(seqs, n, zero=True)
return Matrix(k, k, f).det()
Given linear difference operator L of order 'k' and homogeneous
       equation Ly = 0 we want to compute kernel of L, which is a set
       of 'k' sequences: a(n), b(n), ... z(n).

       Solutions of L are linearly independent iff their Casoratian,
       denoted as C(a, b, ..., z), do not vanish for n = 0.

       Casoratian is defined by k x k determinant::

                  +  a(n)     b(n)     . . . z(n)     +
                  |  a(n+1)   b(n+1)   . . . z(n+1)   |
                  |    .         .     .        .     |
                  |    .         .       .      .     |
                  |    .         .         .    .     |
                  +  a(n+k-1) b(n+k-1) . . . z(n+k-1) +

       It proves very useful in rsolve_hyper() where it is applied
       to a generating set of a recurrence to factor out linearly
       dependent solutions and return a basis:

       >>> from sympy import Symbol, casoratian, factorial
       >>> n = Symbol('n', integer=True)

       Exponential and factorial are linearly independent:

       >>> casoratian([2**n, factorial(n)], n) != 0
       True
Given linear difference operator L of order 'k' and homogeneous
       equation Ly = 0 we want to compute kernel of L, which is a set
       of 'k' sequences: a(n), b(n), ... z(n).
Given linear difference operator L of order k and homogeneous equation Ly = 0 we want to compute kernel of L which is a set of k sequences : a ( n ) b ( n ) ... z ( n ) .
def casoratian(seqs, n, zero=True):
    """Given linear difference operator L of order 'k' and homogeneous
       equation Ly = 0 we want to compute kernel of L, which is a set
       of 'k' sequences: a(n), b(n), ... z(n).

       Solutions of L are linearly independent iff their Casoratian,
       denoted as C(a, b, ..., z), do not vanish for n = 0.

       Casoratian is defined by k x k determinant::

                  +  a(n)     b(n)     . . . z(n)     +
                  |  a(n+1)   b(n+1)   . . . z(n+1)   |
                  |    .         .     .        .     |
                  |    .         .       .      .     |
                  |    .         .         .    .     |
                  +  a(n+k-1) b(n+k-1) . . . z(n+k-1) +

       It proves very useful in rsolve_hyper() where it is applied
       to a generating set of a recurrence to factor out linearly
       dependent solutions and return a basis:

       >>> from sympy import Symbol, casoratian, factorial
       >>> n = Symbol('n', integer=True)

       Exponential and factorial are linearly independent:

       >>> casoratian([2**n, factorial(n)], n) != 0
       True

    """
    from .dense import Matrix

    seqs = list(map(sympify, seqs))

    if not zero:
        f = lambda i, j: seqs[j].subs(n, n + i)
    else:
        f = lambda i, j: seqs[j].subs(n, i)

    k = len(seqs)

    return Matrix(k, k, f).det()
def casoratian ( seqs , n , zero = True ) : from . dense import Matrix seqs = list ( map ( sympify , seqs ) ) if not zero : f = lambda i , j : seqs [ j ] . subs ( n , n + i ) else : f = lambda i , j : seqs [ j ] . subs ( n , i ) k = len ( seqs ) return Matrix ( k , k , f ) . det ( )