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%I #19 Aug 22 2022 14:31:44
%S 0,3,45,630,8925,128898,1891890,28113228,421717725,6372696330,
%T 96865261350,1479397776948,22684103302770,348986220258900,
%U 5384358887026500,83278084352019480,1290810308357922525,20045524792117559130,311819274557663622750,4857816066846093281700
%N a(n) = sum of numbers which in base 2 contain exactly n digits 1 and not more than n digits 0.
%H Michael De Vlieger, <a href="/A131568/b131568.txt">Table of n, a(n) for n = 0..832</a>
%F a(n) = (2*n)!*(2^(2*n)-1)/(2*n!*n!). [Corrected by _Michel Marcus_, Aug 22 2022]
%F From _Peter Luschny_, Aug 22 2022: (Start)
%F a(n) = (4^n - 1)*binomial(2*n, n) / 2.
%F a(n) = [x^n] ((1 - 16*x)^(-1/2) - (1 - 4*x)^(-1/2)) / 2. (End)
%e For n=2, we have these integers in base 2: 11, 101, 110, 1001, 1010, and 1100; so a(2) = 3+5+6+9+10+12 = 45.
%t Array[(4^# - 1)*Binomial[2 #, #]/2 &, 20, 0] (* _Michael De Vlieger_, Aug 22 2022 *)
%K nonn,base,easy
%O 0,2
%A Philippe LALLOUET (philip.lallouet(AT)wanadoo.fr), Aug 27 2007
%E More terms from _Peter Luschny_, Aug 22 2022
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