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%I #6 Apr 29 2022 17:31:03
%S 5,9,8,13,10,17,17,17,33,7,21,19,49,18,37,25,26,65,24,73,18,29,28,81,
%T 26,109,19,65,33,35,97,32,145,30,129,80,37,37,113,43,181,31,193,82,
%U 101,41,44,129,49,217,48,257,161,201,3,45,46,145,51,253,50,321,163
%N Square array read by downward antidiagonals: A(n, k) = k-th Wieferich base of n, i.e., k-th b > 1 such that b^(n-1) == 1 (mod n^2).
%e The array starts as follows:
%e 5, 9, 13, 17, 21, 25, 29, 33, 37, 41, 45
%e 8, 10, 17, 19, 26, 28, 35, 37, 44, 46, 53
%e 17, 33, 49, 65, 81, 97, 113, 129, 145, 161, 177
%e 7, 18, 24, 26, 32, 43, 49, 51, 57, 68, 74
%e 37, 73, 109, 145, 181, 217, 253, 289, 325, 361, 397
%e 18, 19, 30, 31, 48, 50, 67, 68, 79, 80, 97
%e 65, 129, 193, 257, 321, 385, 449, 513, 577, 641, 705
%e 80, 82, 161, 163, 242, 244, 323, 325, 404, 406, 485
%e 101, 201, 301, 401, 501, 601, 701, 801, 901, 1001, 1101
%e 3, 9, 27, 40, 81, 94, 112, 118, 120, 122, 124
%e 145, 289, 433, 577, 721, 865, 1009, 1153, 1297, 1441, 1585
%o (PARI) row(n, terms) = my(i=0); for(b=2, oo, if(i>=terms, print(""); break, if(Mod(b, n^2)^(n-1)==1, print1(b, ", "); i++)))
%o array(rows, cols) = for(x=2, rows+1, row(x, cols))
%o array(6, 5) \\ Print initial 6 rows and 5 columns of array
%o (Python)
%o def T(n, k):
%o j, n2, c = 2, n*n, 0
%o while c != k:
%o if pow(j, n-1, n2) == 1: c += 1
%o j += 1
%o return j-1
%o def auptodiag(maxd):
%o return [T(d+2-j, j) for d in range(1, maxd+1) for j in range(d, 0, -1)]
%o print(auptodiag(11)) # _Michael S. Branicky_, Apr 29 2022
%Y Cf. A185103 (column 1), A353600 (column 2).
%K nonn,tabl
%O 2,1
%A _Felix Fröhlich_, Apr 29 2022