A093684 In binary representation: number of occurrences of n in n!.

1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 0, 3, 0, 1, 0, 2, 0, 1, 1, 1, 1, 2, 3, 2, 2, 2, 1, 1, 1, 3, 0, 2, 1, 3, 1, 1, 0, 1, 2, 2, 3, 3, 1, 1, 1, 1, 2, 2, 4, 3, 3, 3, 2, 2, 0, 3, 1, 5, 5, 6, 4, 1, 5, 2, 3, 2, 2, 4, 1, 1, 1, 4, 1, 1, 1, 2, 3, 3, 4, 5, 0, 3, 2, 1, 4, 3, 4, 5, 3, 2, 1, 2, 3, 3, 3, 3, 6, 2, 3, 4, 4, 2

Offset: 1

Author

Reinhard Zumkeller, Apr 10 2004

Keywords

• nonn

Comments

a(A093685(n)) = 0, a(A093686(n)) > 0.

Examples

			n=12->'1100', 12!=479001600->'11100100011001111110000000000' with three occurrences of '1100': '.1100....1100....1100........', therefore a(12)=3.
		

Links

• Robert Israel, Table of n, a(n) for n = 1..10000
• Index entries for sequences related to factorial numbers.
• Index entries for sequences related to binary expansion of n

Crossrefs

Cf. A036603, A007088, A000142, A092601.

Programs

Maple

f:= proc(n) local L,Lf;
  L:= convert(convert(n,binary),string);
  Lf:= convert(convert(n!,binary),string);
  nops([StringTools:-SearchAll(L,Lf)])
end proc:
map(f, [$1..100]); # Robert Israel, May 20 2016

Mathematica

non[n_]:=Module[{b=IntegerDigits[n,2],f=IntegerDigits[n!,2]}, Length[ Select[ Partition[ f,Length[b],1],#==b&]]]; Array[non,110] (* Harvey P. Dale, Jun 04 2014 *)