Tom,

The following scheme has every pair playing together at least once and at most twice. To do this I took the solution to a previous problem which was for 16 golfers playing 5 rounds. I numbered the golfers from 0 to 15 and for your problem I discarded numbers 0 and 15. Round 2 in the previous solution had 0 and 15 playing together and I modified this round in two ways to get your required 6 rounds. These are day 1 and day 6 in the list below. For day 1 I took player 1 from his 4-some and put him with 5 and 10 and for day 6 I put player 3 with 5 and 10, and placed 1 back in his foursome.

Day 1
{1, 5, 10}, {4, 11, 14}, {2, 7, 8, 13}, {3, 6, 9, 12}

Day 2
{4, 8, 12}, {1, 5, 9, 13}, {2, 6, 10, 14}, {3, 7, 11}

Day 3
{6, 11, 13}, {1, 7, 10, 12}, {2, 4, 9}, {3, 5, 8, 14}

Day 4
{7, 9, 14}, {1, 6, 8}, {2, 5, 11, 12}, {3, 4, 10, 13}

Day 5
{1, 2, 3}, {4, 5, 6, 7}, {8, 9, 10, 11}, {12, 13, 14}

Day 6
{3, 5, 10}, {1, 4, 11, 14}, {2, 7, 8, 13}, {6, 9, 12}

Chris