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%I #23 Mar 12 2024 02:41:59
%S 1,1250,2343750,5126953125,12176513671875,30441284179687500,
%T 78821182250976562500,209368765354156494140625,
%U 567040406167507171630859375,1559361116960644721984863281250,4341403109719976782798767089843750,12210196246087434701621532440185546875
%N Expansion of (1 - (1 - 3125*x)^(1/5)) / (625*x).
%C In general, if k > 1 and g.f. = (1 - (1 - k^k * x)^(1/k)) / (k^(k-1) * x), then a(n) ~ k^(k*n) / (Gamma((k-1)/k) * n^((k+1)/k)).
%H G. C. Greubel, <a href="/A254287/b254287.txt">Table of n, a(n) for n = 0..250</a>
%F G.f.: (1 - (1 - 3125*x)^(1/5)) / (625*x).
%F a(n) ~ 3125^n / (Gamma(4/5) * n^(6/5)).
%F Recurrence: (n+1)*a(n) = 625*(5*n-1)*a(n-1).
%F a(n) = 5^(5*n) * Gamma(n+4/5) / (Gamma(4/5) * Gamma(n+2)).
%F E.g.f.: hypergeom([4/5], [2], 3125*x). - _Vaclav Kotesovec_, Jan 28 2015
%F From _Peter Bala_, Sep 01 2017: (Start)
%F a(n) = (-1)^n*binomial(1/5, n+1)*5^(5*n+1). Cf. A000108(n) = (-1)^n*binomial(1/2, n+1)*2^(2*n+1).
%F a(n) = 125^n*A025748(n+1). (End)
%t CoefficientList[Series[(1-(1-3125*x)^(1/5)) / (625*x),{x,0,20}],x]
%t CoefficientList[Series[Hypergeometric1F1[4/5,2,3125*x],{x,0,20}],x] * Range[0,20]! (* _Vaclav Kotesovec_, Jan 28 2015 *)
%o (Magma) [Round(5^(5*n)*Gamma(n+4/5)/(Gamma(4/5)*Gamma(n+2))): n in [0..30]]; // _G. C. Greubel_, Aug 10 2022
%o (SageMath) [5^(5*n)*rising_factorial(4/5, n)/factorial(n+1) for n in (0..30)] # _G. C. Greubel_, Aug 10 2022
%Y Cf. A000108 (k=2), A254282 (k=3), A254286 (k=4).
%Y Cf. A008546, A025748.
%K nonn,easy
%O 0,2
%A _Vaclav Kotesovec_, Jan 27 2015
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