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 A051775 Table T(n,m) = Nim-product of n and m, read by antidiagonals, for n >= 0, m >= 0. 26
 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 (list; table; graph; refs; listen; history; text; internal format)
 OFFSET 0,8 COMMENTS Note on an algorithm, R. J. Mathar, May 29 2011: (Start) Let N* denote the Nim-product and N+ the Nim-sum (A003987) of two numbers, and let * and + denote the usual multiplication and addition. To compute n N* m, write n and m separately as Nim-sums 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 Nim-summation is the same as the binary XOR-function, the + may then be replaced by N+ in both sums: n = Nim-sum_i 2^a(i) and m = Nim-sum_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 double-Nim-sum over Nim-products: n N* m = Nim-sum_{i,j} 2^a(i) N* 2^b(j) . What remains is to compute the Nim-products 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 Nim-product 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 A-sequence and in the B-sequence. If the bit is set in both sequences, use that the Nim-square 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 Nim-product of distinct Fermat numbers is the ordinary product. Due to the potential presence of the Nim-squares, this leaves in general a Nim-product which is treated by recursion. This algorithm is implemented in the Maple program in the b-file. nimprodP2() calculates the Nim-product 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 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 Nim-multiplication 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 a-1 do for j from 0 to b-1 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] <> i-1 then j := i-1; break; fi; od; MT[a, b] := j; MT[b, a] := j; od; od; CROSSREFS Cf. A051776, A003987. Sequence in context: A322403 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

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Last modified November 21 14:18 EST 2019. Contains 329371 sequences. (Running on oeis4.)