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7-fold convolution of A000302 (powers of 4).
5

%I #25 Mar 25 2022 09:14:51

%S 1,28,448,5376,53760,473088,3784704,28114944,196804608,1312030720,

%T 8396996608,51908706304,311452237824,1820797698048,10404558274560,

%U 58265526337536,320460394856448,1734256254517248,9249366690758656,48680877319782400,253140562062868480

%N 7-fold convolution of A000302 (powers of 4).

%C With a different offset, number of n-permutations (n>=6) of 5 objects: u, v, z, x, y with repetition allowed, containing exactly six (6) u's. Example: a(1)=28 because we have uuuuuuv, uuuuuvu, uuuuvuu, uuuvuuu, uuvuuuu, uvuuuuu, vuuuuuu, uuuuuuz, uuuuuzu, uuuuzuu, uuuzuuu, uuzuuuu, uzuuuuu, zuuuuuu, uuuuuux, uuuuuxu, uuuuxuu, uuuxuuu, uuxuuuu, uxuuuuu, xuuuuuu, uuuuuuy, uuuuuyu, uuuuyuu, uuuyuuu, uuyuuuu, uyuuuuu, yuuuuuu. - _Zerinvary Lajos_, Jun 16 2008

%H Vincenzo Librandi, <a href="/A054337/b054337.txt">Table of n, a(n) for n = 0..400</a>

%F a(n) = binomial(n+6, 6)*4^n.

%F G.f.: 1/(1 - 4*x)^7.

%F a(n) = A054335(n+13, 13).

%F E.g.f.: (45 + 1080*x + 5400*x^2 + 9600*x^3 + 7200*x^4 + 2304*x^5 + 256*x^6)*exp(4*x)/45. - _G. C. Greubel_, Jul 21 2019

%F From _Amiram Eldar_, Mar 25 2022: (Start)

%F Sum_{n>=0} 1/a(n) = 8394/5 - 5832*log(4/3).

%F Sum_{n>=0} (-1)^n/a(n) = 75000*log(5/4) - 83674/5. (End)

%p seq(seq(binomial(i, j)*4^(i-6), j =i-6), i=6..36); # _Zerinvary Lajos_, Dec 03 2007

%p seq(binomial(n+6,6)*4^n,n=0..30); # _Zerinvary Lajos_, Jun 16 2008

%t Table[4^n*Binomial[n+6,6], {n,0,30}] (* _G. C. Greubel_, Jul 21 2019 *)

%o (Sage) [lucas_number2(n, 4, 0)*binomial(n,6)/2^12 for n in range(6, 36)] # _Zerinvary Lajos_, Mar 11 2009

%o (Magma) [4^n*Binomial(n+6, 6): n in [0..30]]; // _Vincenzo Librandi_, Oct 15 2011

%o (PARI) vector(30, n, n--; 4^n*binomial(n+6,6) ) \\ _G. C. Greubel_, Jul 21 2019

%o (GAP) List([0..30], n-> 4^n*Binomial(n+6,6)); # _G. C. Greubel_, Jul 21 2019

%Y Cf. A000302, A054335.

%Y Cf. A038231.

%K easy,nonn

%O 0,2

%A _Wolfdieter Lang_, Mar 13 2000