A277676 Numbers k such that d(k+2) > d(k+1), where d(m) is the number of digits in the base-m representation of m!.

2, 4, 5, 6, 8, 9, 10, 12, 13, 14, 16, 17, 18, 20, 21, 22, 23, 25, 26, 27, 29, 30, 31, 32, 34, 35, 36, 37, 39, 40, 41, 42, 44, 45, 46, 47, 49, 50, 51, 52, 53, 55, 56, 57, 58, 60, 61, 62, 63, 65, 66, 67, 68, 69, 71, 72, 73, 74, 76, 77, 78, 79, 80, 82, 83, 84

Offset: 1

Author

Clark Kimberling, Oct 26 2016

Keywords

• nonn
• base
• easy

Comments

It has been checked for k = 2..50000 that if d(k+1) > d(k), then d(k+1) = d(k) + 1.

Links

• Clark Kimberling, Table of n, a(n) for n = 1..1000

Crossrefs

Cf. A000142, A060151, A277674, A277675.

Programs

Mathematica

d = Differences@Array[Floor@Log[#, #!] &, 10000, 2]; (* Robert G. Wilson v, Oct 27 2016 *)
u = Flatten[Position[d, 0]]; (* A277675 *)
v = Flatten[Position[d, 1]]; (* A277676 *)

PARI

d(n) = #digits(n!, n);
isok(n) = d(n+2) > d(n+1); \\ Michel Marcus, Oct 29 2016

Formula

(See A277674.)