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A(n, k) is the k-th number b > 1 such that b^(prime(n+i)-1) == 1 (mod prime(n+i)^2) for each i = 0..6, with k running over the positive integers; square array, read by antidiagonals, downwards.
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%I #26 Oct 27 2019 11:25:01

%S 4486949,4651993,20950343,4941649,21184318,23250274,5571593,33538051,

%T 163075007,741652533,11903257,78868324,189850207,882345432,710808570,

%U 19397501,86892632,230695118,1528112512,5126829291,2380570527,19841257,111899224,421883318,1701241810

%N A(n, k) is the k-th number b > 1 such that b^(prime(n+i)-1) == 1 (mod prime(n+i)^2) for each i = 0..6, with k running over the positive integers; square array, read by antidiagonals, downwards.

%e The array starts as follows:

%e 4486949, 4651993, 4941649, 5571593, 11903257, 19397501, 19841257

%e 20950343, 21184318, 33538051, 78868324, 86892632, 111899224, 126664001

%e 23250274, 163075007, 189850207, 230695118, 421883318, 422771099, 497941351

%e 741652533, 882345432, 1528112512, 1701241810, 1986592318, 2005090271, 2596285385

%e 710808570, 5126829291

%e 2380570527

%t rows = 6; t = 6;T = Table[lst = {}; b = 2;

%t While[Length[lst] < rows - n + 1,

%t p = Prime[n + Range[0, t]];

%t If[AllTrue[PowerMod[b, (p - 1), p^2], # == 1 &],

%t AppendTo[lst, b]]; b++];

%t lst, {n, rows}];

%t T // TableForm (* Print the A(n,k) table *)

%t Flatten[Table[T[[j, i - j + 1]], {i, 1, rows}, {j, 1, i}]] (* _Robert Price_, Oct 03 2019 *)

%o (PARI) printrow(n, terms) = my(c=0); for(b=2, oo, my(j=0); for(i=0, 6, my(p=prime(n+i)); if(Mod(b, p^2)^(p-1)==1, j++)); if(j==7, print1(b, ", "); c++); if(c==terms, break))

%o array(rows, cols) = for(x=1, rows, printrow(x, cols); print(""))

%o array(5, 7) \\ print initial 5 rows and 7 columns of array

%Y Cf. A244249, A256236.

%Y Cf. analog for i = 0..t: A319059 (t=1), A319060 (t=2), A319061 (t=3), A319062 (t=4), A319063 (t=5), A319065 (t=7).

%K nonn,tabl

%O 1,1

%A _Felix Fröhlich_, Sep 11 2018