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%I #30 Sep 08 2022 08:45:24
%S 1,2,4,10,34,162,1090,10370,139522,2654722,71435266,2718435330,
%T 146299424770,11134711111682,1198484887715842,182431106853797890,
%U 39271952800672710658,11955805018770498519042,5147453397489773531365378
%N Row sums of triangle A117401: a(n) = Sum_{k=0..n} 2^((n-k)*k) for n>=0.
%C a(n) is the number of 2-colored labeled graphs (as in A047863) such that the black nodes are labeled with {1,2,...,k} where k, 0<=k<=n, is the number of black nodes and the white nodes are labeled with {k+1,k+2,...,n}. These graphs form the desired binomial poset (for the case q=2) in the "task left to the reader" in the Stanley reference below. - _Geoffrey Critzer_, May 31 2020
%D R. P. Stanley, Enumerative Combinatorics, Volume I, Second Edition, Cambridge, 2012, Example 3.18.3 e, page 323.
%H Robert Israel, <a href="/A117402/b117402.txt">Table of n, a(n) for n = 0..116</a>
%F G.f.: A(x) = Sum_{n>=0} x^n/(1-2^n*x).
%F Let E(x) = Sum_{n>=0} x^n/2^C(n,2). Then E(x)^2 = Sum_{n>=0} a(n)*x^n/2^C(n,2). - _Geoffrey Critzer_, May 31 2020
%F a(n) ~ c * 2^(n^2/4), where c = EllipticTheta[3, 0, 1/2] = JacobiTheta3(0,1/2) = 2.128936827211877158669458548544951324612516539940878092889... if n is even and c = EllipticTheta[2, 0, 1/2] = JacobiTheta2(0,1/2) = 2.128931250513027558591613402575350180853805396958448940969... if n is odd. - _Vaclav Kotesovec_, Jun 28 2021
%e A(x) = 1/(1-x) + x/(1-2x) + x^2/(1-4x) + x^3/(1-8x) + ...
%p N:= 25:
%p G:= series(add(x^n/(1-2^n*x),n=0..N),x,N+1):
%p seq(coeff(G,x,n),n=0..N)); # _Robert Israel_, Dec 11 2018
%t a[n_]:= Sum[2^((n-k)*k), {k,0,n}]; Array[a, 20, 0] (* _Amiram Eldar_, Dec 12 2018 *)
%o (PARI) a(n)=sum(k=0,n,2^((n-k)*k))
%o (Magma) [(&+[2^(k*(n-k)): k in [0..n]]): n in [0..20]]; // _G. C. Greubel_, Jun 28 2021
%o (Sage) [sum(2^(k*(n-k)) for k in (0..n)) for n in (0..20)] # _G. C. Greubel_, Jun 28 2021
%Y Cf. A117401 (triangle), A117403 (antidiagonal sums).
%K nonn
%O 0,2
%A _Paul D. Hanna_, Mar 12 2006
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