(* derived from an email question by Dr. Sloane
about deriving recursions from known triangle sequences*)
Table[
 FindSequenceFunction[Table[Binomial[n, m], {m, 0, n}], k], {n, 3, 10}]
{-3 + 5 k - k^2, FindSequenceFunction[{1, 4, 6, 4, 1}, k], 
 1/4 (40 - 84 k + 61 k^2 - 14 k^3 + k^4), 
 DifferenceRoot[
   Function[{\[FormalY], \[FormalN]}, {(-7 + \[FormalN]) \[FormalY][\
\[FormalN]] + \[FormalN] \[FormalY][1 + \[FormalN]] == 
      0, \[FormalY][1] == 1}]][k], 
 1/36 (-1260 + 3114 k - 2857 k^2 + 1287 k^3 - 274 k^4 + 27 k^5 - k^6),
  DifferenceRoot[
   Function[{\[FormalY], \[FormalN]}, {(-9 + \[FormalN]) \[FormalY][\
\[FormalN]] + \[FormalN] \[FormalY][1 + \[FormalN]] == 
      0, \[FormalY][1] == 1}]][k], 
 1/576 (72576 - 198000 k + 214020 k^2 - 120824 k^3 + 39493 k^4 - 
    7436 k^5 + 790 k^6 - 44 k^7 + k^8), 
 DifferenceRoot[
   Function[{\[FormalY], \[FormalN]}, {(-11 + \[FormalN]) \[FormalY][\
\[FormalN]] + \[FormalN] \[FormalY][1 + \[FormalN]] == 
      0, \[FormalY][1] == 1}]][k]}
Table[Table[
  DifferenceRoot[
    Function[{y, n}, {(-(2*m + 1) + n) y[n] + n y[1 + n] == 0, 
      y[1] == 1}]][k], {k, 1, m}], {m, 1, 10}]
{{1}, {1, 4}, {1, 6, 15}, {1, 8, 28, 56}, {1, 10, 45, 120, 210}, {1, 
  12, 66, 220, 495, 792}, {1, 14, 91, 364, 1001, 2002, 3003}, {1, 16, 
  120, 560, 1820, 4368, 8008, 11440}, {1, 18, 153, 816, 3060, 8568, 
  18564, 31824, 43758}, {1, 20, 190, 1140, 4845, 15504, 38760, 77520, 
  125970, 167960}}
Flatten[%]
{1, 1, 4, 1, 6, 15, 1, 8, 28, 56, 1, 10, 45, 120, 210, 1, 12, 66, \
220, 495, 792, 1, 14, 91, 364, 1001, 2002, 3003, 1, 16, 120, 560, \
1820, 4368, 8008, 11440, 1, 18, 153, 816, 3060, 8568, 18564, 31824, \
43758, 1, 20, 190, 1140, 4845, 15504, 38760, 77520, 125970, 167960}
Table[Sum[
  DifferenceRoot[
    Function[{y, n}, {(-(2*m + 1) + n) y[n] + n y[1 + n] == 0, 
      y[1] == 1}]][k], {k, 1, m}], {m, 1, 10}]
{1, 5, 22, 93, 386, 1586, 6476, 26333, 106762, 431910}