import math

def po2(num):#is a number a power of 2
	return num != 0 and ((num & (num - 1)) == 0)

#the below section gives the terms of A060802
max_term = [1,2,2,3,3,3,4]
for x in range(8,1025):
    low = 2**(int(math.log(x,2))-2)+2
    high = 2**(int(math.log(x,2))-1)+2
    start = 2**(int(math.log(x,2)))
    stop = 2**(int(math.log(x*2,2)))
    mid = (start + stop)/2+1
    if x < mid and x-3 < start:#The triplicates
        max_term.append(low)
    elif x < mid and max_term[-1] == max_term[-2]:#increase by one
        max_term.append(max_term[-1]+1)
    elif x < mid and max_term[-1] != max_term[-2]:#duplicate the prev. term
        max_term.append(max_term[-1])
    else:#the second half run up to the subsequent power of 2
        max_term.append(max_term[-1]+1)

f = open("b060802.txt","w")
q = 1
for x in max_term:
    j = str(q)+" " + str(x)
    f.write(j)
    f.write('\n')
    q += 1
f.close()

terms = [(1),(1,2),(1,2),(1,2,3),(1,2,3),(1,2,3),(1,2,4)]
g = open("Results.txt","w")
count = 1
for x in terms:
    g.write(str(count))
    g.write(", ")
    g.write(str(x))
    g.write('\n')
    count += 1
for n in range(8,1025):#This term generates one possible set of weights
                       #per the proof given by Jon E. Schoenfield
    term = [1,2]
    k = int(1+math.floor(math.log(n,2)))
    if po2(n):
        term.append(2**(k-3)+2)
    elif n < (3*2**(k-2)):
        term.append(n+1-2**(k-1)+int(math.ceil((3*2**(k-2)-n)/2.0)))
    else:
        term.append(int(n+1-2**(k-1)))
    for a in reversed(range(k-3)):#the terms 1 and 2 and the one just appended
        q = sum(term[2:])
        term.append(max_term[n-q-1])
    term.sort()
    j = '('
    j += str(n)
    j += ", ("
    for x in term:
        j += str(x)
        j += ", "
    j = j[0:-2]
    j += ')'
    g.write(j)
    g.write('\n')
g.close()