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%I #19 Dec 08 2021 10:49:59
%S 1,2,3,7,16,37,89,216,529,1307,3248,8111,20339,51176,129143,326717,
%T 828374,2104361,5354979,13647682,34830191,89000157,227674188,
%U 583017657,1494365341,3833592212,9842373849,25287895051,65016153154,167264946727
%N Main diagonal of number square A081206.
%H G. C. Greubel, <a href="/A081207/b081207.txt">Table of n, a(n) for n = 0..1000</a>
%F a(n) = Sum_{k=0..n} (binomial(floor((n+k)/2), k))^2.
%F G.f.: (1+x)/sqrt(1-2x-x^2-2x^3+x^4) - _Paul Barry_, Jun 04 2005
%F Conjecture: n*(n-2)*a(n) +(-2*n^2+5*n-1)*a(n-1) +(-n^2+3*n-4)*a(n-2) +(-2*n^2+7*n-4)*a(n-3) +(n-1)*(n-3)*a(n-4)=0. - _R. J. Mathar_, Nov 12 2012
%F a(n) ~ (5-sqrt(5)) * ((3+sqrt(5))/2)^n / (2*sqrt(14*sqrt(5)-30) * sqrt(Pi*n)). - _Vaclav Kotesovec_, Feb 04 2014
%F Equivalently, a(n) ~ 5^(1/4) * phi^(2*n + 1) / (2 * sqrt(Pi*n)), where phi = A001622 is the golden ratio. - _Vaclav Kotesovec_, Dec 08 2021
%t Table[Sum[Binomial[Floor[(n+k)/2],k]^2,{k,0,n}],{n,0,30}] (* _Harvey P. Dale_, Oct 02 2011 *)
%t CoefficientList[Series[(1+x)/Sqrt[1-2x-x^2-2x^3+x^4], {x, 0, 20}], x] (* _Vaclav Kotesovec_, Feb 04 2014 *)
%o (PARI) for(n=0,25, print1(sum(k=0,n, (binomial(floor((n+k)/2), k))^2), ", ")) \\ _G. C. Greubel_, Feb 16 2017
%o (PARI) x='x+O('x^25); Vec((1+x)/sqrt(1-2*x-x^2-2*x^3+x^4)) \\ _G. C. Greubel_, Feb 16 2017
%Y Cf. A051286, A081206.
%K easy,nonn
%O 0,2
%A _Paul Barry_, Mar 11 2003
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