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a(0)=1, a(m+1) = Sum_{k=0..m}[a(k)^2 * a(m-k)^2].
1

%I #21 Mar 09 2024 11:02:39

%S 1,1,2,9,170,57978,6722955416,90396259057770600841,

%T 16342967303279146791599896178138243550346

%N a(0)=1, a(m+1) = Sum_{k=0..m}[a(k)^2 * a(m-k)^2].

%C a(9) = 5.341...*10^80 is too large to include in the data section. - _Amiram Eldar_, Oct 02 2021

%H Michael De Vlieger, <a href="/A053294/b053294.txt">Table of n, a(n) for n = 0..12</a>

%H Jean-Marc Luck, <a href="https://arxiv.org/abs/2403.00432">Revisiting log-periodic oscillations</a>, arXiv:2403.00432 [cond-mat.stat-mech], 2024. See p. 7.

%e a(4)= a(0)^2*a(3)^2 + a(1)^2*a(2)^2 + a(2)^2*a(1)^2 + a(3)^2*a(0)^2 = 1^2*9^2 + 1^2*2^2 + 2^2*1^2 + 9^2*1^2 = 170.

%t a[0] = 1; a[n_] := a[n] = Sum[a[k]^2*a[n - 1 - k]^2, {k, 0, n - 1}]; Array[a, 9, 0] (* _Amiram Eldar_, Oct 02 2021 *)

%K nonn

%O 0,3

%A _Leroy Quet_, Mar 04 2000