# 2 

# seq(nops(LT(i, i)), i = 1..6) is not in the oeis, 
# nor is seq( i! * nops(LT(i, i)), i = 1 .. 6) which accounts for 
# all choices of the bottom row.

# 3 
# seq(nops(LT(i,2)),i=2..7) is A127548 in the oeis
# This sequence has generating function Sum_{n>=0} n!*(x/(1+x)^2)^n



# 4 Not sure if this proof will count for the 10 dollars, as it relies on 
#  some statements found on oeis

# Let x_n be our sequence, the number of n x 2 latin triangles with the 
# bottom row 1..n in order.

# Let y_n be the number of such n x 2 triangles that contain n in
# the 2nd row. There is a bijection from solutions for this problem, to
# solutions of another problem which I will state below.

# Consider n heterosexual couples, who will sit in 2*n chairs that are placed 
# in a line. From each couple place a man in an odd indexed chair. In how 
# many ways can we place the women so that no woman is next to her partner?
# The solution to this problem is given by A000271, as described in a comment.

# For the bijection from y_n to this sequence, consider the 1..n in the 
# bottom row of the triangle to be the men. Then each entry in the 2nd row
# can be thought of as a chair for the women, and the 2 women that cannot sit
# in it are the entries diagonally below the spot. This works for the first 
# (n-1) women chairs, but what of the last women chair? Given a solution to the 
# trapezoid, take the number that was not present in the 2nd row. Since we are 
# working with y_n, this number cannot be n. Then create a dummy chair next to 
# the n in the bottom row and place this missing number in it.
# The map works in reverse as well, any chair setup can be reduced to a n x 2
# trapezoid by removing the last chair and placing the chairs in a zigzag pattern.

# Now it only remains to count the n x 2 trapezoids which do not contain n in
# the 2nd row. Each such trapezoid has a sub-trapezoid by removing n on the 
# bottom row and the last number of the top row. This sub-trapezoid will be 
# (n-1) x 2 and must contain an (n-1) in the top row. If it didn't than there 
# are no possibities for the number that we removed from the top row to make 
# this sub-trapezoid. Thus the number of such trapezoids is y_(n-1) !

# This shows that x_n = y_n + y_(n-1).
# It says in a comment for OEIS page of A127548 that
# a(n) = A000271(n) + A000271(n-1)
# Thus in fact x_n is given by A127548.