A185352 The "smallest countdown" numbers are the smallest positive integer that cannot be made using the numbers n through 1, in order, using the operations +, -, *, /, and parentheses.

2, 4, 8, 17, 39, 92, 275, 922, 2894, 10843, 35944

Offset: 1

Author

Peter Boothe and Abraham Asfaw, Feb 08 2012

Keywords

• nonn
• hard
• more

Comments

Inspired by a now-lost blog post in which someone discussed a "new year's countdown" equation for 2012, e.g., 10 * (9 + ((8 * (((7 + (6 / (5 * 4))) * 3) + 2)) + 1)) = 2012. This sequence has been "verified" by two independently created programs.

Examples

			for n = 3, a(3) = 8, because 3*2+1=7, and 3*(2+1)=9, but there is no equation with 3,2,and 1 in order that equals 8. Note that if we allow the order to change, we can make 8, because 2*(3+1)=8, but reordering is not allowed.
		

Crossrefs

Related to A060315, which is the smallest number that cannot be made with the numbers 1 to n, in any order.

Programs

Python

from fractions import Fraction
def genAllTrees(l):
    if len(l) == 0:
        return
    elif len(l) == 1:
        yield l[0], str(l[0])
    else:
        for middle in range(len(l)):
            for lval, leqn in genAllTrees(l[:middle]):
                for rval, reqn in genAllTrees(l[middle:]):
                    yield lval+rval, ("(" + leqn + " + " + reqn + ")")
                    yield lval-rval, ("(" + leqn + " - " + reqn + ")")
                    yield lval*rval, ("(" + leqn + " * " + reqn + ")")
                    if rval != Fraction(0):
                        yield lval/rval, ("(" + leqn + " / " + reqn + ")")
def findSmallestIntNotPresent(n):
    vals = {}
    for val, eqn in genAllTrees([Fraction(i) for i in range(n, 0, -1)]):
        if val.denominator == 1:
            val = val.numerator
            if val not in vals:
                vals[val] = eqn
    i = 1
    while i in vals:
        i += 1
    return i
for i in range(1, 11):
    print(i, findSmallestIntNotPresent(i))

Extensions

a(10)-a(11) from Hiroaki Yamanouchi, Oct 04 2014