A205509 Hamming distance between (n-1)! and n!.

0, 2, 1, 4, 2, 4, 4, 6, 4, 9, 8, 15, 12, 16, 14, 12, 16, 23, 26, 23, 21, 29, 31, 34, 31, 33, 33, 44, 32, 38, 42, 46, 52, 51, 45, 55, 55, 59, 55, 59, 51, 82, 65, 83, 74, 75, 80, 80, 80, 74, 87, 104, 86, 91, 98, 90, 81, 103, 104, 98, 112, 104, 111, 116, 111, 132

Offset: 1

Author

Vladimir Shevelev, Jan 28 2012

Keywords

• nonn
• base

Comments

Problem: To find a better lower estimate for a(n) than the trivial one, which is a(n) >= A000120(floor(log_2(n))).

Note that this trivial estimate yields unboundedness of the sequence.

Examples

			Since 5!=(0001111000)_2 and 6!=(1011010000)_2, then the number of different binary digits is 4. Therefore, a(6)=4.
		

Links

• Alois P. Heinz, Table of n, a(n) for n = 1..1000

Crossrefs

Cf. A001511.

Programs

Maple

read("transforms") :
Hamming := proc(a,b)
        XORnos(a,b) ;
        wt(%) ;
end proc:
A205509 := proc(n)
        Hamming((n-1)!,n!) ;
end proc: # R. J. Mathar, Apr 02 2012

Mathematica

nn = 100; Table[b2 = IntegerDigits[n!, 2]; b1 = IntegerDigits[(n - 1)!, 2, Length[b2]]; Total[Abs[b1 - b2]], {n, nn}] (* T. D. Noe, Jan 31 2012 *)

Python

from math import factorial
def A205509(n): return ((f:=factorial(n-1))^f*n).bit_count() # Chai Wah Wu, Jul 13 2022

Sage

def A205509(n) :
    f = bin(factorial(n)).lstrip("0b")
    g = bin(factorial(n-1)).lstrip("0b")
    h = "".zfill(len(f)-len(g)) + g
    return sum(a != b for a, b in zip(f, h))
[A205509(k) for k in (1..66)] # Peter Luschny, Jan 31 2012