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A341242
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Numbers whose binary representation encodes a subset S of the natural numbers such that the XOR of the binary representations of all s in S gives 0.
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0
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0, 1, 14, 15, 50, 51, 60, 61, 84, 85, 90, 91, 102, 103, 104, 105, 150, 151, 152, 153, 164, 165, 170, 171, 194, 195, 204, 205, 240, 241, 254, 255, 770, 771, 780, 781, 816, 817, 830, 831, 854, 855, 856, 857, 868, 869, 874, 875, 916, 917, 922, 923, 934, 935, 936
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listen;
history;
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internal format)
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OFFSET
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1,3
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COMMENTS
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The numbers for which the set S of positions of bits 1 in the binary representation, interpreted as a set of distinct-sized Nim heaps (including a possible uninteresting size 0 heap for the least significant bit) is losing for the player to move.
Viewing the list as a set of valid code words, every natural number N can be "corrected" to a valid code word by changing exactly one bit, in exactly one way. The position of that bit is found by computing for N the XOR of its raised-bit positions of the title (if the result is 0, then N is already valid but flipping the irrelevant bit 0 makes it valid again).
The "error correcting" interpretation, applied to 64-bit numbers interpreted as orientation of 64 coins, corresponds to a solution of the "coins on a chessboard" puzzle described in the Nick Berry's blog, and also mentioned at A253315.
Numbers 2*n and 2*n+1 for n = A075926(m).
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LINKS
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FORMULA
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PROG
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(C++) (first 2^12 terms)
for (int i=0; i<65536; ++i) {
int sum=0;
for (int n=i, count=0; n>0; n>>=1, ++count)
if ((n&1)!=0)
sum ^= count;
if (sum==0)
std::cout << i << ", ";
}
(Python)
def ok(n):
xor, b = 0, (bin(n)[2:])[::-1]
for i, c in enumerate(b):
if c == '1': xor ^= i
return xor == 0
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CROSSREFS
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KEYWORD
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nonn,base
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AUTHOR
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STATUS
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approved
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