# A358629 - a(n) is the number of signed permutations W of V = (1, 2, ..., n) such that the dot product V*W = 0.
# Python program by Bert Dobbelaere

# This function returns a map with the number of solutions for each sum using the remaining numbers marked in the mask
# Only positive sums are stored at the cost of some extra complexity, but it saves memory which is the limiting factor
def solve(lvl, mask):
	global cache
	if lvl==0:
		return {0:1}
	if mask in cache:
		return cache[mask] # mask alone is a unique key, lvl is implied
	vals={}
	for b in range(N):
		if (mask>>b)&1:
			halfstats=solve(lvl-1,mask & ~(1<<b))
			# Unpack 'positive only' solution
			stats={}
			for x in halfstats:
				stats[x]=halfstats[x]
				stats[-x]=halfstats[x]
			delta = lvl*(b+1)
			for x in stats:
				# Symmetry: equal amount for opposite sign, lets store only the positive solutions
				if (x+delta)>=0:
					if not (x+delta) in vals:
						vals[x+delta]=0
					vals[x+delta] += stats[x]
				if (x-delta)>=0:
					if not (x-delta) in vals:
						vals[x-delta]=0
					vals[x-delta] += stats[x]
	cache[mask]=vals
	return vals

for N in range(1,20+1):
	cache={}
	results=solve(N, 2**N-1)
	if not 0 in results:
		results[0]=0
	print("a(%d) = %d"%(N,results[0]))