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A006042
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The nim-square of n.
(Formerly M2251)
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6
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0, 1, 3, 2, 6, 7, 5, 4, 13, 12, 14, 15, 11, 10, 8, 9, 24, 25, 27, 26, 30, 31, 29, 28, 21, 20, 22, 23, 19, 18, 16, 17, 52, 53, 55, 54, 50, 51, 49, 48, 57, 56, 58, 59, 63, 62, 60, 61, 44, 45, 47, 46, 42, 43, 41, 40, 33, 32, 34, 35, 39, 38, 36, 37, 103, 102, 100, 101, 97, 96, 98, 99
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OFFSET
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0,3
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COMMENTS
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This is a permutation of the natural numbers; A160679 is the inverse permutation. - Jianing Song, Aug 10 2022
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REFERENCES
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J. H. Conway, On Numbers and Games. Academic Press, NY, 1976, pp. 51-53.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
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LINKS
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FORMULA
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If n = Sum_j 2^e(j), then a(n) is the XOR of A006017(e(j))'s. Proof: let N+ = XOR and N* denote the nim addition and the nim multiplication, then n N* n = (Sum_j 2^e(j)) N* (Sum_j 2^e(j)) = (Nim-sum_j 2^e(j)) N* (Nim-sum_j 2^e(j)) = (Nim-sum_j (2^e(j) N* 2^e(j))) N+ (Nim-sum_{i<j} ((2^e(i) N* 2^e(j)) N+ (2^e(j) N* 2^e(i)))) = (Nim-sum_j (2^e(j) N* 2^e(j))) N+ (Nim-sum_{i<j} 0) = Nim-sum_j (2^e(j) N* 2^e(j)).
For example, for n = 11 = 2^0 + 2^1 + 2^3, a(11) = A006017(0) XOR A006017(1) XOR A006017(3) = 1 XOR 3 XOR 13 = 15.
More generally, if n = Sum_j 2^e(j), k is a power of 2, then the nim k-th power of n is the XOR of (nim k-th power of 2^e(j))'s. (End)
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MAPLE
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read("transforms") ;
# insert source of nimprodP2() and A051775() from the b-file at A051776 here...
end proc:
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CROSSREFS
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KEYWORD
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AUTHOR
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EXTENSIONS
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a(1)-a(49) confirmed, a(50)-a(71) added by John W. Layman, Nov 05 2010
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STATUS
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approved
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