A159861 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).

1, 1, 1, 11, 1, 1, 1111, 6, 1, 1, 11111111, 58, 4, 1, 1, 1111111111111111, 5829, 38, 3, 1, 1, 11111111111111111111111111111111, 58292915, 3813, 29, 3, 1, 1, 1111111111111111111111111111111111111111111111111111111111111111, 5829291479146458, 38127938, 2833, 23, 2, 1, 1

Offset: 1

Author

Eric Angelini and Alois P. Heinz, Apr 24 2009

Keywords

• easy
• nonn
• tabl
• base

Examples

			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.
Square array A(m,n) begins:
  1,  1, 11, 1111, 11111111, 1111111111111111,  ...
  1,  1,  6,   58,     5829,         58292915,  ...
  1,  1,  4,   38,     3813,         38127938,  ...
  1,  1,  3,   29,     2833,         28323209,  ...
  1,  1,  3,   23,     2265,         22646453,  ...
  1,  1,  2,   19,     1870,         18698645,  ...
		

Links

• Alois P. Heinz, Antidiagonals n = 1..12, flattened

Crossrefs

Row m=2 gives: A156147.

Main diagonal gives: A159862.

Cf. A156146, A010783.

Programs

Maple

R:= (S,m)-> iquo(S+m-1, m):
A:= proc(m, n) option remember; `if`(n=1, 1,
      R(parse(cat(seq(A(m, j), j=1..n-1))), m))
    end:
seq(seq(A(m, d-m), m=1..d-1), d=1..10);

Mathematica

R[S_, m_] := Quotient[S + m - 1, m];
A[m_, n_] := If[n == 1, 1, R[ToExpression@StringJoin[ToString /@ Table[A[m, j], {j, 1, n - 1}]], m]];
Table[Table[A[m, d - m], {m, 1, d - 1}], {d, 1, 10}] // Flatten (* Jean-François Alcover, Feb 13 2023, after Maple code *)