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%I #34 Sep 11 2019 20:23:10
%S 1,14,540,40824,5103000,953029440,248619571200,86355926616960,
%T 38528927611574400,21473732319740064000,14620825330739032204800,
%U 11941607887300551753216000,11523529003703200697461248000,12970646659082235068963297280000
%N Number of sequences over the alphabet of n symbols of length 2n which have n distinct symbols. Also number of placements of 2n balls into n cells where no cell is empty.
%H Vincenzo Librandi, <a href="/A210029/b210029.txt">Table of n, a(n) for n = 1..200</a>
%F a(n) = Sum_{v=0..n-1}( (-1)^v * binomial(n,v) * (n-v)^(2n) ).
%F a(n) = n! * S2(2*n,n), where S2(n,k) = A008277(n,k) are the Stirling numbers of the second kind. - _Paul D. Hanna_, Oct 26 2012 [Also the central column of A131689 (suggesting a(0) = 1). - _Peter Luschny_, Sep 11 2019]
%F E.g.f.: Sum_{n>=1} (n^2)^n * exp(-n^2*x) * x^n/n! = Sum_{n>=1} a(n)*x^n/n!. - _Paul D. Hanna_, Oct 26 2012
%F a(n) ~ n^(2*n)*(2/(exp(c)*(2-c)))^n / sqrt(1-c), where c = -LambertW(-2/exp(2)) = 0.406375739959959907676958... - _Vaclav Kotesovec_, Jan 02 2013
%F O.g.f.: Sum_{n>=1} n^(2*n) * x^n / (1 + n^2*x)^(n+1). - _Paul D. Hanna_, Feb 24 2013
%F a(n) = [x^n] P(2*n) where P(n) = Sum_{k=1..n} binomial(n, k)*P(n-k)*x based in P(0) = 1. - _Peter Luschny_, Sep 11 2019
%e a(2) = 14 because the 2^4 sequences on 2 symbols of length 4 can be represented by 0000, 0001, 0010, 0011, 0100, 0101, 0110, 0111, 1000, 1001, 1010, 1011, 1100,1110, and 1111. Only two of them do not have n distinct symbols.
%e a(10)= 21473732319740064000 since all digits appear in 21473732319740064000 nonnegative integers with 20 digits.
%e O.g.f.: A(x) = 1 + 14*x + 540*x^2 + 40824*x^3 + 5103000*x^4 + ... where
%e A(x) = x/(1+x)^2 + 2^4*x^2/(1+4*x)^3 + 3^6*x^3/(1+9*x)^4 + 4^8*x^4/(1+16*x)^5 + 5^10*x^5/(1+25*x)^6 +... - _Paul D. Hanna_, Feb 24 2013
%e E.g.f.: E(x) = 1 + 14*x + 540*x^2/2! + 40824*x^3/3! + 5103000*x^4/4! + ... where
%e E(x) = exp(-x)*x + 2^4*exp(-4*x)*x^2/2! + 3^6*exp(-9*x)*x^3/3! + 4^8*exp(-16*x)*x^4/4! + 5^10*exp(-25*x)*x^5/5! +... - _Paul D. Hanna_, Feb 24 2013
%p P := proc(n) option remember; if n = 0 then return 1 fi;
%p expand(add(binomial(n,k)*P(n-k)*x, k=1..n)) end:
%p a := n -> coeff(P(2*n), x, n); # _Peter Luschny_, Sep 11 2019
%t Table[Sum[((-1)^v*Binomial[n, v]*(n - v)^(2 n)), {v, 0, n - 1}], {n, 20}] (* _T. D. Noe_, Mar 16 2012 *)
%o (PARI) {a(n)=n!*polcoeff(sum(m=1, n, (m^2)^m*exp(-m^2*x+x*O(x^n))*x^m/m!), n)} \\ _Paul D. Hanna_, Oct 26 2012
%o (PARI) {a(n)=polcoeff(sum(k=1, n, (k^2)^k*x^k/(1+k^2*x +x*O(x^n))^(k+1)), n)} for(n=0, 20, print1(a(n), ", ")) \\ _Paul D. Hanna_, Feb 24 2013
%Y Cf. A131689, A210024, A209899, A209900, A007820, A008277.
%K nonn
%O 1,2
%A _Washington Bomfim_, Mar 16 2012
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