%I #8 Mar 18 2021 08:23:18
%S 0,2,1,0,3,0,1,0,3,0,1,2,3,2,1,2,3,0,1,2,1,0,3,0,1,0,1,2,1,0,1,2,1,0,
%T 3,2,1,0,3,0,1,2,3,0,1,0,3,0,3,0,1,0,3,2,1,0,3,2,1,0,1,2,3,0,3,2,3,0,
%U 1,2,3,2,1,2,3,0,3,2,3,2,3,0,1,2,3,0
%N a(n) = (prime(n) - a(n-1)) mod 4; a(0)=0.
%H Simon Strandgaard, <a href="/A340867/a340867.png">Visualization</a>
%F a(n) = A008347(n) mod 4.
%e a(1) = ( 2 - 0) mod 4 = 2,
%e a(2) = ( 3 - 2) mod 4 = 1,
%e a(3) = ( 5 - 1) mod 4 = 0,
%e a(4) = ( 7 - 0) mod 4 = 3,
%e a(5) = (11 - 3) mod 4 = 0.
%t a[0] = 0; a[n_] := a[n] = Mod[Prime[n] - a[n - 1], 4]; Array[a, 100, 0] (* _Amiram Eldar_, Jan 30 2021 *)
%o (Ruby) require 'prime'
%o values = [0]
%o Prime.first(50).each do |prime|
%o values << (prime-values[-1]) % 4
%o end
%o p values
%Y Cf. A008347, A339448.
%K nonn
%O 0,2
%A _Simon Strandgaard_, Jan 24 2021