OFFSET
1,1
COMMENTS
Numbers k such that the highest power of 12 dividing n! is determined by the highest power of 4 dividing n!.
Note that A054861 and A090616 are both asymptotic to a(n) = n/2 + O(log(n)), nevertheless, it seems that the number of k such that A090616(k) is bigger predominates. Conjecture: the ratio of k <= N such that A090616(k) > A054861(k) tends to 1 as N tends to infinity, while the ratio of k <= N such that A090616(k) < A054861(k) and A090616(k) = A054861(k) both tend to 0.
10^2...............38........................26........................37
10^3..............344.......................228.......................429
10^4.............2703......................2227......................5071
10^5............23003.....................19892.....................57106
10^6...........203478....................185152....................611371
10^7..........1762288...................1726062...................6511651
LINKS
Jianing Song, Table of n, a(n) for n = 1..2227 (all terms <= 10000)
EXAMPLE
The highest power of 3 dividing 9! is 3^4, while the highest power of 4 dividing 9! is 4^3, so 9 is a term, and the highest power of 12 dividing 9! is 12^3.
The highest power of 3 dividing 15! is 3^6, while the highest power of 4 dividing 15! is 4^5, so 15 is a term, and the highest power of 12 dividing 15! is 12^5.
PROG
(PARI) isA319316(n)=(n-vecsum(digits(n, 2)))\2<(n-vecsum(digits(n, 3)))\2
CROSSREFS
KEYWORD
nonn
AUTHOR
Jianing Song, Sep 17 2018
STATUS
approved