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%I #16 Dec 22 2022 15:22:21
%S 1,19,331,5419,87211,1397419,22366891,357903019,5726579371,
%T 91625794219,1466014804651,23456245263019,375299957762731,
%U 6004799458421419,96076791871613611,1537228672093301419,24595658762082757291,393530540227683855019,6296488643780380633771,100743818301035845954219
%N a(n) = (q^2-q+1)/3 where q = 2^(2*n+1) = A004171(n).
%H Peter Cameron, <a href="https://cameroncounts.wordpress.com/2021/05/31/a-little-problem/">A little problem</a>, May 31 2021.
%H Volkan Yildiz, <a href="https://arxiv.org/abs/2212.08814">Some divisibility properties of Jacobsthal numbers</a>, arXiv:2212.08814 [math.CO], 2022.
%F a(n) = A002061(A004171(n))/3.
%F a(n) = (A060869(n) + 1)/4. - _Hugo Pfoertner_, Jun 30 2021
%p a:= n-> (q-> (q^2-q+1)/3)(2^(2*n+1)):
%p seq(a(n), n=0..20); # _Alois P. Heinz_, Jun 30 2021
%t Table[(2^(4*n + 2) - 2^(2*n + 1) + 1)/3, {n, 0, 19}] (* _Amiram Eldar_, Jun 30 2021 *)
%o (PARI) a(n) = my(q=2^(2*n+1)); (q^2-q+1)/3;
%Y Cf. A002061, A004171.
%K nonn
%O 0,2
%A _Michel Marcus_, Jun 30 2021