Find sequence of sequences?

FindSequenceFunction can find the each individual sequence sec[k] in terms of a recursion with polynomial coefficients:

FindSequenceFunction[PadRight[sec[5], 20]]
(*
  DifferenceRoot[
   Function[{\[FormalY], \[FormalN]}, {(-66 + 23 \[FormalN] - 
          2 \[FormalN]^2) \[FormalY][\[FormalN]] + \[FormalN] (-1 + 
          2 \[FormalN]) \[FormalY][1 + \[FormalN]] == 
      0, \[FormalY][1] == 1}]]
*)

Therefore, it can be used again to try to find the sequence of the coefficients as functions of k, which it can do in this case.

k0 = 3;
{y0coeff, y1coeff} = FindSequenceFunction /@ Transpose@
    Table[
     Normal@Last@CoefficientArrays[
        First[FindSequenceFunction@PadRight[sec[k], 20]][y, n] // 
         First,
        {y[n], y[1 + n]}],
     {k, k0 + 1, 8}] /. n -> #2
(*
{-28 + 15 #2 - 2 #2^2 - 15 #1 + 4 #2 #1 - 2 #1^2 &,
 #2 (-1 + 2 #2) &}
*)

The function secFN[k, n] can be used to generate sec[k]:

secFN = Function[{k, n},
   DifferenceRoot[
     Function[{\[FormalY], \[FormalN]},
      {y0coeff[k - k0, \[FormalN]] \[FormalY][\[FormalN]] + 
         y1coeff[k - k0, \[FormalN]] \[FormalY][1 + \[FormalN]] == 0,
       \[FormalY][1] == 1}
      ]
     ][n + 1]
   ];

Example:

Table[secFN[5, n], {n, 0, 5}]
(*  {1, 45, 210, 210, 45, 1}  *)