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%I #20 Feb 16 2023 16:07:34
%S 1,1,1,11,1,1,1111,6,1,1,11111111,58,4,1,1,1111111111111111,5829,38,3,
%T 1,1,11111111111111111111111111111111,58292915,3813,29,3,1,1,
%U 1111111111111111111111111111111111111111111111111111111111111111,5829291479146458,38127938,2833,23,2,1,1
%N Square array A(m,n), m>=1, n>=1, read by antidiagonals: A(m,1)=1, A(m,n) is the rank with respect to m of the concatenation of all preceding terms in row m, and the rank of S with respect to m is floor ((S+m-1)/m).
%H Alois P. Heinz, <a href="/A159861/b159861.txt">Antidiagonals n = 1..12, flattened</a>
%e A(3,4) = 38, because A(3,1).A(3,2).A(3,3) = 114, and the rank of 114 with respect to 3 is floor(116/3) = 38.
%e Square array A(m,n) begins:
%e 1, 1, 11, 1111, 11111111, 1111111111111111, ...
%e 1, 1, 6, 58, 5829, 58292915, ...
%e 1, 1, 4, 38, 3813, 38127938, ...
%e 1, 1, 3, 29, 2833, 28323209, ...
%e 1, 1, 3, 23, 2265, 22646453, ...
%e 1, 1, 2, 19, 1870, 18698645, ...
%p R:= (S,m)-> iquo(S+m-1, m):
%p A:= proc(m, n) option remember; `if`(n=1, 1,
%p R(parse(cat(seq(A(m, j), j=1..n-1))), m))
%p end:
%p seq(seq(A(m, d-m), m=1..d-1), d=1..10);
%t R[S_, m_] := Quotient[S + m - 1, m];
%t A[m_, n_] := If[n == 1, 1, R[ToExpression@StringJoin[ToString /@ Table[A[m, j], {j, 1, n - 1}]], m]];
%t Table[Table[A[m, d - m], {m, 1, d - 1}], {d, 1, 10}] // Flatten (* _Jean-François Alcover_, Feb 13 2023, after Maple code *)
%Y Row m=2 gives: A156147.
%Y Main diagonal gives: A159862.
%Y Cf. A156146, A010783.
%K easy,nonn,tabl,base
%O 1,4
%A _Eric Angelini_ and _Alois P. Heinz_, Apr 24 2009
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