title: Sequences Given Length Every Element Equal Twice Previous 1
signature: sequencesGivenLengthEveryElementEqualTwicePrevious1 :: Integer -> Integer -> Integer
  sequencesGivenLengthEveryElementEqualTwicePrevious1 m n = 
test_code: import Text.Printf
  import Control.Exception (catch, SomeException)
  import Data.Typeable
  import qualified Data.List as List
  import qualified Data.Map as Map
  import qualified Data.Set as Set
  
  -- SOLUTION CODE
  -- ============================================
  
  PLACEHOLDER_CODE_BODY
  
  -- TESTING CODE 
  -- ============================================
  validateSolution :: Integer -> Integer -> Bool
  validateSolution actual expected = do
      actual == expected
  
  
  driver :: Integer -> Integer -> Integer -> String
  driver m n expected = do
      if (validateSolution (PLACEHOLDER_FN_NAME m n) expected)
          then "PASSED"
      else "FAILED"
  
  main = do
      printf "TEST-0...%s\n" (driver (10) (4) (4) )
  
      printf "TEST-1...%s\n" (driver (5) (2) (6) )
  
      printf "TEST-2...%s\n" (driver (2) (8) (0) )
  
      printf "TEST-3...%s\n" (driver (83) (7) (330) )
  
      printf "TEST-4...%s\n" (driver (91) (0) (0) )
  
      printf "TEST-5...%s\n" (driver (18) (53) (0) )
  
      printf "TEST-6...%s\n" (driver (83) (41) (0) )
  
      printf "TEST-7...%s\n" (driver (98) (53) (0) )
  
      printf "TEST-8...%s\n" (driver (43) (37) (0) )
  
      printf "TEST-9...%s\n" (driver (31) (20) (0) )
  
source_py: def sequences_given_length_every_element_equal_twice_previous_1(m, n):
      T = [[0 for i in range((n + 1))] for i in range((m + 1))]
      for i in range((m + 1)):
          for j in range((n + 1)):
              if ((i == 0) or (j == 0)):
                  T[i][j] = 0
              elif (i < j):
                  T[i][j] = 0
              elif (j == 1):
                  T[i][j] = i
              else:
                  T[i][j] = (T[(i - 1)][j] + T[(i // 2)][(j - 1)])
      return T[m][n]
source_cpp: int sequences_given_length_every_element_equal_twice_previous_1(int m, int n) {
    int T [ m + 1 ] [ n + 1 ];
    for ( int i = 0;
    i < m + 1;
    i ++ ) {
      for ( int j = 0;
      j < n + 1;
      j ++ ) {
        if ( i == 0 || j == 0 ) T [ i ] [ j ] = 0;
        else if ( i < j ) T [ i ] [ j ] = 0;
        else if ( j == 1 ) T [ i ] [ j ] = i;
        else T [ i ] [ j ] = T [ i - 1 ] [ j ] + T [ i / 2 ] [ j - 1 ];
      }
    }
    return T [ m ] [ n ];
  }