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%I #27 Apr 26 2020 21:52:09
%S 1,1,2,1,1,3,0,0,0,2,1,1,1,2,4,0,0,0,0,0,3,1,1,1,1,1,2,5,0,0,0,0,0,0,
%T 0,2,0,0,0,0,0,0,0,2,4,0,0,0,0,0,0,0,0,0,3,1,1,1,1,1,1,1,2,2,3,6,0,0,
%U 0,0,0,0,0,0,0,0,0,3,0,0,0,0,0,0,0,0,0,0,0,2,5,0,0,0,0,0,0,0,0,0,0,0,0,0,4,1,1,1,1,1,1,1,1,1,1,1,2,2,3,7
%N Triangle of regions and partitions of integers (see Comments lines for definition).
%C Let r = T(n,k) be a record in the sequence. The consecutive records "r" are the natural numbers A000027. Consider the first n rows; the triangle T(n,k) has the property that the columns, without the zeros, from k..1, are also the partitions of r in juxtaposed reverse-lexicographical order, so k is also A000041(r), the number of partitions of r. Note that a record r is always the final term of a row if such row contains 1’s. The number of positive integer a(1)..r is A006128(r). The sums a(1)..r is A066186(r). Here the set of positive integers in every row (from 1 to n) is called a “region” of r. The number of regions of r equals the number of partitions of r. If T(n,1) = 1 then the row n is formed by the smallest parts, in nondecreasing order, of all partitions of T(n,n).
%H Robert Price, <a href="/A186114/b186114.txt">Table of n, a(n) for n = 1..196878, rows 1-627.</a>
%H Omar E. Pol, <a href="http://www.polprimos.com/imagenespub/polpar02.jpg">Illustration of the seven regions of 5</a>
%F T(n,1) = A167392(n).
%F T(n,k) = A141285(n), if k = n.
%e Triangle begins:
%e 1,
%e 1, 2,
%e 1, 1, 3,
%e 0, 0, 0, 2,
%e 1, 1, 1, 2, 4,
%e 0, 0, 0, 0, 0, 3,
%e 1, 1, 1, 1, 1, 2, 5,
%e 0, 0, 0, 0, 0, 0, 0, 2,
%e 0, 0, 0, 0, 0, 0, 0, 2, 4,
%e 0, 0, 0, 0, 0, 0, 0, 0, 0, 3,
%e 1, 1, 1, 1, 1, 1, 1, 2, 2, 3, 6
%e ...
%e The row n = 11 contains the 6th record in the sequence: a(66) = T(11,11) = 6, then consider the first 11 rows of triangle. Note that the columns, from k = 11..1, without the zeros, are also the 11 partitions of 6 in juxtaposed reverse-lexicographical order: [6], [3, 3], [4, 2], [2, 2, 2], [5, 1], [3, 2, 1], [4, 1, 1], [2, 2, 1, 1], [3, 1, 1, 1], [2, 1, 1, 1, 1], [1, 1, 1, 1, 1, 1]. See A026792.
%t A206437 = Cases[Import["https://oeis.org/A206437/b206437.txt",
%t "Table"], {_, _}][[All, 2]];
%t A194446 = Cases[Import["https://oeis.org/A194446/b194446.txt",
%t "Table"], {_, _}][[All, 2]];
%t f[n_] := Module[{v},
%t v = Take[A206437, A194446[[n]]];
%t A206437 = Drop[A206437, A194446[[n]]];
%t Reverse[PadRight[v, n]]];
%t Table[f[n], {n, PartitionsP[20]}] // Flatten (* _Robert Price_, Apr 26 2020 *)
%Y Mirror of triangle A193870. Column 1 gives A167392. Right diagonal gives A141285.
%Y Cf. A000041, A135010, A138121, A183152, A186412, A187219, A194436-A194439, A194446-A194448, A206437.
%K nonn,tabl
%O 1,3
%A _Omar E. Pol_, Aug 08 2011
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