%I
%S 53,67,78,105,106,114,156,160,161,164,192,205,215,220,222,225,236,243,
%T 251,257,268,278,279,282,287,306,320,338,343,345,353,367,377,386,395,
%U 398,408,409,420,423,434,443,444,448,451,455,458,459,463,467,468,469
%N Numbers k such that k! contains a pandigital substring.
%e 53! = 42748832840600255(6429801375)3389399649690343788366813724672000000000000.
%t Select[Range[500],Count[Sort/@Partition[IntegerDigits[#!],10,1], Range[ 0,9]]>0&] (* _Harvey P. Dale_, May 23 2017 *)
%o (Python)
%o from math import factorial
%o def haspan(s): return any(len(set(s[i:i+10]))==10 for i in range(len(s)9))
%o print([m for m in range(500) if haspan(str(factorial(m)))]) # _Michael S. Branicky_, Feb 28 2021
%Y Cf. A115933, A115934, A115935, A115937, A115938.
%K nonn,base
%O 1,1
%A _Giovanni Resta_, Feb 06 2006
