Irregular triangle of the offsets used in S000064 and S000065.
1, 1, 1, 3, 1, 3, 1, 5, 1, 3, 1, 3, 7, 1, 5, 7, 1, 3, 7, 9, 1, 3, 7, 9, 1, 5, 7, 11, 1, 3, 7, 9, 1, 3, 9, 13, 1, 7, 11, 13, 1, 5, 7, 11, 13, 1, 3, 7, 9, 13, 1, 5, 7, 11, 13, 17, 1, 5, 7, 11, 13, 17, 1, 7, 11, 13, 17, 19, 1, 5, 11, 13, 19, 1, 3, 9, 13, 15, 19, 21
2
Note that the last term s(22) = {1, 3, 9, 13, 15, 19, 21} has three twin primes. The prime numbers are found using the calculation 22 * S000065(22) + s(22) = 22 * 867709 + {1, 3, 9, 13, 15, 19, 21} = {19089599, 19089601, 19089607, 19089611, 19089613, 19089617, 19089619}.
T. D. Noe, Plot of irregular rows 2 to 22
The triangle begins
{1},
{1},
{1, 3},
{1, 3},
{1, 5},
{1, 3},
{1, 3, 7},
{1, 5, 7},
{1, 3, 7, 9},
{1, 3, 7, 9},
{1, 5, 7, 11},
{1, 5, 7, 11},
{1, 3, 9, 13},
{1, 7, 11, 13},
{1, 3, 7, 13, 15},
{1, 3, 9, 13, 15},
{1, 5, 7, 11, 13, 17},
{1, 5, 7, 11, 13, 17},
{1, 3, 7, 9, 13, 19},
{1, 5, 11, 17, 19},
{1, 3, 9, 13, 15, 19, 21}
(Mma) Table[mx = {}; Do[s = Select[n*i + Range[n], PrimeQ]; If[Length[s] > Length[mx] || (Length[s] == Length[mx] && Min[s - n*i] < Min[mx]), mx = s - n*i], {i, 10^6}]; mx, {n, 2, 22}]
nonn,hard,tabf
T. D. Noe, May 30 2014