A186284 - OEIS

%I #15 Aug 14 2018 23:38:17

%S 1,2,48,1704,71490,3291780,160844160,8189867280,429832053840,

%T 23088359467040,1263134996327680,70138971602098560,

%U 3942799810867610280,223942062435751452240,12831882367225056387840,740872398293620831990080

%N Self-convolution square equals A127776.

%H G. C. Greubel, <a href="/A186284/b186284.txt">Table of n, a(n) for n = 0..500</a>

%F Self-convolution 4th power equals A002897.

%F G.f.: sqrt( K(k)/(Pi/2) ) in powers of (kk'/4)^2, where K(k) is complete elliptic integral of first kind evaluated at modulus k. [From a formula by _Michael Somos_ in A002897]

%F G.f.: sqrt( 1/AGM(1, (1-16x)^(1/2)) ) in powers of x(1-16x) where AGM() is the arithmetic-geometric mean. [From a formula by _Michael Somos_ in A004981]

%F a(n) ~ Pi^(3/4) * 2^(6*n - 1/2) / (Gamma(1/4)^3 * n^(3/2)). - _Vaclav Kotesovec_, Apr 10 2018

%e G.f.: A(x) = 1 + 2*x + 48*x^2 + 1704*x^3 + 71490*x^4 + 3291780*x^5 +...

%e Related expansions.

%e The g.f. of A127776 equals A(x)^2:

%e A(x)^2 = 1 + 4*x + 100*x^2 + 3600*x^3 + 152100*x^4 + 7033104*x^5 +...+ A004981(n)^2*x^n +...

%e The g.f. of A002897 equals A(x)^4:

%e A(x)^4 = 1 + 8*x + 216*x^2 + 8000*x^3 + 343000*x^4 + 16003008*x^5 +...+ A000984(n)^3*x^n +...

%e The g.f. of A004981 begins:

%e 1/(1-8*x)^(1/4) = 1 + 2*x + 10*x^2 + 60*x^3 + 390*x^4 + 2652*x^5 +...

%e where A004981(n) = (2^n/n!)*Product_{k=0..n-1} (4k + 1).

%e The g.f. of A000984 begins:

%e 1/(1-4*x)^(1/2) = 1 + 2*x + 6*x^2 + 20*x^3 + 70*x^4 + 252*x^5 +...

%e where A000984(n) = (2n)!/(n!)^2 forms the central binomial coefficients.

%t nmax = 20; CoefficientList[Series[Sqrt[Hypergeometric2F1[ 1/4, 1/4, 1, 64*x]], {x, 0, nmax}], x] (* _Vaclav Kotesovec_, Apr 10 2018 *)

%o (PARI) {a(n)=local(A004981=1/(1-8*x+x*O(x^n))^(1/4),A=sum(m=0,n,polcoeff(A004981,m)^2*x^m+x*O(x^n))^(1/2));polcoeff(A,n)}

%o (PARI) {a(n)=local(A000984=1/(1-4*x+x*O(x^n))^(1/2),A=sum(m=0,n,polcoeff(A000984,m)^3*x^m+x*O(x^n))^(1/4));polcoeff(A,n)}

%Y Cf. A004981, A000984, A127776, A002897.

%K nonn

%O 0,2

%A _Paul D. Hanna_, Feb 16 2011