Displaying 1-10 of 18 results found.
The nim-square of n.
(Formerly M2251)
+10
6
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
COMMENTS
This is a permutation of the natural numbers; A160679 is the inverse permutation. - Jianing Song, Aug 10 2022
REFERENCES
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).
FORMULA
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)
MAPLE
read("transforms") ;
# insert source of nimprodP2() and A051775() from the b-file at A051776 here... end proc:
EXTENSIONS
a(1)-a(49) confirmed, a(50)-a(71) added by John W. Layman, Nov 05 2010
0, 1, 1, 1, 14, 13, 8, 10, 14, 10, 13, 8, 14, 8, 10, 13, 152, 145, 133, 141, 189, 182, 167, 173, 203, 199, 212, 217, 224, 238, 248, 247, 152, 141, 145, 133, 224, 247, 238, 248, 189, 173, 182, 167, 203, 217, 199, 212, 152, 133, 141, 145, 203, 212, 217, 199, 224
PROG
(PARI) See Links section.
0, 1, 1, 1, 8, 10, 13, 14, 8, 14, 10, 13, 8, 13, 14, 10, 203, 199, 217, 212, 248, 247, 238, 224, 182, 189, 167, 173, 141, 133, 152, 145, 203, 212, 199, 217, 141, 145, 133, 152, 248, 224, 247, 238, 182, 173, 189, 167, 203, 217, 212, 199, 182, 167, 173, 189, 141
0, 1, 3, 2, 2, 2, 3, 3, 1, 2, 1, 2, 3, 1, 1, 3, 72, 76, 65, 69, 88, 93, 83, 86, 106, 108, 96, 102, 122, 125, 114, 117, 196, 207, 200, 195, 229, 239, 235, 225, 244, 253, 251, 242, 213, 221, 216, 208, 140, 130, 138, 132, 191, 176, 187, 180, 159, 147, 154, 150
0, 1, 2, 3, 11, 9, 15, 12, 13, 4, 14, 5, 6, 10, 8, 7, 118, 113, 32, 34, 105, 111, 40, 42, 78, 74, 55, 52, 58, 57, 95, 90, 155, 51, 146, 48, 31, 175, 30, 165, 190, 63, 181, 60, 135, 24, 143, 25, 237, 16, 17, 227, 201, 46, 44, 197, 37, 250, 245, 39, 215, 20, 21
0, 1, 2, 3, 5, 4, 7, 6, 10, 11, 8, 9, 15, 14, 13, 12, 21, 20, 23, 22, 16, 17, 18, 19, 31, 30, 29, 28, 26, 27, 24, 25, 42, 43, 40, 41, 47, 46, 45, 44, 32, 33, 34, 35, 37, 36, 39, 38, 63, 62, 61, 60, 58, 59, 56, 57, 53, 52, 55, 54, 48, 49, 50, 51, 87, 86, 85, 84
0, 1, 3, 2, 7, 6, 4, 5, 14, 15, 13, 12, 9, 8, 10, 11, 31, 30, 28, 29, 24, 25, 27, 26, 17, 16, 18, 19, 22, 23, 21, 20, 58, 59, 57, 56, 61, 60, 62, 63, 52, 53, 55, 54, 51, 50, 48, 49, 37, 36, 38, 39, 34, 35, 33, 32, 43, 42, 40, 41, 44, 45, 47, 46, 123, 122, 120
n^n using Nim multiplication.
+10
2
1, 1, 3, 1, 5, 2, 13, 12, 14, 13, 1, 6, 13, 8, 13, 1, 17, 8, 158, 155, 72, 170, 198, 48, 145, 208, 165, 25, 55, 205, 171, 206, 55, 158, 6, 140, 151, 53, 113, 252, 191, 254, 228, 26, 116, 130, 146, 243, 145, 118, 72, 14, 75, 115, 20, 69, 60, 177, 121, 99, 171, 169, 170
PROG
(PARI) See Links section.
Square root of n under Nim (or Conway) multiplication.
+10
2
0, 1, 3, 2, 7, 6, 4, 5, 14, 15, 13, 12, 9, 8, 10, 11, 30, 31, 29, 28, 25, 24, 26, 27, 16, 17, 19, 18, 23, 22, 20, 21, 57, 56, 58, 59, 62, 63, 61, 60, 55, 54, 52, 53, 48, 49, 51, 50, 39, 38, 36, 37, 32, 33, 35, 34, 41, 40, 42, 43, 46, 47, 45, 44, 124, 125, 127, 126, 123, 122, 120
COMMENTS
Because Conway's field On2 (endowed with Nim-multiplication and [bitwise] Nim-addition) has characteristic 2, the Nim-square function ( A006042) is an injective field homomorphism (i.e., the square of a sum is the sum of the squares). Thus the square function is a bijection within any finite additive subgroup of On2 (which is a fancy way to say that an integer and its Nim-square have the same bit length). Therefore the Nim square-root function is also a field homomorphism (the square-root of a Nim-sum is the Nim-sum of the square roots) which can be defined as the inverse permutation of A006042 (as such, it preserves bit-length too).
FORMULA
Letting NIM (= XOR) TIM and RIM denote respectively the sum, product and square root in Conway's Nim-field On2, we see that the bit-length of NIM(x,TIM(x,x)) is less than that of the positive integer x. This remark turns the following relations into an effective recursive definition of a(n) = RIM(n) which uses the fact that RIM is a field homomorphism in On2:
a(0) = 0
a(n) = NIM(n, a(NIM(n, a(n, TIM(n,n)) )
For 0 <= n <= 2^2^k-1, a(n) = A335162(n, 2^(2^k-1)). This is because {0,1,...,2^2^k-1} together with the nim operations makes a field isomorphic to GF(2^2^k).
EXAMPLE
a(2) = 3 because TIM(3,3) = 2
More generally, a(x)=y because A006042(y)=x.
a(n) is the greatest nim-power of n.
+10
1
1, 1, 3, 3, 15, 15, 15, 15, 14, 15, 14, 15, 15, 14, 14, 15, 252, 252, 255, 255, 252, 252, 255, 255, 252, 252, 255, 255, 255, 255, 252, 252, 255, 255, 255, 255, 252, 255, 252, 255, 255, 255, 255, 255, 255, 252, 255, 252, 255, 252, 252, 255, 255, 255, 255, 255
COMMENTS
For any n >= 0, a(n) is the greatest term in n-th row of A335162.
EXAMPLE
For n = 10:
- the first nim-powers of 10 are: 1, 10, 14, 13, 8, 1, ...
- so a(10) = 14.
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