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

Author

Paul Barry, Mar 08 2003

Keywords

• easy
• nonn

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).

Crossrefs

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

Programs

Magma

[5^(n-3) * Binomial(n, 3): n in [0..25]]; // Vincenzo Librandi, Aug 06 2013

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 *)

PARI

vector(31, n, my(m=n-1); 5^(m-3)*binomial(m,3)) \\ G. C. Greubel, Mar 05 2020

Sage

[lucas_number2(n, 5, 0)*binomial(n,3)/5^3 for n in range(0, 22)] # Zerinvary Lajos, Mar 12 2009

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)