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Numbers k such that the arithmetic mean of the digits of k! is an integer.
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%I #21 Dec 21 2021 06:34:31

%S 0,1,2,3,4,5,6,12,26,28,32,59,262,391,533,579

%N Numbers k such that the arithmetic mean of the digits of k! is an integer.

%C A heuristic argument suggests that this short list is complete. By Stirling's approximation, n! has order n*log(n) digits of which n/4 are terminal zeros. If the remaining digits are random, the mean will be just below 4.5. For n > 6, n! and also its digits sum are divisible by 9. 12! is the only factorial with 9 digits. The others have 27, 30, 36, 81, 522, 846, 1224, and 1350 digits, respectively.

%e 4 is a term because 4! = 24 and (2+4)/2 = 3 is an integer.

%p q:= n-> (f-> (add(i, i=convert(f, base, 10))/length(f))::integer)(n!):

%p select(q, [$0..1000])[]; # _Alois P. Heinz_, Dec 19 2021

%t Do[If[IntegerQ[Mean[IntegerDigits[n!]]], Print[n, " ", Mean[IntegerDigits[n!]]]], {n, 1, 100000}]

%o (PARI) isok(k) = my(d=digits(k!)); (vecsum(d) % #d) == 0; \\ _Michel Marcus_, Dec 19 2021

%Y Cf. A000142, A004152, A034886, A058814, A061383.

%K nonn,base,more

%O 1,3

%A _Zachary M Franco_, Dec 19 2021