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A360099
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To get A(n,k), replace 0's in the binary expansion of n with (-1) and interpret the result in base k; square array A(n,k), n>=0, k>=0, read by antidiagonals.
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9
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0, 0, 1, 0, 1, -1, 0, 1, 0, 1, 0, 1, 1, 2, -1, 0, 1, 2, 3, -1, 1, 0, 1, 3, 4, 1, 1, -1, 0, 1, 4, 5, 5, 3, 1, 1, 0, 1, 5, 6, 11, 7, 5, 3, -1, 0, 1, 6, 7, 19, 13, 11, 7, -2, 1, 0, 1, 7, 8, 29, 21, 19, 13, 1, 0, -1, 0, 1, 8, 9, 41, 31, 29, 21, 14, 3, 0, 1, 0, 1, 9, 10, 55, 43, 41, 31, 43, 16, 5, 2, -1
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history;
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OFFSET
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0,14
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COMMENTS
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The empty bit string is used as binary expansion of 0, so A(0,k) = 0.
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LINKS
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FORMULA
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G.f. for column k satisfies g_k(x) = k*(x+1)*g_k(x^2) + x/(1+x).
A(n,k) = k*A(floor(n/2),k)+2*(n mod 2)-1 for n>0, A(0,k)=0.
A(n,k) mod 2 = A057427(n) if k is even.
A(n,k) mod 2 = A030300(n) if k is odd and n>=1.
A(2^(n+1),1) + n = 0.
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EXAMPLE
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Square array A(n,k) begins:
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, ...
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, ...
-1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, ...
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, ...
-1, -1, 1, 5, 11, 19, 29, 41, 55, 71, 89, ...
1, 1, 3, 7, 13, 21, 31, 43, 57, 73, 91, ...
-1, 1, 5, 11, 19, 29, 41, 55, 71, 89, 109, ...
1, 3, 7, 13, 21, 31, 43, 57, 73, 91, 111, ...
-1, -2, 1, 14, 43, 94, 173, 286, 439, 638, 889, ...
1, 0, 3, 16, 45, 96, 175, 288, 441, 640, 891, ...
-1, 0, 5, 20, 51, 104, 185, 300, 455, 656, 909, ...
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MAPLE
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A:= proc(n, k) option remember; local m;
`if`(n=0, 0, k*A(iquo(n, 2, 'm'), k)+2*m-1)
end:
seq(seq(A(n, d-n), n=0..d), d=0..12);
# second Maple program:
A:= (n, k)-> (l-> add((2*l[i]-1)*k^(i-1), i=1..nops(l)))(Bits[Split](n)):
seq(seq(A(n, d-n), n=0..d), d=0..12);
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CROSSREFS
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Rows n=0-10 give: A000004, A000012, A023443, A000027(k+1), A165900, A002061, A165900(k+1), A002061(k+1), A083074, A152618, A062158.
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KEYWORD
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AUTHOR
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STATUS
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approved
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