It's easy to miss. The square numbers are 1, 4, 9, 16, 25, 36, 49 and so on but it's not obvious that the consecutive integers 27423, 27424 and 27425 are divisible by consecutive square numbers. Thus we have:$$ \begin{align} 27423 &= 3^2 \cdot 11 \cdot 277 \text{ divisible by }9=3^2\\27424 &= 2^5 \cdot 857 \text{ divisible by }16=4^2\\27425 &= 5^2 \cdot 1097 \text{ divisible by }25=5^2 \end{align}$$I only noticed this fact because my diurnal age today is 27423 and this number is a member of OEIS A178919:
A178919 | Smallest of three consecutive integers divisible respectively by three consecutive squares greater than 1. |
Membership of this sequence does not come easy and can be seen in the list of its initial members (permalink):
2223, 5823, 9423, 13023, 16623, 20223, 23823, 27423, 31023, 32975, 34623, 38223, 41823, 45423, 49023, 52623, 56223, 59823, 63423, 67023, 70623, 74223, 77075, 77823, 81423, 85023, 88623, 92223, 95823, 99423, 103023, 106623, 110223
Not surprisingly membership in the equivalent sequence of two consecutive integers divisible by two consecutive squares is a lot easier. This sequence is OEIS A178918. The natural question to ask is whether there are groups of four consecutive integers divisible by four consecutive squares. Testing up in the range up to ten million, we find no such groups. However, they may well exist further out.
What about cubes? Can we find groups of three consecutive integers that are divisible by three consecutive cubes greater than 1. Indeed we can and, up one million, the sequence of the smallest members of these trios is (permalink):
106623, 322623, 538623, 754623, 970623 (not listed in the OEIS)
Let's look at the first member of the sequence where we find:$$\begin{align} 106623 &= 3^3 \cdot 11 \cdot 359 \text{ divisible by } 27 =3^3\\106624 &= 2^7 \cdot 7^2 \cdot 17 \text{ divisible by }64 =4^3\\106625 &= 5^3 \cdot 853 \text{ divisible by }125 =5^3 \end{align}$$What's interesting about sequences like this is that the numbers derive their membership via the groups to which they belong. For convenience, as in the case of OEIS A178919, only the first number in the group is listed. It is the relationship between the numbers in the group that are important. In the case of OEIS A178919 the numbers form a group of three that are consecutive and divisible by consecutive squares. Thus we have in the case of 27423:$$ \text{consecutive integers -->}\\ \frac{27423}{9} \, \frac{27424}{16} \, \frac{27425}{25} \\ \text{consecutive squares -->} $$or in the case of 106623:$$ \text{consecutive integers -->}\\ \frac{106623}{27} \, \frac{106624}{64} \, \frac{106625}{125} \\ \text{consecutive cubes -->} $$It would be interesting to explore divisibility using criteria other than divisibility by consecutive squares or cubes. What about divisibility of three consecutive integers by three consecutive fibonacci numbers (0, 1, 1, 2, 3, 5, 8, ...)? Well, if we ignore 0 and 1 and start with 2, it turns out that a great many groups of three qualify, most of which are divisible by 2, 3 and 5. The first of these begins with 8:$$ \begin{align} 8 &= 2^3 \text{ divisible by fibonacci number }2\\9 &= 3^2 \text{ divisible by fibonacci number } 3\\10 &= 2 \cdot 5 \text{ divisible by fibonacci number } 5 \end{align}$$There are 4417 such groups of three in the range up to 100,000, so they are very common. If we exclude 2, 3 and 5 and begin instead with 8, then the groupings of three become far less common (only 60 in the range up 100,000). The first of these begins with 376 (permalink):$$ \begin{align} 376 &= 2^3 \cdot 47 \text{ divisible by fibonacci number } 8\\377 &= 13 \cdot 29 \text{ divisible by fibonacci number }13\\378 &= 2 \cdot 3^3 \cdot 7 \text{ divisible by fibonacci number } 21 \end{align}$$This is clearly a topic worthy of further research.
