A “block” generalization of factorions

Question 1

I’ve been exploring a generalization of factorions—that is, numbers equal to the sum of factorials of their digits—and wanted some feedback on the concept.

Let$n \ge 1$ where$n$ is an integer. A number$N$ is an$n$-factorion if you can split$N$ into consecutive groups of$n$ digits each, take the factorial of these$n$-digit blocks and sum them, and they all sum to$N$. Keep in mind this does require$N$ to have a number of digits divisible by$n$. By this logic, a typical factorion would be considered a 1-factorion.

For example, consider 2,763 as a candidate 2-factorion. It would turn into$27! + 63!$ which is equal to approximately$1.98 \times 10^{88}$ which is pretty obviously way off. As you can see, 2-factorions alone would be monstrously big, the smallest candidate possible being 3,628,800 (10!). Clearly the smallest one would have to be hundreds or thousands of digits.

I personally conjecture that there is no upper bound to how large$n$ can be. The only thing holding regular factorions back was the fact that 9! can only work for so many digits, however we can keep theoretically increasing up to 99!, 999!, etc.

I’d love feedback on whether this generalization is interesting, or if there are known results related to this topic. Are there any theoretical approaches for estimating or bounding the size of$n$-factorials without brute-force computation?

Question 2

I think $27!+63! \approx 1.9826\cdot{10}^{87}$. Also, about "Keep in mind this does require $N$ to have a number of digits divisible by $n$." "As you can see, 2-factorions alone would be monstrously big, the smallest candidate possible being 3,628,800 (10!)." -- I do not understand these parts of your question. The added restriction on the number of digits seems to conflict with the given example of a candidate: $10! = 3{,}628{,}800$ has 7 digits (in base 10) and 7 is not divisible by 2.

Question 3

Let$n\ge 1$ where$n$ is an integer. A number$N$ is an$n$-factorion if you can split$N$ into consecutive groups of$n$ digits each, take the factorial of these$n$-digit blocks and sum them, and they all sum to$N$.

At least without the added restriction "require$N$ to have a number of digits divisible by$n$", I think this is equivalent to considering (usual, "single-digit") factorions in base${10}^n$ instead of base-10 factorions. In base${10}^n$, each "block" becomes a single digit.

As long as the only modification is changing the base, you can use the same approaches for proving bounds. Related questions:

Keep in mind this does require$N$ to have a number of digits divisible by$n$.
As you can see, 2-factorions alone would be monstrously big, the smallest candidate possible being 3,628,800 (10!).

I do not understand these parts of your question. In particular, the added restriction on the number of digits seems to conflict with the given example of a candidate:$10! = 3{,}628{,}800$ has 7 digits (in base 10) and 7 is not divisible by 2.

Clearly the smallest one would have to be hundreds or thousands of digits.

If$N$ is a positive integer whose base-$b$ expansion has$K+1$ digits and the sum of factorials of those digits equals$N$, then$ b^K \le N \le \left(b-1\right)!\cdot\left(K+1\right) $. Setting$b=10$ implies$K \le 6$ (at most 7 digits). I think setting$b=100$ implies$K \le 78$ (at most 79 base-100 digits; at most 158 digits in decimal expansion), and setting$b=1000$ implies$K \le 855$ (at most 856 base-1000 digits; at most 2568 digits in decimal expansion).

I do not know an example of a number$N$ with this property for$b={10}^n$,$n\ge 2$. Related question:

Efficient factorion search in arbitrary base

mezzoctane 1,5636 silver badges17 bronze badges · Accepted Answer · 2025-11-29 00:38:47Z

Let$n\ge 1$ where$n$ is an integer. A number$N$ is an$n$-factorion if you can split$N$ into consecutive groups of$n$ digits each, take the factorial of these$n$-digit blocks and sum them, and they all sum to$N$.

At least without the added restriction "require$N$ to have a number of digits divisible by$n$", I think this is equivalent to considering (usual, "single-digit") factorions in base${10}^n$ instead of base-10 factorions. In base${10}^n$, each "block" becomes a single digit.

As long as the only modification is changing the base, you can use the same approaches for proving bounds. Related questions:

Keep in mind this does require$N$ to have a number of digits divisible by$n$.
As you can see, 2-factorions alone would be monstrously big, the smallest candidate possible being 3,628,800 (10!).

I do not understand these parts of your question. In particular, the added restriction on the number of digits seems to conflict with the given example of a candidate:$10! = 3{,}628{,}800$ has 7 digits (in base 10) and 7 is not divisible by 2.

Clearly the smallest one would have to be hundreds or thousands of digits.

If$N$ is a positive integer whose base-$b$ expansion has$K+1$ digits and the sum of factorials of those digits equals$N$, then$ b^K \le N \le \left(b-1\right)!\cdot\left(K+1\right) $. Setting$b=10$ implies$K \le 6$ (at most 7 digits). I think setting$b=100$ implies$K \le 78$ (at most 79 base-100 digits; at most 158 digits in decimal expansion), and setting$b=1000$ implies$K \le 855$ (at most 856 base-1000 digits; at most 2568 digits in decimal expansion).

I do not know an example of a number$N$ with this property for$b={10}^n$,$n\ge 2$. Related question:

Efficient factorion search in arbitrary base

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