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).
LINKS
FORMULA
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