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%I #31 Aug 27 2018 09:03:04
%S 1,6,54,588,7110,91476,1224636,16849944,236523078,3371140740,
%T 48630906324,708412918824,10403176168476,153813188724552,
%U 2287366047735480,34185974267420208,513159651195396678,7732530110414488932
%N a(n) equals the sum of the squares of the expansion coefficients for (x + y + 2*z)^n.
%C Equivalently, a(n) equals the sum of the squares of the coefficients in any one of the following polynomials: (1 + 2*x^k + x^p)^n, (2 + x^k + x^p)^n, or (1 + x^k + 2*x^p)^n, for all p > n*k and fixed k > 0.
%C Rescaling the g.f. G(x) to T(u)=G(3*u/16) moves the singular point x=1/16 to u=1/3. Period function T(u) measures precession of the J-vector along an algebraic sphere curve with local cyclic C_3 symmetry. For precise definitions, pictures, a proof certificate, and more information, see A318245. - _Bradley Klee_, Aug 22 2018
%H Vincenzo Librandi, <a href="/A186375/b186375.txt">Table of n, a(n) for n = 0..200</a>
%H E. Weisstein, <a href="http://mathworld.wolfram.com/GoursatsSurface.html">Goursat's Surface</a>, Mathworld--A Wolfram Web Resource.
%F (1) a(n) = Sum_{k=0..n} C(n,k)^2*C(2k,k)*4^(n-k).
%F Let g.f. A(x) = Sum_{n>=0} a(n)*x^n/n!^2, then
%F (2) A(x) = B(x)^2 * B(2^2*x)
%F where B(x) = Sum_{n>=0} x^n/n!^2 = BesselI(0, 2*sqrt(x)).
%F Recurrence: n^2*a(n) = 2*(10*n^2-10*n+3)*a(n-1) - 4*(4*n-5)*(4*n-3)*a(n-2). - _Vaclav Kotesovec_, Oct 20 2012
%F a(n) ~ 2^(4*n+1/2)/(Pi*n). - _Vaclav Kotesovec_, Oct 20 2012
%F a(n) = 4^n*hypergeom([1/2,-n,-n], [1,1], 1). - _Peter Luschny_, May 24 2017
%F G.f.: G(x)=Sum_{n>=0}a(n)x^n, 6*(10*x-1)*G + (192*x^2-40*x+1)*G' + x*(16*x-1)*(4*x-1)*G''=0. - _Bradley Klee_, Aug 22 2018
%e G.f.: A(x) = 1 + 6*x + 54*x^2/2!^2 + 588*x^3/3!^2 + 7110*x^4/4!^2 + ...
%e The g.f. may be expressed as:
%e A(x) = [Sum_{n>=0} x^n/n!^2]^2 *[Sum_{n>=0} (4x)^n/n!^2] where
%e [Sum_{n>=0} x^n/n!^2]^2 = 1 + 2*x + 6*x^2/2!^2 + 20*x^3/3!^2 + 70*x^4/4!^2 + ... + (2n)!/n!^2 *x^n/n!^2 + ...
%e a(4) = 256
%e + 1024 + 1024
%e + 576 + 2304 + 576
%e + 64 + 576 + 576 + 64
%e + 1 + 16 + 36 + 16 + 1 = 7110.
%p A186375 := n -> 4^n*hypergeom([1/2,-n,-n], [1,1], 1):
%p seq(simplify(A186375(n)), n=0..17); # _Peter Luschny_, May 24 2017
%t Table[Sum[Binomial[n,k]^2*Binomial[2k,k]*4^(n-k),{k,0,n}],{n,0,20}] (* _Vaclav Kotesovec_, Oct 20 2012 *)
%t (* From _Bradley Klee_, Aug 22 2018: Start *)PyramidLevel[n_]:=If[n==0, {{1}}, Table[Coefficient[(2*x+y+z)^n,x^j*y^k*z^(n-j-k)]^2, {j,0,n}, {k,0,n-j}]]; a1[n_]:= Total[Flatten[PyramidLevel[n]]];
%t a1 /@ Range[0, 10]
%t RecurrenceTable[{4*(4*n-5)*(4*n-3)*a[n-2]-2*(10*n^2-10*n+3)*a[n-1]+n^2*a[n]==0, a[0]==1, a[1]==6},a,{n,0,1000}] (* End *)
%t a[ n_] := If[ n < 0, 0, Block[ {x, y, z}, Expand[ (x + y + 2 z)^n] /. {t_Integer -> t^2, x -> 1, y -> 1, z -> 1}]]; (* _Michael Somos_, Aug 27 2018 *)
%o (PARI) {a(n)=sum(k=0,n,binomial(n,k)^2*binomial(2*(n-k),n-k)*4^k)}
%o (PARI) {a(n)=n!^2*polcoeff(sum(m=0,n,x^m/m!^2)^2*sum(m=0,n,(2^2*x)^m/m!^2),n)}
%o (PARI) {a(n)=local(V=Vec((1+2*x+x^(n+2))^n));V*V~}
%Y Cf. A046816, A186376, A186377, A186378. Periods: A318245, A318417.
%K nonn
%O 0,2
%A _Paul D. Hanna_, Feb 19 2011
%E Name edited by _Bradley Klee_, Aug 22 2018
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