Displaying 1-6 of 6 results found.
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0, 1, 10, 11, 100, 101, 110, 111, 1000, 1010, 1100, 1110, 10000, 10001, 10100, 10101, 11000, 11100, 100000, 100001, 100010, 100011, 101000, 101010, 110000, 110001, 111000, 1000000, 1000001, 1000010, 1000011, 1000100, 1000101
Numbers n such that binary expansion matches ((0)*00(1?)1)*(0*).
+10
13
0, 1, 2, 3, 4, 6, 8, 9, 12, 16, 17, 18, 19, 24, 25, 32, 33, 34, 35, 36, 38, 48, 49, 50, 51, 64, 65, 66, 67, 68, 70, 72, 73, 76, 96, 97, 98, 99, 100, 102, 128, 129, 130, 131, 132, 134, 136, 137, 140, 144, 145, 146, 147, 152, 153, 192, 193, 194, 195, 196, 198, 200, 201
COMMENTS
If bit i is 1, then bits i+-2 must be 0. All terms satisfy A048725(n) = 5*n.
It appears that n is in the sequence if and only if C(5n,n) is odd (cf. A003714). - Benoit Cloitre, Mar 09 2003
Yes, as remarked in A048715, "This is easily proved using the well-known result that the multiplicity with which a prime p divides C(n+m,n) is the number of carries when adding n+m in base p." - Jason Kimberley, Dec 21 2011
MATHEMATICA
Reap[Do[If[OddQ[Binomial[5n, n]], Sow[n]], {n, 0, 400}]][[2, 1]]
(* Second program: *)
filterQ[n_] := With[{bb = IntegerDigits[n, 2]}, MatchQ[bb, {0}|{1}|{1, 1}|{___, 0, _, 1, ___}|{___ 1, _, 0, ___}] && !MatchQ[bb, {___, 1, _, 1, ___}]];
PROG
(PARI) list(lim)=my(v=List(), n, t); while(n<=lim, t=bitand(n, n>>2); if(t, n+=1<<valuation(t, 2), listput(v, n); n++)); Vec(v) \\ Charles R Greathouse IV, Oct 22 2021
AUTHOR
Antti Karttunen, Mar 30 1999
Differences between the ordinary multiplication table A004247 and the carryless multiplication table for GF(2)[X] polynomials A048720, i.e., the effect of the carry bits in binary multiplication.
+10
12
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 4, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 8, 0, 0, 8, 0, 0, 0, 0, 0, 0, 12, 0, 8, 0, 12, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 8, 16, 8, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 24, 24, 0, 0, 0
EXAMPLE
The top left 12 X 12 corner of the table:
| 0 1 2 3 4 5 6 7 8 9 10 11
------+------------------------------------------------
0 | 0 0 0 0 0 0 0 0 0 0 0 0
1 | 0 0 0 0 0 0 0 0 0 0 0 0
2 | 0 0 0 0 0 0 0 0 0 0 0 0
3 | 0 0 0 4 0 0 8 12 0 0 0 4
4 | 0 0 0 0 0 0 0 0 0 0 0 0
5 | 0 0 0 0 0 8 0 8 0 0 16 16
6 | 0 0 0 8 0 0 16 24 0 0 0 8
7 | 0 0 0 12 0 8 24 28 0 0 16 28
8 | 0 0 0 0 0 0 0 0 0 0 0 0
9 | 0 0 0 0 0 0 0 0 0 16 0 16
10 | 0 0 0 0 0 16 0 16 0 0 32 32
11 | 0 0 0 4 0 16 8 28 0 16 32 52
(End)
CROSSREFS
Main diagonal divided by 4: A213673.
Smallest k such that n XOR n*2^k = n*(2^k + 1).
+10
9
0, 1, 1, 2, 1, 1, 2, 3, 1, 1, 1, 4, 2, 4, 3, 4, 1, 1, 1, 2, 1, 1, 4, 5, 2, 2, 4, 5, 3, 5, 4, 5, 1, 1, 1, 2, 1, 1, 2, 6, 1, 1, 1, 6, 4, 4, 5, 6, 2, 2, 2, 2, 4, 6, 5, 6, 3, 6, 5, 6, 4, 6, 5, 6, 1, 1, 1, 2, 1, 1, 2, 3, 1, 1, 1, 4, 2, 5, 6, 7, 1, 1, 1, 7, 1, 1, 6, 7, 4, 5, 4, 7, 5, 5, 6, 7, 2, 2, 2, 2, 2, 7, 2, 7, 4
MATHEMATICA
a[n_] := Module[{k}, For[k = 0, True, k++,
If[BitXor[n, n*2^k] == n*(2^k+1), Return[k]]]];
PROG
(Python)
from itertools import count
def A116361(n): return next(k for k in count(0) if n^(m:=n<<k)==m+n) # Chai Wah Wu, Jul 19 2024
Integers i such that 17*i = 17 X i, i.e., 16*i XOR i = 17*i.
+10
8
0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 18, 20, 22, 24, 26, 28, 30, 32, 33, 36, 37, 40, 41, 44, 45, 48, 52, 56, 60, 64, 65, 66, 67, 72, 73, 74, 75, 80, 82, 88, 90, 96, 97, 104, 105, 112, 120, 128, 129, 130, 131, 132, 133, 134, 135, 144, 146, 148, 150
COMMENTS
Here * stands for ordinary multiplication and X means carryless (GF(2)[X]) multiplication ( A048720).
CROSSREFS
Cf. A115848 shows this sequence in binary. Complement of A115849. Differs from A032966 for the first time at n=25, where A032966(25)=34 while a(25)=33.
Numbers m such that m XOR 32*m = 33*m.
+10
2
0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 34, 36, 38, 40, 42, 44, 46, 48, 50, 52, 54, 56, 58, 60, 62, 64, 65, 68, 69, 72, 73, 76, 77, 80, 81, 84, 85, 88, 89, 92, 93, 96, 100, 104, 108, 112, 116, 120
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