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

1,2

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.

Links

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

Example

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

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 *)

Prog

(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
(Python 3.10+)
from math import factorial
def A205509(n): return ((f:=factorial(n-1))^f*n).bit_count() # Chai Wah Wu, Jul 13 2022

Crossrefs

Cf. A001511.

Sequence in context: A261211 A233521 A035685 * A118736 A201161 A105474

Adjacent sequences: A205506 A205507 A205508 * A205510 A205511 A205512

Keyword

nonn,base

Author

Vladimir Shevelev, Jan 28 2012

Status

approved