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Square array T(n, k), n, k >= 0, read by antidiagonals; T(n, k) = g(f(n) + f(k)) where g maps the complement, say s, of a finite set of nonnegative integers to the value Sum_{e >= 0 not in s} 2^e, f is the inverse of g, and "+" denotes the Minkowski sum.
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%I #12 Mar 20 2021 14:37:21

%S 0,1,1,0,3,0,3,1,1,3,0,7,2,7,0,1,1,3,3,1,1,0,3,0,15,0,3,0,7,1,5,3,3,5,

%T 1,7,0,15,2,7,0,7,2,15,0,1,1,7,3,1,1,3,7,1,1,0,3,0,31,4,11,4,31,0,3,0,

%U 3,1,1,3,7,5,5,7,3,1,1,3,0,7,2,7,0,15,6,15,0,7,2,7,0

%N Square array T(n, k), n, k >= 0, read by antidiagonals; T(n, k) = g(f(n) + f(k)) where g maps the complement, say s, of a finite set of nonnegative integers to the value Sum_{e >= 0 not in s} 2^e, f is the inverse of g, and "+" denotes the Minkowski sum.

%C In other words:

%C - we consider the set S of sets s of nonnegative integers whose complement is finite,

%C - the function g encodes the "missing integers" in binary:

%C g(A001477 \ {1, 4}) = 2^1 + 2^4 = 18

%C - the function f is the inverse of g:

%C f(42) = f(2^1 + 2^3 + 2^5) = A001477 \ {1, 3, 5},

%C - the Minkowski sum of two sets, say U and V, is the set of sums u+v where u belongs to U and v belongs to V,

%C - the Minkowski sum is stable over S,

%C - and T provides an encoding for this operation.

%C This sequence has connections with A067138; here we consider complements of finite sets of nonnegative integers, there finite sets of nonnegative integers.

%H Rémy Sigrist, <a href="/A342639/b342639.txt">Table of n, a(n) for n = 0..10010</a>

%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Minkowski_addition">Minkowski addition</a>

%F T(n, k) = T(k, n).

%F T(m, T(n, k)) = T(T(m, n), k).

%F T(n, 0) = A135481(n).

%F T(n, 1) = A038712(n+1).

%F T(2^n-1, 2^k-1) = 2^(n+k)-1.

%F T(n, n) = A342640(n).

%e Array T(n, k) begins:

%e n\k| 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

%e ---+------------------------------------------------------------------

%e 0| 0 1 0 3 0 1 0 7 0 1 0 3 0 1 0 15

%e 1| 1 3 1 7 1 3 1 15 1 3 1 7 1 3 1 31

%e 2| 0 1 2 3 0 5 2 7 0 1 2 11 0 5 2 15

%e 3| 3 7 3 15 3 7 3 31 3 7 3 15 3 7 3 63

%e 4| 0 1 0 3 0 1 4 7 0 1 0 3 0 9 4 15

%e 5| 1 3 5 7 1 11 5 15 1 3 5 23 1 11 5 31

%e 6| 0 1 2 3 4 5 6 7 0 9 2 11 4 13 6 15

%e 7| 7 15 7 31 7 15 7 63 7 15 7 31 7 15 7 127

%e 8| 0 1 0 3 0 1 0 7 0 1 0 3 0 1 8 15

%e 9| 1 3 1 7 1 3 9 15 1 3 1 7 1 19 9 31

%e 10| 0 1 2 3 0 5 2 7 0 1 10 11 0 5 10 15

%e 11| 3 7 11 15 3 23 11 31 3 7 11 47 3 23 11 63

%e 12| 0 1 0 3 0 1 4 7 0 1 0 3 8 9 12 15

%e 13| 1 3 5 7 9 11 13 15 1 19 5 23 9 27 13 31

%e 14| 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

%e 15| 15 31 15 63 15 31 15 127 15 31 15 63 15 31 15 255

%o (PARI) T(n,k) = { my (v=0); for (x=0, #binary(n)+#binary(k), my (f=0); for (y=0, x, if (!bittest(n,y) && !bittest(k,x-y), f=1; break)); if (!f, v+=2^x)); return (v) }

%Y Cf. A038712, A067138, A133457, A135481, A342640, A342641, A342642.

%K nonn,tabl,base

%O 0,5

%A _Rémy Sigrist_, Mar 17 2021