A252652 - OEIS

%I #22 Oct 25 2023 20:33:38

%S 0,7,12,22,107,264,812,4919,12154,24612,75705,101805,236441,1946174

%N a(n) is the smallest nonnegative integer such that a(n)! contains a string of exactly n consecutive 0's, not including trailing 0's.

%e a(0) = 0 since 0! = 1, which does not contain a 0.

%e a(1) = 7 since 7! = 5040, which contains a 0 other than the trailing 0, and no integer smaller than 7 satisfies this requirement. (a(1) is not 5; 5! = 120, which has no 0 digits other than the trailing 0.)

%e a(2) = 12 since 12! = 479001600; discarding the trailing 0's leaves 4790016, which contains a string of exactly two consecutive 0's, and no integer smaller than 12 satisfies this requirement.

%t A252652[n_] := Module[{m = 0, s, t},

%t    If[n == 0, While[MemberQ[IntegerDigits[m!], 0], m++]; m,

%t     t = Table[0, n];

%t     While[s = Split[IntegerDigits[m!]];

%t      If[MemberQ[Last[s], 0], s = Delete[s, -1]]; ! MemberQ[s, t],

%t      m++]; m]];

%t Table[A252652[n], {n, 0, 13}] (* _Robert Price_, Mar 21 2019 *)

%o (Python)

%o import re

%o def A252652(n):

%o     if n == 0:

%o         return 0

%o     f, i, s = 1, 0, re.compile('[0-9]*[1-9]0{'+str(n)+'}[1-9][0-9]*')

%o     while s.match(str(f)) == None:

%o         i += 1

%o         f *= i

%o     return i # _Chai Wah Wu_, Dec 29 2015

%o (PARI) f(k, sz, sz1) = my(f=k!, s=Str(f/10^valuation(f, 10))); #strsplit(s, sz) - #strsplit(s, sz1);

%o a(n) = if (n==0, return(0)); my(sz=concat(vector(n, k, "0")), sz1=concat(sz, "0"), k=1); while (f(k, sz, sz1) != 1, k++); k; \\ _Michel Marcus_, Oct 25 2023

%Y Cf. A254042, A254447, A254448, A254449, A254500, A254501, A254502, A254716, A254717.

%K nonn,base,more

%O 0,2

%A _Jon E. Schoenfield_, Mar 22 2015, at the suggestion of _Martin Y. Champel_

%E a(13) from _Lars Blomberg_, Apr 05 2015