A081144 Starting at 1, four-fold convolution of A000400 (powers of 6).

0, 0, 0, 1, 24, 360, 4320, 45360, 435456, 3919104, 33592320, 277136640, 2217093120, 17293326336, 132058128384, 990435962880, 7313988648960, 53287631585280, 383670947414016, 2733655500324864, 19296391766999040, 135074742368993280

Offset: 0

Author

Paul Barry, Mar 08 2003

Keywords

• nonn
• easy

Comments

With a different offset, number of n-permutations (n=4) of 7 objects: t, u, v, w, z, x, y with repetition allowed, containing exactly three u's. Example: a(4)=24 because we have uuut, uutu, utuu, tuuu, uuuv, uuvu, uvuu, vuuu, uuuw, uuwu, uwuu, wuuu, uuuz, uuzu, uzuu, zuuu, uuux, uuxu, uxuu, xuuu, uuuy, uuyu, uyuu, yuuu. - Zerinvary Lajos, Jun 03 2008

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..400
• Index entries for linear recurrences with constant coefficients, signature (24,-216,864,-1296).

Crossrefs

Cf. A081143, A081136.

Cf. A038255.

Programs

GAP

List([-3..18],n->Binomial(n+3,3)*6^n); # Muniru A Asiru, Feb 19 2018

Magma

[6^n* Binomial(n+3, 3): n in [-3..20]]; // Vincenzo Librandi, Oct 16 2011

Maple

seq(seq(binomial(i+2, j)*6^(i-1), j =i-1), i=-2..19); # Zerinvary Lajos, Dec 30 2007
seq(binomial(n+3,3)*6^n,n=-3..18); # Zerinvary Lajos, Jun 03 2008

Sage

[lucas_number2(n, 6, 0)*binomial(n,3)/6^3 for n in range(0, 22)] # Zerinvary Lajos, Mar 13 2009

Formula

G.f.: x^3/(1 - 6*x)^4.

a(n) = 24*a(n-1) - 216*a(n-2) + 864*a(n-3) - 1296*a(n-4) for n > 3, a(0) = a(1) = a(2) = 0, a(3) = 1.

a(n) = 6^(n - 3)*binomial(n, 3).

From Amiram Eldar, Jan 04 2022: (Start)

Sum_{n>=3} 1/a(n) = 450*log(6/5) - 81.

Sum_{n>=3} (-1)^(n+1)/a(n) = 882*log(7/6) - 135. (End)