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A297705 a(n) = Sum_{k=0..n} binomial(n, k)*hypergeom([k - n, n + 1], [k + 2], -4).

%I A297705 #36 Jan 15 2024 16:19:59
%S A297705 1,6,56,636,8036,108516,1533316,22389396,335177396,5116746276,
%T A297705 79350018276,1246583463156,19797057247956,317304181980036,
%U A297705 5126097354722436,83384214787592916,1364582474360361716,22450780862515853796,371129420131691349796,6161232472210183370676
%N A297705 a(n) = Sum_{k=0..n} binomial(n, k)*hypergeom([k - n, n + 1], [k + 2], -4).
%H A297705 Robert Israel, <a href="/A297705/b297705.txt">Table of n, a(n) for n = 0..800</a>
%F A297705 From _Robert Israel_, Jan 08 2018: (Start)
%F A297705 G.f.: (1 - 3*x - sqrt(1 - 18*x + x^2))/(6*x + 4*x^2).
%F A297705 (-2*n-4)*a(n+1) + (33*n+120)*a(n+2) + (52*n+179)*a(n+3) + (-3*n-15)*a(n+4) = 0.
%F A297705 (End) [Conjectures verified with Maple's FormalPowerSeries Module. - _Peter Luschny_, Nov 10 2022]
%F A297705 a(n) ~ 5^(1/4) * phi^(6*n - 3) / (sqrt(2*Pi) * (349 - 156*sqrt(5)) * n^(3/2)), where phi = A001622 = (1 + sqrt(5))/2 is the golden ratio. - _Vaclav Kotesovec_, Jul 05 2018
%F A297705 O.g.f. A(x) = (1/x) * series reversion of x*(1 - 3*x)/((1 + x)*(1 + 2*x)). Cf. A114710. - _Peter Bala_, Nov 08 2022
%F A297705 a(n) = Sum_{k=0..n} (Sum_{j=k..n} 4^(j - k)*(k + 1)*binomial(n + j - k, 2*j - k)*binomial(2*j - k, j - k)/(j + 1)). - _Detlef Meya_, Jan 15 2024
%p A297705 f:= n -> simplify(add(binomial(n,k)*hypergeom([k-n,n+1],[k+2],-4),k=0..n)):
%p A297705 map(f, [$0..40]); # _Robert Israel_, Jan 08 2018
%p A297705 a := proc(n) option remember; if n < 4 then return [1,6,56,636][n + 1] fi;
%p A297705 ((-26*n^2 + 130*n - 156)*a(n - 4) + (423*n^2 - 1431*n + 1170)*a(n - 3) + (775*n^2 - 2441*n + 2106)*a(n - 2) + n*9*(13*n - 1)*a(n - 1))/(9*(n + 1)*n) end:
%p A297705 seq(a(n), n = 0..19); # _Peter Luschny_, Nov 10 2022
%t A297705 a[n_] := Sum[Binomial[n, k] Hypergeometric2F1[k - n, n + 1, k + 2, -4], {k, 0, n}]; Table[a[n], {n, 0, 17}]
%t A297705 a[n_] := Sum[Sum[4^(j - k)*(k + 1)*Binomial[n + j - k, 2*j - k]*Binomial[2*j - k, j - k]/(j + 1), {j, k, n}], {k, 0, n}]; Flatten[Table[a[n], {n, 0, 19}]] (* _Detlef Meya_, Jan 15 2024 *)
%o A297705 (PARI) a(n) = sum(k=0, n, sum(j=k, n, 4^(j - k)*(k + 1)*binomial(n + j - k, 2*j - k)*binomial(2*j - k, j - k)/(j + 1))) \\ _Andrew Howroyd_, Jan 15 2024
%Y A297705 Cf. A001622, A114710.
%K A297705 nonn,easy
%O A297705 0,2
%A A297705 _Peter Luschny_, Jan 08 2018