S(n,k)=S(n-1,k-1)+k*S(n-1,k) is a well-known recursion, and I tried to find out which of the two terms dominate. So, I looked for the values n & k where S(n-1,k-1)<=k*S(n-1,k). Going down diagonals in the triangular table of S(n,k) using Table[k = 1; While[StirlingS2[n - 1 + k, k] <= (1 + k)*StirlingS2[n - 1 + k, k + 1] && k < 1000, k++]; k, {n, 256}] and I got: {1, 3, 5, 8, 10, 13, 15, 18, 21, 23, 26, 28, 31, 33, 36, 39, 41, 44, 46, 49, 52, 54, 57, 59, 62, 65, 67, 70, 72, 75, 77, 80, 83, 85, 88, 90, 93, 96, 98, 101, 103, 106, 109, 111, 114, 116, 119, 122, 124, 127, 129, 132, 134, 137, 140, 142, 145, 147, 150, 153, 155, 158, 160, 163, ... The first differences are 2 (106 times) or 3 (149 times). No regularity in sight. Picking out the positions of the 2's, I got: {1, 2, 4, 6, 9, 11, 13, 16, 18, 21, 23, 26, 28, 30, 33, 35, 38, 40, 43, 45, \ 48, 50, 52, 55, 57, 60, 62, 65, 67, 69, 72, 74, 77, 79, 82, 84, 86, 89, 91, \ 94, 96, 99, 101, 103, 106, 108, 111, 113, 116, 118, 120, 123, 125, 128, 130, \ 133, 135, 137,... Again taking first differences, {1, 2, 2, 3, 2, 2, 3, 2, 3, 2, 3, 2, 2, 3, 2, 3, 2, 3, 2, 3, 2, 2, 3, 2, 3, \ 2, 3, 2, 2, 3, 2, 3, 2, 3, 2, 2, 3, 2, 3, 2, 3, 2, 2, 3, 2, 3, 2, 3, 2, 2, 3, \ 2, 3, 2, 3, 2, 2, 3, 2, 3, 2, 3, 2, 2, 3, 2, 3, 2, 3, 2, 3, 2, 2, 3, 2, 3, 2, \ 3, 2, 2, 3, 2, 3, 2, 3, 2, 2, 3, 2, 3, 2, 3, 2, 2, 3, 2, 3, 2, 3, 2, 2, 3, 2, \ 3, 2} with the 3's at positions {4, 7, 9, 11, 14, 16, 18, 20, 23, 25, 27, 30, 32, 34, 37, 39, 41, 44, 46, 48, \ 51, 53, 55, 58, 60, 62, 65, 67, 69, 71, 74, 76, 78, 81, 83, 85, 88, 90, 92, \ 95, 97, 99, 102, 104} This seems convoluted beyond avail. Why? Wouter.