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a(n) = 1/Integral_{x=0..1} (x^5 - x^6)^n.
4

%I #23 Sep 08 2022 08:45:12

%S 1,42,858,15504,265650,4417686,72068304,1160068104,18490100706,

%T 292486494300,4599035681526,71963547329856,1121519754006288,

%U 17419158268943970,269767427275060200,4167406330765934256

%N a(n) = 1/Integral_{x=0..1} (x^5 - x^6)^n.

%H G. C. Greubel, <a href="/A090969/b090969.txt">Table of n, a(n) for n = 0..845</a>

%F a(n) = A016921(n)*A004355(n). - _R. J. Mathar_, Jun 21 2009

%F a(n) = 1/B(5*n+1,n+1) = (6*n+1)!/(n! * (5*n)!), where B(p,q) is Euler's beta function (basically identical with _R. J. Mathar_'s comment). - _Emeric Deutsch_, Jun 29 2009

%F a(n) ~ 2^(6*n+1) * 3^(6*n+3/2) * sqrt(n) / (sqrt(Pi) * 5^(5*n+1/2)). - _Vaclav Kotesovec_, Aug 15 2017

%p seq(factorial(6*n+1)/(factorial(n)*factorial(5*n)), n = 0 .. 16); # _Emeric Deutsch_, Jun 29 2009

%t Table[1/Beta[5*n+1, n+1], {n,0,20}] (* _G. C. Greubel_, Feb 03 2019 *)

%o (PARI) vector(20, n, n--; (6*n+1)!/(n!*(5*n)!)) \\ _G. C. Greubel_, Feb 03 2019

%o (Magma) [Factorial(6*n+1)/(Factorial(n)*Factorial(5*n)): n in [0..20]]; // _G. C. Greubel_, Feb 03 2019

%o (Sage) [1/beta(5*n+1,n+1) for n in range(20)] # _G. C. Greubel_, Feb 03 2019

%Y Cf. A002457, A004355, A016921, A090816, A090957.

%K nonn

%O 0,2

%A Al Hakanson (hawkuu(AT)excite.com), Feb 29 2004

%E Extended by _Emeric Deutsch_, Jun 29 2009