|
%I #19 Sep 27 2019 17:09:24
%S 1,1,81,625,21025,314721,8071281,155975121,3685097025,79065004225,
%T 1832690505361,41201108978481,955863424195681,22002282841672225,
%U 513656632397480625,11987246142597870225,281793325930991748225,6636532731958908324225,156986802654088221500625,3721955785026092495600625,88512056987508595608293025
%N a(n) = A084605(n)^2, the square of the central coefficient in (1 + x + 4*x^2)^n.
%C The g.f. of A084605 is 1/sqrt(1 - 2*x - 15*x^2).
%C Ignoring initial term, a(n) equals the logarithmic derivative of A322241.
%H Seiichi Manyama, <a href="/A322240/b322240.txt">Table of n, a(n) for n = 0..717</a>
%F G.f.: 1 / AGM(1 + 15*x, sqrt((1 - 9*x)*(1 - 25*x)) ), where AGM(x,y) = AGM((x+y)/2, sqrt(x*y)) is the arithmetic-geometric mean.
%F G.f.: 1 / AGM((1-3*x)*(1-5*x), (1+3*x)*(1+5*x)) = Sum_{n>=0} a(n) * x^(2*n).
%F a(n) = A084605(n)^2, where A084605(n) = Sum_{k=0..n} (-3)^(n-k) * 2^k * binomial(n,k)*binomial(2*k,k).
%F a(n) ~ 5^(2*n+1) / (8*Pi*n). - _Vaclav Kotesovec_, Sep 27 2019
%e G.f.: A(x) = 1 + x + 81*x^2 + 625*x^3 + 21025*x^4 + 314721*x^5 + 8071281*x^6 + 155975121*x^7 + 3685097025*x^8 + 79065004225*x^9 + ... + A084605(n)^2*x^n + ...
%e that is,
%e A(x) = 1 + x + 9^2*x^2 + 25^2*x^3 + 145^2*x^4 + 561^2*x^5 + 2841^2*x^6 + 12489^2*x^7 + 60705^2*x^8 + 281185^2*x^9 + ... + A084605(n)^2*x^n + ...
%t a[n_] := Sum[(-3)^(n-k) * 2^k * Binomial[n,k] * Binomial[2k,k], {k, 0, n}]^2; Array[a, 20, 0] (* _Amiram Eldar_, Dec 13 2018 *)
%o (PARI) /* a(n) = A084605(n)^2 */
%o {a(n)=polcoeff(1/sqrt(1 - 2*x - 15*x^2 +x*O(x^n)), n)^2}
%o for(n=0, 20, print1(a(n), ", "))
%o (PARI) /* Using AGM: */
%o {a(n)=polcoeff( 1 / 1 / agm(1 + 3*5*x, sqrt((1 - 3^2*x)*(1 - 5^2*x) +x*O(x^n))), n)}
%o for(n=0, 20, print1(a(n), ", "))
%o (PARI) /* Square of binomial sum */
%o {a(n) = sum(k=0,n, (-3)^(n-k)*2^k*binomial(n,k)*binomial(2*k,k))^2}
%o for(n=0, 20, print1(a(n), ", "))
%Y Cf. A322241, A084605.
%K nonn
%O 0,3
%A _Paul D. Hanna_, Dec 08 2018
|
