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a(0) = a(1) = a(2) = 1; a(n) = (1/2) * Sum_{k=0..n-2} binomial(n-2,k) * a(k) * a(n-k-2).
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%I #19 Jan 21 2021 13:50:49

%S 1,1,1,1,2,4,9,24,73,241,881,3526,15252,70989,354454,1887349,10674773,

%T 63936181,404240736,2690174976,18798077227,137612318374,1053181689299,

%U 8410678868749,69966341141373,605313810207096,5438284743262816,50668461879077851

%N a(0) = a(1) = a(2) = 1; a(n) = (1/2) * Sum_{k=0..n-2} binomial(n-2,k) * a(k) * a(n-k-2).

%H Vaclav Kotesovec, <a href="/A336023/b336023.txt">Table of n, a(n) for n = 0..500</a>

%F E.g.f. A(x) satisfies: A(x) = 1 + x + x^2/4 + (1/2) * Integral( Integral A(x)^2 dx) dx.

%F a(n) ~ n! * c * d^n * n, where d = 0.3327497845052484431663758951399704951054382840130942017786917698589406... is the root of the equation WeierstrassP(1/d, -1/12, 1/432) = 1/12 and c = 1.3286690290594713229820603959597228439699874045711069972066245497... - _Vaclav Kotesovec_, Jul 05 2020, updated Nov 28 2020

%F E.g.f.: 12^(1/3) * WeierstrassP((x + c)/12^(1/3), {-12^(1/3), 1/3}), where c = 6.010522299732561104963296256980218197216667207108031458739075310344... is the root of the equation WeierstrassP(c/12^(1/3), {-12^(1/3), 1/3}) = 12^(-1/3). - _Vaclav Kotesovec_, Jul 05 2020

%t a[0] = a[1] = a[2] = 1; a[n_] := a[n] = (1/2) Sum[Binomial[n - 2, k] a[k] a[n - k - 2], {k, 0, n - 2}]; Table[a[n], {n, 0, 27}]

%t nmax = 27; A[_] = 0; Do[A[x_] = 1 + x + x^2/4 + 1/2 Integrate[Integrate[A[x]^2, x], x] + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x] Range[0, nmax]!

%t Table[SeriesCoefficient[1 + (72 (WeierstrassP[x, {-1/12, 1/432}] - WeierstrassPPrime[x, {-1/12, 1/432}]))/(1 - 12 WeierstrassP[x, {-1/12, 1/432}])^2, {x, 0, k}] k!, {k, 0, 30}] (* _Jan Mangaldan_, Nov 27 2020 *)

%t a[ n_] := Module[{b, b0, b1, b2}, b[0]=b0; b[m_] := b[n] = Expand[Dt[b[m-1]] /. {Dt[b0]->b1, Dt[b1]->b2, Dt[b2]->b0*b1}]; b[n] /. {b0->1, b1->1, b2->1}]; (* _Michael Somos_, Jan 21 2021 *)

%o (PARI) {a(n) = my(t, b0='b0, b1='b1, b2='b2); t = b0; for(k=1, n, t = deriv(t, b0) * b1 + deriv(t, b1) * b2 + deriv(t, b2) * b0*b1); substvec(t, [b0, b1, b2], [1, 1, 1])}; /* _Michael Somos_, Jan 21 2021 */

%Y Cf. A000111, A007558.

%K nonn

%O 0,5

%A _Ilya Gutkovskiy_, Jul 05 2020