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
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
Comments