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A182703
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Triangle read by rows: T(n,k) = number of occurrences of k in the last section of the set of partitions of n.
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100
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1, 1, 1, 2, 0, 1, 3, 2, 0, 1, 5, 1, 1, 0, 1, 7, 4, 2, 1, 0, 1, 11, 3, 2, 1, 1, 0, 1, 15, 8, 3, 3, 1, 1, 0, 1, 22, 7, 6, 2, 2, 1, 1, 0, 1, 30, 15, 6, 5, 3, 2, 1, 1, 0, 1, 42, 15, 10, 5, 4, 2, 2, 1, 1, 0, 1, 56, 27, 14, 10, 5, 5, 2, 2, 1, 1, 0, 1
(list;
table;
graph;
refs;
listen;
history;
text;
internal format)
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OFFSET
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1,4
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COMMENTS
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For the definition of "section" of the set of partitions of n see A135010.
Also, column 1 gives the number of partitions of n-1. For k >= 2, row n lists the number of k's in all partitions of n that do not contain 1 as a part.
It appears that reversed rows converge to A002865.
It appears that row n is also the base of an isosceles triangle in which the column sums give the partition numbers A000041 in descending order starting with p(n-1) = A000041(n-1). Example for n = 7:
.
. 1,
. 1, 0, 1,
. 4, 2, 1, 0, 1,
11, 3, 2, 1, 1, 0, 1,
---------------------
11, 7, 5, 3, 2, 1, 1,
.
It appears that in row n starts an infinite trapezoid in which column sums always give the number of partitions of n-1. Example for n = 7:
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11, 3, 2, 1, 1, 0, 1,
. 8, 3, 3, 1, 1, 0, 1,
. 6, 2, 2, 1, 1, 0, 1,
. 5, 3, 2, 1, 1, 0, 1,
. 4, 2, 2, 1, 1, 0, 1,
. 5, 2, 2, 1, 1, 0,...
. 4, 2, 2, 1, 1,...
. 4, 2, 2, 1,...
. 4, 2, 2,...
. 4, 2,...
. 4,...
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The sum of any column is always p(7-1) = p(6) = A000041(6) = 11.
It appears that the first term of row n is one of the vertices of an infinite isosceles triangle in which column sums give the partition numbers A000041 in ascending order starting with p(n-1) = A000041(n-1). Example for n = 7:
11,
. 8,
. 7, 6,
. 6, 5,
. 10, 5, ...
. 10, ...
. 10, ...
-------------------
11, 15, 22, 30, ...
(End)
It appears that row n lists the first differences of the row n of triangle A207031 together with 1 (as the final term of row n). - Omar E. Pol, Feb 26 2012
More generally T(n,k) is the number of occurrences of k in the n-th section of the set of partitions of any integer >= n. - Omar E. Pol, Oct 21 2013
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LINKS
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FORMULA
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It appears that T(n,k) = A207032(n,k) - A207032(n,k+2). - Omar E. Pol, Feb 26 2012
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EXAMPLE
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Illustration of three arrangements of the last section of the set of partitions of 7, or more generally the 7th section of the set of partitions of any integer >= 7:
. _ _ _ _ _ _ _
. (7) (7) |_ _ _ _ |
. (4+3) (4+3) |_ _ _ _|_ |
. (5+2) (5+2) |_ _ _ | |
. (3+2+2) (3+2+2) |_ _ _|_ _|_ |
. (1) (1) | |
. (1) (1) | |
. (1) (1) | |
. (1) (1) | |
. (1) (1) | |
. (1) (1) | |
. (1) (1) | |
. (1) (1) | |
. (1) (1) | |
. (1) (1) | |
. (1) (1) |_|
. ----------------
. 19,8,5,3,2,1,1 --> Row 7 of triangle A207031.
. |/|/|/|/|/|/|
. 11,3,2,1,1,0,1 --> Row 7 of this triangle.
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Note that the "head" of the last section is formed by the partitions of 7 that do not contain 1 as a part. The "tail" is formed by A000041(7-1) parts of size 1. The number of rows (or zones) is A000041(7) = 15. The last section of the set of partitions of 7 contains eleven 1's, three 2's, two 3's, one 4, one 5, there are no 6's and it contains one 7. So, for k = 1..7, row 7 gives: 11, 3, 2, 1, 1, 0, 1.
Triangle begins:
1;
1, 1;
2, 0, 1;
3, 2, 0, 1;
5, 1, 1, 0, 1;
7, 4, 2, 1, 0, 1;
11, 3, 2, 1, 1, 0, 1;
15, 8, 3, 3, 1, 1, 0, 1;
22, 7, 6, 2, 2, 1, 1, 0, 1;
30, 15, 6, 5, 3, 2, 1, 1, 0, 1;
42, 15, 10, 5, 4, 2, 2, 1, 1, 0, 1;
56, 27, 14, 10, 5, 5, 2, 2, 1, 1, 0, 1;
...
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MAPLE
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p:= (f, g)-> zip((x, y)-> x+y, f, g, 0):
b:= proc(n, i) option remember; local g;
if n=0 then [1]
elif n<2 or i<2 then [0]
else g:= `if`(i>n, [0], b(n-i, i));
p(p([0$j=2..i, g[1]], b(n, i-1)), g)
fi
end:
h:= proc(n) option remember;
`if`(n=0, 1, b(n, n)[1]+h(n-1))
end:
T:= proc(n) h(n-1), b(n, n)[2..n][] end:
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MATHEMATICA
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p[f_, g_] := Plus @@ PadRight[{f, g}]; b[n_, i_] := b[n, i] = Module[{g}, Which[n == 0, {1}, n<2 || i<2, {0}, True, g = If [i>n, {0}, b[n-i, i]]; p[p[Append[Array[0&, i-1], g[[1]]], b[n, i-1]], g]]]; h[n_] := h[n] = If[n == 0, 1, b[n, n][[1]] + h[n-1]]; t[n_] := {h[n-1], Sequence @@ b[n, n][[2 ;; n]]}; Table[t[n], {n, 1, 20}] // Flatten (* Jean-François Alcover, Jan 16 2014, after Alois P. Heinz's Maple code *)
Table[{PartitionsP[n-1]}~Join~Table[Count[Flatten@Cases[IntegerPartitions[n], x_ /; Last[x] != 1], k], {k, 2, n}], {n, 1, 12}] // Flatten (* Robert Price, May 15 2020 *)
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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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