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A255400
a(n) is the smallest nonnegative integer such that a(n)! contains a string of exactly n consecutive 0's.
1
0, 5, 10, 15, 20, 264, 25, 30, 35, 40, 45, 101805, 50, 55, 60, 65, 70
OFFSET
0,2
COMMENTS
Most multiples of 5 belong to the sequence (if not all).
All terms whose indices are included in A000966 are far bigger than their neighboring terms whose indices are multiples of 5.
a(11) is a multiple of 5, we can verify a(11) = a(25448).
EXAMPLE
a(0) = 0 as 0! = 1 does not contain '0'.
a(1) = 5 as 5! = 120 contains '0'.
a(2) = 10 as 10! = 3628800 contains '00' and 10 is the smallest integer for which the condition is met.
PROG
(Python 2.7)
from math import factorial as fct
def trailing_zero(n):
k=0
while n!=0:
n/=5
k+=n
return k
def A255400():
index = 1
f = 1
while True:
if trailing_zero(f) == index:
print("A255400("+str(index)+") = " +str(f))
index += 1
elif trailing_zero(f) > index:
while True:
clnzer = str(fct(f))[:-trailing_zero(f)]
if index*'0' in clnzer and (index+1)*'0' not in clnzer:
print("A255400("+str(index)+") = " +str(f))
index += 1
f = 0
break
f +=1
f +=1
return
(Python)
import re
def A255400(n):
f, i, s = 1, 0, re.compile('[0-9]*[1-9]0{'+str(n)+'}[1-9][0-9]*')
while s.match(str(f)+'1') is None:
i += 1
f *= i
return i # Chai Wah Wu, Apr 02 2015
(PARI) \\ uses is() from A000966
f(k, special, sz, sz1) = my(f=k!); if (special, s=Str(f/10^valuation(f, 10)), s=Str(k!)); #strsplit(s, sz) - #strsplit(s, sz1);
a(n) = if (n==0, return(0)); my(sz= concat(vector(n, k, "0")), sz1=concat(sz, "0"), k=1, special=is(n)); while (f(k, special, sz, sz1) != 1, k++); k; \\ Michel Marcus, Oct 25 2023
KEYWORD
nonn,base,more
AUTHOR
Martin Y. Champel, Feb 22 2015
STATUS
approved