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T(m,n) = T(m-1,n) + T(floor(m/2), n-1)

Base conditions
T(m,n) = 1 when n = 0
T(m,n) = 0 when m < n

Edited: Below is the code for which I want to know the time complexity in terms of m and n.

#Python3 program to count total number of
#special sequences of length n where
#Recursive function to find the number of
# special sequences
def getTotalNumberOfSequences(m,n):

#A special sequence cannot exist if length
#n is more than the maximum value m.
if m<n:
return 0

#If n is 0, found an empty special sequence
if n==0:
return 1

#There can be two possibilities : (1) Reduce
#last element value (2) Consider last element
#as m and reduce number of terms
res=(getTotalNumberOfSequences(m-1,n)+
getTotalNumberOfSequences(m//2,n-1))
return res

#Driver Code
if __name__=='__main__':
m=10
n=4
print('Total number


of possible sequences:',getTotalNumberOfSequences(m,n))

T(m,n) = T(m-1,n) + T(m/2, n-1)

Base conditions
T(m,n) = 1 when n = 0
T(m,n) = 0 when m < n

T(m,n) = T(m-1,n) + T(floor(m/2), n-1)

Base conditions
T(m,n) = 1 when n = 0
T(m,n) = 0 when m < n

Edited: Below is the code for which I want to know the time complexity in terms of m and n.

#Python3 program to count total number of
#special sequences of length n where
#Recursive function to find the number of
# special sequences
def getTotalNumberOfSequences(m,n):

#A special sequence cannot exist if length
#n is more than the maximum value m.
if m<n:
return 0

#If n is 0, found an empty special sequence
if n==0:
return 1

#There can be two possibilities : (1) Reduce
#last element value (2) Consider last element
#as m and reduce number of terms
res=(getTotalNumberOfSequences(m-1,n)+
getTotalNumberOfSequences(m//2,n-1))
return res

#Driver Code
if __name__=='__main__':
m=10
n=4
print('Total number


of possible sequences:',getTotalNumberOfSequences(m,n))

1

# How to solve 2 variable recursion?

T(m,n) = T(m-1,n) + T(m/2, n-1)

Base conditions
T(m,n) = 1 when n = 0
T(m,n) = 0 when m < n