A254042 a(n) is the smallest nonnegative integer such that a(n)! contains a string of exactly n consecutive 1's.

2, 0, 22, 47, 38, 436, 736, 2322, 3912, 47262, 123398, 263600, 679530, 725244, 8118161, 5690326

Offset

0,1

Links

Table of n, a(n) for n=0..15.

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

Cf. A254447, A254448, A254449, A254500, A254501, A254502, A254716, A254717, A252652.

Sequence in context: A156438 A282134 A293938 * A368014 A009378 A106708

Adjacent sequences: A254039 A254040 A254041 * A254043 A254044 A254045

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