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A214676
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A(n,k) is n represented in bijective base-k numeration; square array A(n,k), n>=1, k>=1, read by antidiagonals.
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14
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1, 1, 11, 1, 2, 111, 1, 2, 11, 1111, 1, 2, 3, 12, 11111, 1, 2, 3, 11, 21, 111111, 1, 2, 3, 4, 12, 22, 1111111, 1, 2, 3, 4, 11, 13, 111, 11111111, 1, 2, 3, 4, 5, 12, 21, 112, 111111111, 1, 2, 3, 4, 5, 11, 13, 22, 121, 1111111111
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OFFSET
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1,3
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COMMENTS
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The digit set for bijective base-k numeration is {1, 2, ..., k}.
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LINKS
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Eric Weisstein's World of Mathematics, Zerofree
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EXAMPLE
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Square array A(n,k) begins:
: 1, 1, 1, 1, 1, 1, 1, 1, ...
: 11, 2, 2, 2, 2, 2, 2, 2, ...
: 111, 11, 3, 3, 3, 3, 3, 3, ...
: 1111, 12, 11, 4, 4, 4, 4, 4, ...
: 11111, 21, 12, 11, 5, 5, 5, 5, ...
: 111111, 22, 13, 12, 11, 6, 6, 6, ...
: 1111111, 111, 21, 13, 12, 11, 7, 7, ...
: 11111111, 112, 22, 14, 13, 12, 11, 8, ...
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MAPLE
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A:= proc(n, b) local d, l, m; m:= n; l:= NULL;
while m>0 do d:= irem(m, b, 'm');
if d=0 then d:=b; m:=m-1 fi;
l:= d, l
od; parse(cat(l))
end:
seq(seq(A(n, 1+d-n), n=1..d), d=1..12);
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MATHEMATICA
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A[n_, b_] := Module[{d, l, m}, m = n; l = Nothing; While[m > 0, {m, d} = QuotientRemainder[m, b]; If[d == 0, d = b; m--]; l = {d, l}]; FromDigits @ Flatten @ l];
Table[A[n, d-n+1], {d, 1, 12}, {n, 1, d}] // Flatten (* Jean-François Alcover, May 28 2019, from Maple *)
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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