A212551
Number of partitions T(n,k) of n containing at least one other part m-k if m is the largest part; triangle T(n,k), n>=0, 0<=k<=n.
Original entry on oeis.org
1, 0, 0, 1, 0, 0, 1, 1, 0, 0, 2, 1, 1, 0, 0, 2, 3, 1, 1, 0, 0, 4, 3, 3, 1, 1, 0, 0, 4, 6, 4, 3, 1, 1, 0, 0, 7, 7, 7, 4, 3, 1, 1, 0, 0, 8, 11, 9, 8, 4, 3, 1, 1, 0, 0, 12, 13, 15, 10, 8, 4, 3, 1, 1, 0, 0, 14, 20, 18, 17, 11, 8, 4, 3, 1, 1, 0, 0
Offset: 0
T(4,0) = 2: [1,1,1,1], [2,2].
T(4,1) = 1: [2,1,1].
T(5,1) = 3: [2,1,1,1], [2,2,1], [3,2].
T(6,2) = 3: [3,1,1,1], [3,2,1], [4,2].
T(7,2) = 4: [3,1,1,1,1], [3,2,1,1], [3,3,1], [4,2,1].
T(8,4) = 3: [5,1,1,1], [5,2,1], [6,2].
Triangle T(n,k) begins:
1;
0, 0;
1, 0, 0;
1, 1, 0, 0;
2, 1, 1, 0, 0;
2, 3, 1, 1, 0, 0;
4, 3, 3, 1, 1, 0, 0;
4, 6, 4, 3, 1, 1, 0, 0;
7, 7, 7, 4, 3, 1, 1, 0, 0;
Columns k=0-10 give:
A002865,
A083751(n+1),
A119907,
A212543,
A212544,
A212545,
A212546,
A212547,
A212548,
A212549,
A212550.
Row sums give
A000070(n-2) for n>1.
-
b:= proc(n, i) option remember;
`if`(n=0 or i=1, 1, b(n, i-1)+`if`(i>n, 0, b(n-i, i)))
end:
T:= (n, k)-> `if`(n=0 and k=0, 1,
add(b(n-2*m-k, min(n-2*m-k, m+k)), m=1..(n-k)/2)):
seq(seq(T(n, k), k=0..n), n=0..14);
-
b[n_, i_] := b[n, i] = If[n == 0 || i == 1, 1, b[n, i-1] + If[i > n, 0, b[n-i, i]]]; t[n_, k_] := If[n == 0 && k == 0, 1, Sum[b[n-2*m-k, Min[n-2*m-k, m+k]], {m, 1, (n-k)/2}]]; Table[Table[t[n, k], {k, 0, n}], {n, 0, 14}] // Flatten (* Jean-François Alcover, Dec 13 2013, translated from Maple *)
A210945
Triangle read by rows: T(n,k) = number of parts in the k-th column of the mirror of the last shell of the partitions of n.
Original entry on oeis.org
1, 2, 3, 5, 1, 7, 1, 11, 3, 1, 15, 3, 1, 22, 6, 3, 1, 30, 7, 4, 1, 42, 11, 7, 3, 1, 56, 13, 9, 4, 1, 77, 20, 15, 8, 3, 1, 101, 23, 18, 10, 4, 1, 135, 33, 27, 17, 8, 3, 1, 176, 40, 34, 22, 11, 4, 1, 231, 54, 47, 33, 18, 8, 3, 1, 297, 65, 58, 42, 24, 11, 4, 1
Offset: 1
For n = 7 the illustration shows two arrangements of the last shell of the partitions of 7:
.
. (7) (7)
. (4+3) (3+4)
. (5+2) (2+5)
. (3+2+2) (2+2+3)
. (1) (1)
. (1) (1)
. (1) (1)
. (1) (1)
. (1) (1)
. (1) (1)
. (1) (1)
. (1) (1)
. (1) (1)
. (1) (1)
. (1) (1)
. --------
. 15,3,1
.
We can see that in the right hand picture (the mirror) the number of part for columns 1..3 are 15, 3, 1 therefore row 7 lists 15, 3, 1.
Written as a triangle begins:
1;
2;
3;
5, 1;
7, 1;
11, 3, 1;
15, 3, 1;
22, 6, 3, 1;
30, 7, 4, 1;
42, 11, 7, 3, 1;
56, 13, 9, 4, 1;
77, 20, 15, 8, 3, 1;
101, 23, 18, 10, 4, 1;
135, 33, 27, 17, 8, 3, 1;
176, 40, 34, 22, 11, 4, 1;
231, 54, 47, 33, 18, 8, 3, 1;
297, 65, 58, 42, 24, 11, 4, 1;
A362548
Number of partitions of n with at least three parts larger than 1.
Original entry on oeis.org
0, 0, 0, 0, 0, 0, 1, 2, 5, 9, 16, 25, 40, 58, 85, 119, 166, 224, 303, 399, 526, 681, 880, 1122, 1430, 1801, 2266, 2827, 3521, 4354, 5378, 6601, 8092, 9870, 12020, 14576, 17652, 21294, 25653, 30804, 36937, 44162, 52732, 62798, 74690, 88627, 105028, 124201, 146696, 172924, 203600, 239292, 280912
Offset: 0
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Table[PartitionsP[n]-(1 + Floor[n^2/4]),{n,0,30}];
Table[ Count[Partitions[n], pa_ /; Length[DeleteCases[pa, 1]] > 2] , {n,0,30}]
-
from sympy import npartitions
def A362548(n): return npartitions(n)-1-(n**2>>2) # Chai Wah Wu, Apr 27 2023
Showing 1-3 of 3 results.
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