OFFSET
0,1
EXAMPLE
a(0) = 2 since 2! equals 2, which does not contain any '1'.
a(1) = 0 since 0! equals 1, which contains '1' but not '11'.
a(2) = 22 since 22! equals 1124000727777607680000, which contains '11', and 22 is the smallest integer for which this condition is met.
MATHEMATICA
f[n_] := Block[{k = 0, s = ToString[(10^n - 1)/9]}, While[ Length@ StringPosition[ ToString[k!], s] != 1, j = k++]; k]; f[0] = 2; Array[f, 12, 0] (* Robert G. Wilson v, Feb 27 2015 *)
PROG
(Python 2.7)
def A254042():
index = 1
k = 0
f = 1
u = '1'
while True:
sf = str(f)
if u in sf and u+'1' not in sf:
print "A254042("+str(index)+") = " +str(k)
index += 1
k = 0
f = 1
u +='1'
k += 1
f *= k
return
(PARI) a(n)=k=0; while(k<10^4, d=digits(2*10^(#(digits(k!))+1)+10*k!); for(j=1, #d-n+1, c=0; for(i=j, j+n-1, if(d[i]==1, c++); if(d[i]!=1, c=0; break)); if(c==n&&d[j+n]!=1&&d[j-1]!=1, return(k))); if(c==n, return(k)); if(c!=n, k++))
for(n=1, 6, print1(a(n), ", ")) \\ Derek Orr, Jan 29 2015
(PARI) max1s(n)=my(v=digits(n), r, t); for(i=1, #v, if(v[i]==1, t++, r=max(r, t); t=0)); max(t, r)
a(n)=my(m=0); while(max1s(m!)!=n, m++); m \\ Charles R Greathouse IV, Jan 30 2015
CROSSREFS
KEYWORD
nonn,base,more
AUTHOR
Martin Y. Champel, Jan 25 2015
EXTENSIONS
a(11) from Jon E. Schoenfield, Feb 22 2015
a(12), a(13) from Jon E. Schoenfield, Mar 07 2015, Mar 08 2015
a(14)-a(15) from Bert Dobbelaere, Oct 29 2018
STATUS
approved