

A051775


Table T(n,m) = Nimproduct of n and m, read by antidiagonals, for n >= 0, m >= 0.


25



0, 0, 0, 0, 1, 0, 0, 2, 2, 0, 0, 3, 3, 3, 0, 0, 4, 1, 1, 4, 0, 0, 5, 8, 2, 8, 5, 0, 0, 6, 10, 12, 12, 10, 6, 0, 0, 7, 11, 15, 6, 15, 11, 7, 0, 0, 8, 9, 13, 2, 2, 13, 9, 8, 0, 0, 9, 12, 14, 14, 7, 14, 14, 12, 9, 0, 0, 10, 14, 4, 10, 8, 8, 10, 4, 14, 10, 0, 0, 11, 15, 7, 11
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OFFSET

0,8


COMMENTS

Note on an algorithm, R. J. Mathar, May 29 2011: (Start)
Let N* denote the Nimproduct and N+ the Nimsum (A003987) of two numbers, and let * and + denote the usual multiplication and addition.
To compute n N* m, write n and m separately as Nimsums with the aid of the binary representation of n = n0 + n1*2 + n2*4 + n3*8 + n4*16.. and m = m0 + m1*2 + m2*4 + m3*8 + m4*16... . Because Nimsummation is the same as the binary XORfunction, the + may then be replaced by N+ in both sums:
n = Nimsum_i 2^a(i) and m = Nimsum_j 2^b(j) with two integer sequences a(i) and b(j).
Because N+ and N* are the operations in a field, N+ and N* are distributive, which is used to write the product over the sums as a doubleNimsum over Nimproducts:
n N* m = Nimsum_{i,j} 2^a(i) N* 2^b(j) .
What remains is to compute the Nimproducts of powers of 2.
Splitting a(i) and b(j) separately into (ordinary) products of Fermat numbers A001146 (i.e., writing a(i) and b(j) in binary), and noting that the ordinary product of distinct Fermat numbers equals the Nimproduct of distinct Fermat numbers,
2^a(i) N* 2^b(j) = 2^(2^A0) N* 2^(2^A1) N* ... N* 2^(2^B0) N* 2^(2^B1) N* ... for two binary integer sequences A and B.
This finite product is regrouped by pairing the cases for the same bit in the Asequence and in the Bsequence. If the bit is set in both sequences, use that the Nimsquare of a Fermat number is 3/2 times (ordinary multiple of) that Fermat number; if the bit is set only in one of the two sequences, use (again) that the Nimproduct of distinct Fermat numbers is the ordinary product.
Due to the potential presence of the Nimsquares, this leaves in general a Nimproduct which is treated by recursion.
This algorithm is implemented in the Maple program in the bfile. nimprodP2() calculates the Nimproduct of two powers of 2. (End)


REFERENCES

J. H. Conway, On Numbers and Games, Academic Press, p. 52.


LINKS

R. J. Mathar, Table of n, a(n) for n = 0..1890
Tilman Piesk, 256x256 table and dual matrix
Wikipedia, Nimber
Index entries for sequences related to Nimmultiplication


EXAMPLE

Table begins:
0 0 0 0 0 0 0 ...
0 1 2 3 4 5 6 ...
0 2 3 1 8 10 11 ...
0 3 1 2 12 15 13 ...
0 4 8 12 6 2 14 ...


MAPLE

We continue from A003987: to compute a Nimmultiplication table using (a) an addition table AT := array(0..NA, 0..NA) and (b) a nimsum procedure for larger values; MT := array(0..N, 0..N); for a from 0 to N do MT[a, 0] := 0; MT[0, a] := 0; MT[a, 1] := a; MT[1, a] := a; od: for a from 2 to N do for b from a to N do t1 := {}; for i from 0 to a1 do for j from 0 to b1 do u1 := MT[i, b]; u2 := MT[a, j];
if u1<=NA and u2<=NA then u12 := AT[u1, u2]; else u12 := nimsum(u1, u2); fi; u3 := MT[i, j]; if u12<=NA and u3<=NA then u4 := AT[u12, u3]; else u4 := nimsum(u12, u3); fi; t1 := { op(t1), u4}; #t1 := { op(t1), AT[ AT[ MT[i, b], MT[a, j] ], MT[i, j] ] }; od; od;
t2 := sort(convert(t1, list)); j := nops(t2); for i from 1 to nops(t2) do if t2[i] <> i1 then j := i1; break; fi; od; MT[a, b] := j; MT[b, a] := j; od; od;


CROSSREFS

Cf. A051776, A003987.
Sequence in context: A109042 A128540 A160692 * A108036 A190174 A263139
Adjacent sequences: A051772 A051773 A051774 * A051776 A051777 A051778


KEYWORD

tabl,nonn,easy,nice


AUTHOR

N. J. A. Sloane, Dec 19 1999


STATUS

approved



