-150,500,-625).

Formula

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.

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

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

E.g.f.: x^3*exp(x)/6. - G. C. Greubel, Mar 05 2020

From Amiram Eldar, Jan 04 2022: (Start)

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

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

Maple

seq(binomial(n, 3)*5^(n-3), n=0..25); # Zerinvary Lajos, Jun 03 2008

Mathematica

CoefficientList[Series[x^3/(1-5x)^4, {x, 0, 30}], x] (* Vincenzo Librandi, Aug 06 2013 *)
LinearRecurrence[{20, -150, 500, -625}, {0, 0, 0, 1}, 30] (* Harvey P. Dale, Dec 24 2015 *)

Prog

(Sage) [lucas_number2(n, 5, 0)*binomial(n, 3)/5^3 for n in range(0, 22)] # Zerinvary Lajos, Mar 12 2009
(Magma) [5^(n-3) * Binomial(n, 3): n in [0..25]]; // Vincenzo Librandi, Aug 06 2013
(PARI) vector(31, n, my(m=n-1); 5^(m-3)*binomial(m, 3)) \\ G. C. Greubel, Mar 05 2020

Crossrefs

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

Sequence in context: A278673 A055757 A022744 * A218101 A241228 A306290

Adjacent sequences: A081140 A081141 A081142 * A081144 A081145 A081146

Keyword

easy,nonn

Author

Paul Barry, Mar 08 2003

Status

approved