A357272 a(n) is the number of ways n can be calculated with expressions of the form "d1 o1 d2 o2 d3 o3 d4" where d1-d4 are decimal digits (0-9) and o1-o3 are chosen from the four basic arithmetic operators (+, -, *, /).

29235, 12654, 12450, 12425, 12427, 11915, 12419, 11792, 12062, 11725, 8748, 7686, 8180, 6632, 6549, 6077, 5758, 4532, 4915, 3503, 3649, 3451, 2684, 2468, 3253, 2288, 1957, 2347, 2197, 1627, 2028, 1444, 1899, 1439, 1281, 1531, 2080, 1195, 1126, 1147, 1513

Offset: 0

Author

Rod McFarland, Sep 22 2022

Keywords

• nonn
• base

Comments

This sequence is an extension of the "four fours" puzzle.

Expressions follow operator precedence (*/) then (+-), and left to right within the same level of precedence: "5/6/7/8" is ((5/6)/7)/8, "3-4+5*8" is 3-4+(5*8) = (3-4)+40 = -1+40 = 39.

Expressions are treated as ordered, so that 1+2+3+4 is distinct from 1+3+2+4 (but has the same value).

If negative n is allowed, the first nonzero a(n) is n = -729 (0-9*9*9). The last nonzero a(n) is n = 6561 (9*9*9*9).

There are 671 nonzero terms. - Michael S. Branicky, Sep 24 2022

Examples

			a(235) = 9 because 235 may be expressed in nine ways: "3*9*9-8", "5*6*8-5", "5*8*6-5", "6*5*8-5", "6*8*5-5", "8*5*6-5", "8*6*5-5", "9*3*9-8", and "9*9*3-8".
		

Links

• Michael S. Branicky, Table of n, a(n) for n = 0..6561

Programs

Python

from itertools import product
from fractions import Fraction
from collections import Counter
def afull(): # all further terms are 0
    a = Counter()
    for digs in product("0123456789", repeat=4):
        for ops in product("+-*/", repeat=3):
            e = digs[0] + "".join(ops[i] + digs[i+1] for i in range(3))
            if "/0" in e: continue
            if "/" in e:
                for d in set(digs): e = e.replace(d, f"Fraction({d}, 1)")
            t = eval(e)
            if t >= 0 and t.denominator == 1: a[t] += 1
    return [a[n] for n in range(max(a)+1)]
print(afull()[:100]) # Michael S. Branicky, Sep 24 2022