%I #14 Jul 15 2021 16:01:52
%S 1,1,4,1,2,3,1,1,12,7,1,4,9,9,24,1,2,10,17,9,25,1,1,1,11,7,26,11,1,4,
%T 4,6,28,30,1,7,1,2,3,5,25,9,4,28,22,1,1,12,1,13,36,16,6,27,12,1,4,9,4,
%U 23,33,21,24,47,48,9,1,2,10
%N Triangle read by rows: R(n, k) = 2^(2n - 2) mod prime(2k), 1<=k<=n.
%C The leftmost diagonal is all 1s because all even-indexed powers of 2 are congruent to 1 mod 3 (since 3 is the first even-indexed prime).
%e Triangle begins:
%e 1
%e 1, 4
%e 1, 2, 3
%e 1, 1, 12, 7
%e 1, 4, 9, 9, 24
%e 1, 2, 10, 17, 9, 25
%e For example, R(4, 4) = 7 because 2^(2 * 4 - 2) = 2^6 = 64; the (2 * 4)th prime is 19; and 64 divided by 19 leaves a remainder of 7.
%t ColumnForm[Table[Mod[2^(2n - 2), Prime[2k]], {n, 12}, {k, n}], Center]
%t Table[PowerMod[2,2n-2,Prime[2k]],{n,15},{k,n}]//Flatten (* _Harvey P. Dale_, Jul 15 2021 *)
%Y Cf. A000079, A181670.
%K nonn,tabl
%O 1,3
%A _Juri-Stepan Gerasimov_, Dec 06 2010