%I #23 Jan 15 2024 18:18:19
%S 1,10,175,3500,74375,1636250,36815625,841500000,19459687500,
%T 454059375000,10670395312500,252209343750000,5989971914062500,
%U 142837791796875000,3417904303710937500,82029703289062500000
%N Expansion of (1-25*x)^(-2/5).
%H Robert Israel, <a href="/A049380/b049380.txt">Table of n, a(n) for n = 0..716</a>
%F G.f.: (1-25*x)^(-2/5).
%F a(n) = (5^n/n!) * Product_{k=0..n-1} (5*k + 2).
%F a(n) ~ Gamma(2/5)^(-1)*n^(-3/5)*5^(2*n)*{1 - 3/25*n^-1 + ...}. - Joe Keane (jgk(AT)jgk.org), Nov 24 2001
%F a(n+1) = (10 + 25*n)*a(n)/(n+1). - _Robert Israel_, Sep 04 2018
%F a(n) = (-25)^n*binomial(-2/5,n). - _Peter Luschny_, Oct 23 2018
%F From _Peter Bala_, Sep 24 2023: (Start)
%F a(n) = 25^n * binomial(n - 3/5, n).
%F P-recursive: a(n) = 5*(5*n - 3)/n * a(n-1) with a(0) = 1. (End)
%e (1-x)^(-2/5) = 1 + 2/5*x + 7/25*x^2 + 28/125*x^3 + ...
%p f:= gfun:-rectoproc({a(n+1) = (10+25*n)*a(n)/(n+1),a(0)=1},a(n),remember):
%p map(f, [$0..50]); # _Robert Israel_, Sep 04 2018
%t CoefficientList[Series[1/Surd[(1-25x)^2,5],{x,0,20}],x] (* _Harvey P. Dale_, Jan 15 2024 *)
%o (PARI) x='x+O('x^99); Vec((1-25*x)^(-2/5)) \\ _Altug Alkan_, Sep 04 2018
%Y Cf. A034688, A049381, A049382, A049391, A049395, A049382.
%K nonn,easy
%O 0,2
%A Joe Keane (jgk(AT)jgk.org)