A319061 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..3, with k running over the positive integers; square array, read by antidiagonals, downwards.
557, 901, 1207, 1549, 4607, 1451, 2449, 5176, 2774, 13543, 4049, 10124, 8201, 42269, 24675, 5293, 19601, 13543, 91110, 45124, 39016, 5849, 20924, 24482, 91678, 95236, 302947, 217682, 6193, 22049, 30949, 101399, 188872, 387587, 928423, 165407, 7057, 26018
Offset: 1
Examples
The array starts as follows: 557, 901, 1549, 2449, 4049, 5293, 5849, 6193 1207, 4607, 5176, 10124, 19601, 20924, 22049, 26018 1451, 2774, 8201, 13543, 24482, 30949, 31457, 40199 13543, 42269, 91110, 91678, 101399, 132576, 142148, 210258 24675, 45124, 95236, 188872, 236915, 273971, 296449, 298117 39016, 302947, 387587, 609436, 637111, 962525, 1015033, 1074751 217682, 928423, 1546225, 1666084, 1756986, 2105290, 2673538, 2733520 165407, 215029, 1008933, 1370816, 1487743, 1493395, 1624207, 2998943
Crossrefs
Programs
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Mathematica
rows = 8; t = 3; T = Table[lst = {}; b = 2; While[Length[lst] < rows, p = Prime[n + Range[0, t]]; If[AllTrue[PowerMod[b,(p-1) p^2], #==1 &], AppendTo[lst, b]]; b++]; lst, {n, rows}]; T // TableForm (* Print the A(n,k) table *) Flatten[Table[T[[j, i - j + 1]], {i, 1, rows}, {j, 1, i}]] (* Robert Price, Sep 30 2019 *)
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PARI
printrow(n, terms) = my(c=0); for(b=2, oo, my(j=0); for(i=0, 3, my(p=prime(n+i)); if(Mod(b, p^2)^(p-1)==1, j++)); if(j==4, print1(b, ", "); c++); if(c==terms, break)) array(rows, cols) = for(x=1, rows, printrow(x, cols); print("")) array(8, 10) \\ print initial 8 rows and 10 columns of array