A319317 Numbers k such that A090616(k) > A054861(k).

8, 16, 17, 20, 24, 25, 26, 32, 34, 35, 40, 41, 44, 48, 49, 50, 52, 53, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 76, 77, 78, 79, 80, 96, 97, 98, 104, 106, 107, 116, 128, 129, 130, 131, 132, 133, 134, 136, 137, 140, 142, 143, 144, 145, 146, 148, 149, 150, 151

Offset

1,1

Comments

Numbers k such that the highest power of 12 dividing n! is determined by the highest power of 3 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 thatA090616(k) < A054861(k) and A090616(k) = A054861(k) both tend to 0.

Number of k in range [0, N] such that A090616(k) =, < or > A054861(k):

N A090616(k) = A054861(k) A090616(k) < A054861(k) A090616(k) > A054861(k)

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..5071 (all terms <= 10000)

Example

The highest power of 3 dividing 8! is 3^2, while the highest power of 4 dividing 8! is 4^3, so 8 is a term, and the highest power of 12 dividing 8! is 12^2.
The highest power of 3 dividing 16! is 3^6, while the highest power of 4 dividing 16! is 4^7, so 16 is a term, and the highest power of 12 dividing 16! is 12^6.

Prog

(PARI) isA319317(n)=(n-vecsum(digits(n, 2)))\2>(n-vecsum(digits(n, 3)))\2

Crossrefs

Cf. A217445 (k such that A090616(k) = A054861(k)), A319316 (k such that A090616(k) < A054861(k)).

Sequence in context: A204651 A203452 A363982 * A091251 A297264 A296708

Adjacent sequences: A319314 A319315 A319316 * A319318 A319319 A319320

Keyword

nonn

Author

Jianing Song, Sep 17 2018

Status

approved