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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!.
3

%I #14 Jun 29 2019 01:40:15

%S 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,

%T 35,36,37,39,40,41,42,44,45,46,47,49,50,51,52,53,55,56,57,58,60,61,62,

%U 63,65,66,67,68,69,71,72,73,74,76,77,78,79,80,82,83,84

%N 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!.

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

%H Clark Kimberling, <a href="/A277676/b277676.txt">Table of n, a(n) for n = 1..1000</a>

%F (See A277674.)

%t d = Differences@Array[Floor@Log[#, #!] &, 10000, 2]; (* _Robert G. Wilson v_, Oct 27 2016 *)

%t u = Flatten[Position[d, 0]]; (* A277675 *)

%t v = Flatten[Position[d, 1]]; (* A277676 *)

%o (PARI) d(n) = #digits(n!, n);

%o isok(n) = d(n+2) > d(n+1); \\ _Michel Marcus_, Oct 29 2016

%Y Cf. A000142, A060151, A277674, A277675.

%K nonn,base,easy

%O 1,1

%A _Clark Kimberling_, Oct 26 2016