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Integers n such that the number of digits in n! is a cube.
4

%I #37 Jan 04 2020 09:51:23

%S 0,1,2,3,11,26,83,128,186,258,572,875,1494,2029,3859,4810,6497,9274,

%T 18033,19243,24600,26073,30828,32528,34287,41930,48325,96475,103590,

%U 118814,126936,205022,240742,260009,331334,379612,396656,405360,414186

%N Integers n such that the number of digits in n! is a cube.

%C Numbers whose cube is represented by the number of digits of n!: 1, 2, 3, 5, 6, 7, 8, 11, 13, 16, 18, 23, 25, 28, ..., . - _Robert G. Wilson v_, May 14 2014

%C _Ed Pegg Jr_ conjectures that n^3 - n = k! has a solution if and only if n is 2, 3, 5 or 9 (when k is 3, 4, 5 and 6).

%H Robert G. Wilson v, <a href="/A056851/b056851.txt">Table of n, a(n) for n = 1..78</a>

%H <a href="/index/Fa#factorial">Index entries for sequences related to factorial numbers</a>

%F {n: A034886(n) in A000578}. - _R. J. Mathar_, Jan 15 2013

%t Do[ If[ IntegerQ[ RealDigits[ n! ][[ 2 ]]^(1/3) ], Print[ n ]], {n, 0, 53100}]

%t LogBase10Stirling[n_] := Floor[Log[10, 2 Pi n]/2 + n*Log[10, n/E] + Log[10, 1 + 1/(12n) + 1/(288n^2) - 139/(51840n^3) - 571/(2488320n^4) + 163879/(209018880n^5)]]; Select[ Range[ 500000], IntegerQ[ (LogBase10Stirling[ # ] + 1)^(1/3)] & ]

%t Select[Range[0,420000],IntegerQ[Surd[IntegerLength[#!],3]]&] (* _Harvey P. Dale_, Mar 09 2019 *)

%Y Cf. A000142, A000578, A034886, A006488.

%K nonn,base

%O 1,3

%A _Robert G. Wilson v_, Aug 30 2000

%E More terms from _Robert G. Wilson v_, Jun 25 2003