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%I #19 Aug 19 2022 04:16:54
%S 1,7,57,511,4849,47607,477609,4862319,50026977,518839783,5414767897,
%T 56795795679,598213529809,6322787125207,67026654455433,
%U 712352213507151,7587639773475777,80977812878889927,865716569022673401,9269461606674304959,99387936492243451569,1066975862517563301303
%N G.f.: 1/sqrt((1-3*x)*(1-11*x)).
%H Seiichi Manyama, <a href="/A248168/b248168.txt">Table of n, a(n) for n = 0..961</a>
%H Hacène Belbachir, Abdelghani Mehdaoui, László Szalay, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL22/Szalay/szalay42.html">Diagonal Sums in the Pascal Pyramid, II: Applications</a>, J. Int. Seq., Vol. 22 (2019), Article 19.3.5.
%F a(n) equals the central coefficient in (1 + 7*x + 4*x^2)^n, n>=0.
%F a(n) = Sum_{k=0..n} 3^(n-k) * 2^k * C(n,k) * C(2*k,k).
%F a(n) = Sum_{k=0..n} 11^(n-k) * (-2)^k * C(n,k) * C(2*k,k). - _Paul D. Hanna_, Apr 20 2019
%F a(n)^2 = A248167(n), which gives the coefficients in 1 / AGM(1-3*11*x, sqrt((1-3^2*x)*(1-11^2*x))).
%F Equals the binomial transform of 2^n*A026375(n).
%F Equals the second binomial transform of A084771.
%F Equals the third binomial transform of A059304(n) = 2^n*(2*n)!/(n!)^2.
%F a(n) ~ 11^(n+1/2)/(2*sqrt(2*Pi*n)). - _Vaclav Kotesovec_, Oct 03 2014
%F D-finite with recurrence: n*a(n) +7*(-2*n+1)*a(n-1) +33*(n-1)*a(n-2)=0. [Belbachir]
%e G.f.: A(x) = 1 + 7*x + 57*x^2 + 511*x^3 + 4849*x^4 + 47607*x^5 +...
%e where A(x)^2 = 1/((1-3*x)*(1-11*x)):
%e A(x)^2 = 1 + 14*x + 163*x^2 + 1820*x^3 + 20101*x^4 + 221354*x^5 +...
%t CoefficientList[Series[1/Sqrt[(1-3*x)*(1-11*x)], {x, 0, 20}], x] (* _Vaclav Kotesovec_, Oct 03 2014 *)
%o (PARI) {a(n)=polcoeff( 1 / sqrt((1-3*x)*(1-11*x) +x*O(x^n)), n) }
%o for(n=0, 25, print1(a(n), ", "))
%o (PARI) {a(n)=polcoeff( (1 + 7*x + 4*x^2 +x*O(x^n))^n, n) }
%o for(n=0, 25, print1(a(n), ", "))
%o (PARI) {a(n)=sum(k=0,n, 3^(n-k)*2^k*binomial(n,k)*binomial(2*k,k))}
%o for(n=0, 25, print1(a(n), ", "))
%Y Cf. A248167, A084771.
%K nonn
%O 0,2
%A _Paul D. Hanna_, Oct 03 2014
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