

A217445


Numbers n such that n! has the same number of terminating zeros in bases 3 and 4.


3



1, 2, 4, 5, 6, 7, 10, 11, 12, 13, 14, 18, 19, 21, 22, 23, 33, 36, 37, 38, 42, 43, 46, 47, 51, 56, 58, 59, 60, 61, 62, 75, 86, 88, 89, 92, 100, 101, 102, 103, 105, 112, 113, 114, 115, 120, 121, 122, 124, 125, 138, 139, 141, 147, 153, 159, 164, 166, 167, 168
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OFFSET

1,2


COMMENTS

The number of zeros of n! base 3 is approaching n/2 as n grows. Similarly, the number of zeros of n! base 4 is approaching n/2 as n grows. Consequently, this sequence is expected to have high density.
Since typically A000120(n) ~ log_2(n) while typically A053735(n) ~ log_3(n), the density of this sequence should go to 0, contrary to the previous comment. (End)
Appears to be the same as the list of positive numbers n such that the last nonzero digit of n! in base 12 belongs to the set [1, 2, 5, 7, 10, 11].
The first footnote in Deshouillers et al. (2016) says: "if the last nonzero digit of n! in base 12 belongs to {1, 2, 5, 7, 10, 11} then (digitsum of n in base 3)  (digitsum of n in base 2) is <= 1; this seems to occur infinitely many times." Compare A096288. (End)


REFERENCES

JeanMarc Deshouillers, Laurent Habsieger, Shanta Laishram, Bernard Landreau, Sums of the digits in bases 2 and 3, arXiv:1611.08180, 2016


LINKS



EXAMPLE

6! is 222200 in base 3 and 23100 in base 4, both of them have 2 zeros at the end, so 6 is in the sequence.


MAPLE

s2:= n > convert(convert(n, base, 2), `+`):
s3:= n > convert(convert(n, base, 3), `+`):
select(n > s2(n) + (n+s2(n) mod 2) = s3(n), [$1..1000]); # Robert Israel, Jan 19 2017


MATHEMATICA

sntzQ[n_]:=Module[{f=n!}, Last[Split[IntegerDigits[f, 3]]]==Last[ Split[ IntegerDigits[ f, 4]]]]; Select[Range[200], sntzQ] (* Harvey P. Dale, Jul 11 2020 *)


PROG

(PARI) is(n)=my(L=log(n+1)); sum(k=1, L\log(3), n\3^k)==sum(k=1, L\log(2), n>>k)\2 \\ Charles R Greathouse IV, Oct 04 2012


CROSSREFS



KEYWORD

nonn,base


AUTHOR



EXTENSIONS



STATUS

approved



