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%I #38 Nov 01 2023 03:52:48
%S 1,9,121,2025,38025,762129,15912121,341621289,7484845225,166549691025,
%T 3751508008161,85341068948529,1957289174870121,45199191579030225,
%U 1049893021288265625,24510327614556266025,574726636455361317225,13528549573868347823025,319541915502909478890625
%N G.f.: 1 / AGM(1-5*x, sqrt((1-x)*(1-25*x))).
%C In general, the g.f. of the squares of coefficients in g.f. 1/sqrt((1-p*x)*(1-q*x)) is given by
%C 1/AGM(1-p*q*x, sqrt((1-p^2*x)*(1-q^2*x))) = Sum_{n>=0} x^n*[Sum_{k=0..n} p^(n-k)*((q-p)/4)^k*C(n,k)*C(2*k,k)]^2,
%C and consists of integer coefficients when 4|(q-p).
%C Here AGM(x,y) = AGM((x+y)/2,sqrt(x*y)) is the arithmetic-geometric mean.
%H Seiichi Manyama, <a href="/A246467/b246467.txt">Table of n, a(n) for n = 0..717</a>
%F a(n) = A026375(n)^2 = [Sum_{k=0..n} binomial(n,k)*binomial(2*k,k)]^2.
%F G.f.: 1 / AGM((1-x)*(1+5*x), (1+x)*(1-5*x)) = Sum_{n>=0} a(n)*x^(2*n).
%F a(n) ~ 5^(2*n+1) / (4*Pi*n). - _Vaclav Kotesovec_, Dec 10 2018
%e G.f.: A(x) = 1 + 9*x + 121*x^2 + 2025*x^3 + 38025*x^4 + 762129*x^5 +...
%e where the square-root of the terms yields A026375:
%e [1, 3, 11, 45, 195, 873, 3989, 18483, 86515, 408105, ...]
%e the g.f. of which is 1/sqrt((1-x)*(1-5*x)).
%t CoefficientList[Series[1/ArithmeticGeometricMean[1-5x,Sqrt[(1-x)(1-25x)]],{x,0,20}],x] (* _Harvey P. Dale_, Nov 01 2023 *)
%o (PARI) {a(n)=polcoeff( 1 / agm(1-5*x, sqrt((1-x)*(1-25*x) +x*O(x^n))), n)}
%o for(n=0, 20, print1(a(n), ", "))
%o (PARI) {a(n)=sum(k=0,n,binomial(n,k)*binomial(2*k,k))^2}
%o for(n=0, 20, print1(a(n), ", "))
%Y Cf. A026375, A168597, A246876, A246906, A248167.
%K nonn
%O 0,2
%A _Paul D. Hanna_, Sep 06 2014
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