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A056851
Integers n such that the number of digits in n! is a cube.
4
0, 1, 2, 3, 11, 26, 83, 128, 186, 258, 572, 875, 1494, 2029, 3859, 4810, 6497, 9274, 18033, 19243, 24600, 26073, 30828, 32528, 34287, 41930, 48325, 96475, 103590, 118814, 126936, 205022, 240742, 260009, 331334, 379612, 396656, 405360, 414186
OFFSET
1,3
COMMENTS
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
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).
FORMULA
{n: A034886(n) in A000578}. - R. J. Mathar, Jan 15 2013
MATHEMATICA
Do[ If[ IntegerQ[ RealDigits[ n! ][[ 2 ]]^(1/3) ], Print[ n ]], {n, 0, 53100}]
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)] & ]
Select[Range[0, 420000], IntegerQ[Surd[IntegerLength[#!], 3]]&] (* Harvey P. Dale, Mar 09 2019 *)
CROSSREFS
KEYWORD
nonn,base
AUTHOR
Robert G. Wilson v, Aug 30 2000
EXTENSIONS
More terms from Robert G. Wilson v, Jun 25 2003
STATUS
approved