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a(n) = f(prime(n*2+1) mod 3, prime(n*2+2) mod 3) where f(1,1) = 3, f(1,2) = 1, f(2,1) = 2, f(2,2) = 4.
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%I #12 Sep 15 2024 14:40:32

%S 2,2,2,4,3,2,4,2,1,3,4,1,1,1,1,2,2,3,4,2,2,2,3,2,4,1,4,2,1,1,1,1,3,2,

%T 4,3,1,2,2,2,2,1,2,2,4,1,1,4,3,1,4,3,4,2,3,2,1,1,4,3,4,1,1,3,1,3,2,2,

%U 4,1,1,1,1,1,1,1,1,1,1,4,1,4,3,1,2,2,1,1,3,1,1,4,3,1,1,1,4,3,4,2

%N a(n) = f(prime(n*2+1) mod 3, prime(n*2+2) mod 3) where f(1,1) = 3, f(1,2) = 1, f(2,1) = 2, f(2,2) = 4.

%F a(n) = f(prime(n*2+1) mod 3, prime(n*2+2) mod 3) where f(1,1) = 3, f(1,2) = 1, f(2,1) = 2, f(2,2) = 4. - _Jason Yuen_, Sep 15 2024

%t a = Partition[Table[1 + Mod[Prime[n], 3], {n, 3, 203}], 2] /. {2, 3} -> 1 /. {3, 2} -> 2 /. { 2, 2} -> 3 /. {3, 3} -> 4

%o (PARI) a(n) = my(f(i,j)=[3,1,2,4][i*2+j-2]);f(prime(n*2+1)%3,prime(n*2+2)%3) \\ _Jason Yuen_, Sep 15 2024

%K nonn,uned,less,easy

%O 1,1

%A _Roger L. Bagula_, Aug 29 2006

%E New name from _Jason Yuen_, Sep 15 2024