|
%I #7 Apr 14 2015 11:04:25
%S 0,1,4,-2,4,-1,4,9,-4,9,-4,1,9,-4,2,9,-1,9,16,-9,4,-1,16,-9,4,16,-4,
%T 16,-4,1,16,-4,2,16,-1,16,25,-9,1,25,-9,4,-2,25,-9,4,-1,25,-9,4,25,-4,
%U 25,-4,1,25,-4,2,25,-1,25,36,-16,9,-4,1,36,-9,36,-9,1
%N R(k), the minimal alternating squares representation of k, concatenated for k = 0, 1, 2,....
%C Let B(n) be the least square >= n. The minimal alternating squares representation of a nonnegative integer n is defined as the sum B(n) - B(m(1) + B(m(2) + ... + d*B(m(k)) that results from the recurrence R(n) = B(n) - R(B(n) - n), with initial representations R(0) = 0, R(1) = 1, and R(2) = 4 - 2. The sum B(n) + B(m(2)) + ... is the positive part of R(n), and the sum B(m(1)) + B(m(3)) + ... is the nonpositive part of R(n). The last term of R(k) is the trace of n. If b(n) = n(n+1)/2, the n-th triangular number, then the sum R(n) is the minimal alternating triangular-number representation of n.
%C Unlike minimal alternating representations for other bases (e.g., Fibonacci numbers, A256655; binary, A256696, triangular numbers, A244974), the trace of a minimal alternating squares representation is not necessarily a member of the base; specifically, the trace can be -2 or 2, which are not squares.
%H Clark Kimberling, <a href="/A256789/b256789.txt">Table of n, a(n) for n = 0..1000</a>
%e R(0) = 0
%e R(1) = 1
%e R(2) = 4 - 2
%e R(3) = 4 - 1
%e R(4) = 4
%e R(5) = 9 - 4
%e R(6) = 9 - 4 + 1
%e R(7) = 9 - 4 + 2
%e R(89) = 100 - 16 + 9 - 4
%t b[n_] := n^2; bb = Table[b[n], {n, 0, 1000}];
%t s[n_] := Table[b[n], {k, 1, 2 n - 1}];
%t h[1] = {1}; h[n_] := Join[h[n - 1], s[n]];
%t g = h[100]; r[0] = {0}; r[1] = {1}; r[2] = {4, -2};
%t r[n_] := If[MemberQ[bb, n], {n}, Join[{g[[n]]}, -r[g[[n]] - n]]];
%t Table[r[n], {n, 0, 120}] (* A256789, individual representations *)
%t Flatten[Table[r[n], {n, 0, 120}]] (* A256789, concatenated representations *)
%Y Cf. A000290, A256655, A256696, A244974, A256780 (number of terms), A256791 (trace).
%K easy,sign
%O 0,3
%A _Clark Kimberling_, Apr 13 2015
|
