OFFSET
0,2
COMMENTS
Conjecture: the sequence is bounded.
I conjecture the contrary: for every k, there exists n with a(n) > k. See A103670. - Charles R Greathouse IV, Aug 21 2011
For n>0: A103670(n) = smallest m such that a(m)=n;
A103671(n) = smallest m such that in binary representation n! doesn't contain m!;
A103672(n) = greatest m less than n such that in binary representation n! contains m!.
LINKS
EXAMPLE
n=6: 6!=720->'1011010000' contains a(6)=5 factorials: 0!=1->'1', 1!=1->'1', 2!=2->'10', 3!=6->'110' and 6! itself, but not 4!=24->'11000' and 5!=120->'1111000'.
PROG
(PARI) contains(v, u)=for(i=0, #v-#u, for(j=1, #u, if(v[i+j]!=u[j], next(2))); return(1)); 0
a(n)=my(v=binary(n--!)); sum(i=0, n-1, contains(v, binary(i!)))+1 \\ Charles R Greathouse IV, Aug 21 2011
CROSSREFS
KEYWORD
nonn
AUTHOR
Reinhard Zumkeller, Feb 07 2005
STATUS
approved