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a(0) = 1; for n > 0, a(n) = -1 + 4*Product_{i=0..n-1} a(i)^2.
5

%I #25 Apr 16 2023 20:54:35

%S 1,3,35,44099,85762231424099,

%T 630794963141019085083178800095033630804099

%N a(0) = 1; for n > 0, a(n) = -1 + 4*Product_{i=0..n-1} a(i)^2.

%C Sequence designed to show that there are an infinity of primes congruent to 3 modulo 4 (A002145). Terms are not necessarily prime. Their smallest prime factor from A002145 are: 3, 7, 11, 23, 4111, 2809343.

%C Next term is too large to include.

%C Similarly to Sylvester's sequence (A000058), it is unknown if all terms are squarefree (see also MathOverflow link). - _Max Alekseyev_, Mar 26 2023

%C Primes dividing terms of this sequence are listed in A362250. Since terms are pairwise coprime, for each n prime A362250(n) divides exactly one term, whose index is A362251(n). That is, A362250(n) divides a(A362251(n)). - _Max Alekseyev_, Apr 16 2023

%H S. A. Shirali, <a href="http://www.jstor.org/stable/2690862">A family portrait of primes-a case study in discrimination</a>, Math. Mag. Vol. 70, No. 4 (Oct., 1997), pp. 263-272.

%H fredrickmnelson et al., <a href="https://mathoverflow.net/q/443446">Does a(0)=6, a(n+1)=a(n)^3-a(n), define a square-free sequence?</a>, MathOverflow, 2023.

%F For n > 1, a(n) = (a(n-1) + 1) * a(n-1)^2 - 1. - _Max Alekseyev_, Mar 26 2023

%o (PARI) lista(nn) = {a = vector(nn); a[1] = 3; for (n=2, nn, a[n] = 4*prod(i=1, n-1, a[i]^2) - 1;); a;}

%Y Cf. A000058, A002145, A007018, A231830, A362250, A362251.

%K nonn

%O 0,2

%A _Michel Marcus_, Nov 14 2013

%E a(0) = 1 prepended by _Max Alekseyev_, Mar 26 2023