|
%I #7 Jan 05 2015 15:05:07
%S 1,45,1575,49875,1496250,43391250,1229418750,34248093750,941822578125,
%T 25638503515625,692239594921875,18564607318359375,495056195156250000,
%U 13138029794531250000,347219358855468750000
%N Expansion of (1-25*x)^(-9/5).
%H Todd Silvestri, <a href="/A049397/b049397.txt">Table of n, a(n) for n = 0..713</a>
%F G.f.: (1-25*x)^(-9/5).
%F a(n) = 5^n/n! * product[ k=0..n-1 ] (5*k+9).
%F a(n) ~ 5/4*Gamma(4/5)^-1*n^(4/5)*5^(2*n)*{1 + 18/25*n^-1 - ...}. - Joe Keane (jgk(AT)jgk.org), Nov 24 2001
%F a(n) = (25^n*(9/5)_n)/n!, where the rising factorial (c)_n = Gamma(c+n)/Gamma(c). - _Todd Silvestri_, Dec 17 2014. See the a(n) formula above.
%e (1-25*x)^(-9/5) = 1 + 9/5*(5^2*x) + 63/25*(5^2*x)^2 + 399/125*(5^2*x)^3 + ... = 1 + 5*9*x + 63*5^2* x^2 + 399*5^3*x^3 + ... = 1 + 45*x + 1575*x^2 + 49875*x^3 + ...
%t a[n_Integer/;n>=0]:=25^n Pochhammer[9/5,n]/n! (* _Todd Silvestri_, Dec 17 2014 *)
%Y Cf. A049382.
%K nonn,easy
%O 0,2
%A Joe Keane (jgk(AT)jgk.org)
%E Edited: name and example corrected according to G.f. - _Wolfdieter Lang_, Jan 05 2015
| 