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Starting at 1, four-fold convolution of A000400 (powers of 6).
8

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

%S 0,0,0,1,24,360,4320,45360,435456,3919104,33592320,277136640,

%T 2217093120,17293326336,132058128384,990435962880,7313988648960,

%U 53287631585280,383670947414016,2733655500324864,19296391766999040,135074742368993280

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

%C 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

%H Vincenzo Librandi, <a href="/A081144/b081144.txt">Table of n, a(n) for n = 0..400</a>

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (24,-216,864,-1296).

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

%F 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.

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

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

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

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

%p seq(seq(binomial(i+2, j)*6^(i-1), j =i-1), i=-2..19); # _Zerinvary Lajos_, Dec 30 2007

%p seq(binomial(n+3,3)*6^n,n=-3..18); # _Zerinvary Lajos_, Jun 03 2008

%o (Sage)[lucas_number2(n, 6, 0)*binomial(n,3)/6^3 for n in range(0, 22)] # _Zerinvary Lajos_, Mar 13 2009

%o (Magma) [6^n* Binomial(n+3, 3): n in [-3..20]]; // _Vincenzo Librandi_, Oct 16 2011

%o (GAP) List([-3..18],n->Binomial(n+3,3)*6^n); # _Muniru A Asiru_, Feb 19 2018

%Y Cf. A081143, A081136.

%Y Cf. A038255.

%K nonn,easy

%O 0,5

%A _Paul Barry_, Mar 08 2003