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.