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 A205509 Hamming distance between (n-1)! and n!. 18
 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 (list; graph; refs; listen; history; text; internal format)
 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 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

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Last modified May 29 20:42 EDT 2020. Contains 334710 sequences. (Running on oeis4.)