Pronic Pandigital Numbers and Beyond

My previous post on Pandigital Numbers Formed From Squares prompted me to investigate other ways of generating pandigital numbers. In July of 2018, I'd posted about Pandigital Numbers Formed From the Product of a Number and its Reversal. It occurred to me: why not consider pronic pandigital numbers. Pronic numbers are formed by multiplying two consecutive integers and are thus of the form \(n(n+1) \) where \(n\) is any integer.

Let's begin by considering what integers, when multiplied by the next consecutive integer, produce pandigital numbers with digits 1 to 9 occurring only once. It turns out that there are only 11 such numbers:

17846, 19403, 19727, 19871, 24768, 24776, 25568, 28521, 28556, 30878, 31203

Here is a permalink to a SageMath algorithm that will confirm this. This sequence of numbers does not appear in the OEIS and so it afforded me the opportunity to create a new sequence of my own.

S006: Integers \(n\) such that the product of \(n\) and \(n+1\) produce pandigital

numbers in which the digits from 1 to 9 occur only once. These pandigital

numbers are pronic.

If zero is allowed, then there are 52 integers that, multiplied by the next consecutive integer, produce pandigital numbers in which the digits from 0 to 9 occur only once. These numbers are:

38627, 40508, 43065, 44027, 44576, 46565, 48735, 51714, 54269, 54459, 55151, 55152, 55331, 55403, 58454, 59579, 61497, 63072, 65465, 67580, 67662, 70154, 73737, 74906, 75662, 76203, 76337, 76760, 78011, 80631, 82809, 83015, 84555, 86076, 86553, 86688, 86769, 87669, 89064, 90198, 90423, 90909, 91943, 92169, 92268, 93356, 94464, 94617, 96362, 96570, 98702, 99270

Once again, this sequence of numbers does not occur in the OEIS and so I again seized the opportunity to create my own sequence:

S007: Integers \(n\) such that the product of \(n\) and \(n+1\) produce

pandigital numbers in which the digits from 0 to 9 occur only once.

These pandigital numbers are pronic.

What about numbers of the form \(n(n+1)(n+2)\)? These are sphenic numbers consisting of three consecutive integers. It turns out that there are no such numbers if the digits are to range from 1 to 9. However, there are two such numbers if the digits from 0 to 9 are considered. These two numbers are 1267 and 1332. We find that:$$1267 \times 1268 \times 1269 = 2038719564\\1332 \times 1333 \times 1334 =2368591704$$Going a step further and considering numbers of the form \(n(n+1)(n+2)(n+3)\), we find that only 291 satisfies in producing pandigital numbers with digits from 0 to 9:$$291 \times 292 \times 293 \times 294 =7319658024$$There are other variations on this theme. Consider numbers of the form \(n \times \text{ prime}(n) \). Here we find there are two numbers that produce pandigital numbers with digits from 1 to 9: 5499 and 7569$$5499 \times \text{ prime}(5499)=5499 \times 53987=296874513\\7569 \times \text{ prime}(7569)=7569 \times 77017 =582941673$$If the digits are to range from 0 to 9, then we find that there are ten possible numbers viz.$$11376, 14562, 15057, 15723, 16659, 20421, 21330, 24867, 28494, 28746$$The corresponding pandigital numbers are respectively:$$1375028496,2308615794,2476108593,2714308659,3064572981,\\4692357801,5147632890,7094281563,9435702618,9612058734$$The fact that there are more than a couple of suitable numbers here justifies another sequence:

S008: Integers \(n\) such that the product of \(n\) and prime(\(n\)) produce

pandigital numbers in which the digits from 0 to 9 occur only once. 

 

Magic Cubes

I had a dream some nights ago in which I dreamt of the number 42. I hadn't been reading or thinking about The Hitchhiker's Guide to the Galaxy but I was prompted by the dream to investigate some of the properties of this number. As I discovered, it's a pronic number or a product of consecutive digits, in this case 6 and 7. It's also a partition number since it is equal to the number of partitions of 10. For information about its being the sum of three cubes, see this blog post of mine. The number is also associated with the 3 x 3 x 3 magic cube as shown in Figure 1:

Figure 1: dissection of a 3 x 3 x 3 magic cube
Source: https://mathematicscentre.com/taskcentre/174magic.htm

In such a cube, all the rows and columns in each of the three square "slices" shown in Figure 1 add to 42. It would be the same if we sliced in other similar ways. The long diagonals (19, 14, 9 for example) also add to 42 but the diagonals within the slices do not. Figure 2 shows another possible configuration:

Figure 2: source 

9 rows, such as 1 - 17 - 24, sum to 42.
9 columns, such as 1 - 15 - 26, sum to 42.
9 Pillars, such as 1 - 23 - 18, sum to 42.
4 triagonals, such as 26 - 14 - 2 sum to 42. 

Some of the squares may have diagonals summing to 42, but this is not a requirement. In fact, order-8 is the smallest cube for which it is possible for all the diagonals to sum correctly. 

What is required is that the 4 triagonals or 3-agonals, such as 1 - 14 - 27 sum to 42. 

There are four different basic pure (using numbers 1 to 27) magic cubes. Each of these have 48 equivalents due to rotations and/or reflections.

To quote from Wikipedia:

In mathematics, a magic cube is the 3-dimensional equivalent of a magic square, that is, a number of integers arranged in a \(n × n × n \) pattern such that the sums of the numbers on each row, on each column, on each pillar and on each of the four main space diagonals are equal to the same number, the so-called magic constant of the cube, denoted \(M_3(n)\). It can be shown that if a magic cube consists of the numbers \(1, 2, ..., n^3\), then it has magic constant (sequence A027441 in the OEIS):

$$
M_3(n) = \frac{n(n^3+1)}{2}
$$The OEIS sequence runs:

0, 1, 9, 42, 130, 315, 651, 1204, 2052, 3285, 5005, 7326, 10374, 14287, 19215, 25320, 32776, 41769, 52497, 65170, 80010, 97251, 117139, 139932, 165900, 195325, 228501, 265734, 307342, 353655, 405015, 461776, 524304, 592977, 668185, 750330, 839826, 937099

Of course, the formula above for 3 dimensions can be generalised to any number of dimensions. To quote again from Wikipedia:

In mathematics, a magic hypercube is the k-dimensional generalisation of magic squares, magic cubes and magic tesseracts; that is, a number of integers arranged in an \(n × n × n × ... × n\) pattern such that the sum of the numbers on each pillar (along any axis) as well as the main space diagonals is equal to a single number, the so-called magic constant of the hypercube, denoted \(M_k(n)\). It can be shown that if a magic hypercube consists of the numbers \(1, 2, ..., n^k\), then it has magic number: 

$$
M_k(n) = \frac{n(n^k+1)}{2}
$$For \(n=4\), the OEIS sequence A021003 is:

0, 1, 17, 123, 514, 1565, 3891, 8407, 16388, 29529, 50005, 80531, 124422, 185653, 268919, 379695, 524296, 709937, 944793, 1238059, 1600010, 2042061, 2576827, 3218183, 3981324, 4882825, 5940701, 7174467, 8605198, 10255589, 12150015

This whole topic leads in all sort of interesting directions but I'll have to leave off there and pursue some of these directions at another time.