A206563
Triangle read by rows: T(n,k) = number of odd/even parts >= k in all partitions of n, if k is odd/even.
Original entry on oeis.org
1, 2, 1, 5, 1, 1, 8, 4, 1, 1, 15, 5, 3, 1, 1, 24, 11, 5, 3, 1, 1, 39, 15, 9, 4, 3, 1, 1, 58, 28, 13, 9, 4, 3, 1, 1, 90, 38, 23, 12, 8, 4, 3, 1, 1, 130, 62, 33, 21, 12, 8, 4, 3, 1, 1, 190, 85, 51, 29, 20, 11, 8, 4, 3, 1, 1, 268, 131, 73, 48, 28, 20, 11, 8, 4, 3, 1, 1
Offset: 1
Calculation for n = 6. Write the partitions of 6 and below the sums of their columns:
.
. 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
. ------------------------
. 35, 16, 8, 4, 2, 1 --> Row 6 of triangle A181187.
. | /| /| /| /| /|
. | / | / | / | / | / |
. |/ |/ |/ |/ |/ |
. 19, 8, 4, 2, 1, 1 --> Row 6 of triangle A066633.
.
More generally, it appears that the sum of column k is also the total number of parts >= k in all partitions of n. It appears that the first differences of the column sums together with 1 give the number of occurrences of k in all partitions of n.
On the other hand we can see that the partitions of 6 contain:
24 odd parts >= 1 (the odd parts).
11 even parts >= 2 (the even parts).
5 odd parts >= 3.
3 even parts >= 4.
2 odd parts >= 5.
1 even part >= 6.
Then, using the values of the column sums, it appears that:
T(6,1) = 35 - 16 + 8 - 4 + 2 - 1 = 24
T(6,2) = 16 - 8 + 4 - 2 + 1 = 11
T(6,3) = 8 - 4 + 2 - 1 = 5
T(6,4) = 4 - 2 + 1 = 3
T(6,5) = 2 - 1 = 1
T(6,6) = 1 = 1
So the 6th row of our triangle gives 24, 11, 5, 3, 1, 1.
Finally, for all partitions of 6, we can write:
The number of odd parts is equal to T(6,1) = 24.
The number of even parts is equal to T(6,2) = 11.
The number of odd parts >= 3 is equal to T(6,3) = 5.
The number of even parts >= 4 is equal to T(6,4) = 3.
The number of odd parts >= 5 is equal to T(6,5) = 1.
The number of even parts >= 6 is equal to T(6,6) = 1.
More generally, we can write the same properties for any positive integer.
Triangle begins:
1;
2, 1;
5, 1, 1;
8, 4, 1, 1;
15, 5, 3, 1, 1;
24, 11, 5, 3, 1, 1;
39, 15, 9, 4, 3, 1, 1;
58, 28, 13, 9, 4, 3, 1, 1;
90, 38, 23, 12, 8, 4, 3, 1, 1;
130, 62, 33, 21, 12, 8, 4, 3, 1, 1;
A024788
Number of 4's in all partitions of n.
Original entry on oeis.org
0, 0, 0, 1, 1, 2, 3, 6, 8, 13, 18, 28, 38, 55, 74, 105, 139, 190, 250, 336, 436, 575, 740, 963, 1228, 1577, 1995, 2538, 3186, 4013, 5005, 6256, 7751, 9617, 11847, 14605, 17894, 21927, 26730, 32582, 39531, 47942, 57915, 69920, 84114, 101116, 121176, 145095, 173248
Offset: 1
From _Omar E. Pol_, Oct 25 2012: (Start)
For n = 7 we have:
--------------------------------------
. Number
Partitions of 7 of 4's
--------------------------------------
7 .............................. 0
4 + 3 .......................... 1
5 + 2 .......................... 0
3 + 2 + 2 ...................... 0
6 + 1 .......................... 0
3 + 3 + 1 ...................... 0
4 + 2 + 1 ...................... 1
2 + 2 + 2 + 1 .................. 0
5 + 1 + 1 ...................... 0
3 + 2 + 1 + 1 .................. 0
4 + 1 + 1 + 1 .................. 1
2 + 2 + 1 + 1 + 1 .............. 0
3 + 1 + 1 + 1 + 1 .............. 0
2 + 1 + 1 + 1 + 1 + 1 .......... 0
1 + 1 + 1 + 1 + 1 + 1 + 1 ...... 0
------------------------------------
. 7 - 4 = 3
The difference between the sum of the fourth column and the sum of the fifth column of the set of partitions of 7 is 7 - 4 = 3 and equals the number of 4's in all partitions of 7, so a(7) = 3.
(End)
- Alois P. Heinz, Table of n, a(n) for n = 1..1000
- G. E. Andrews and E. Deutsch, A note on a method of Erdos and the Stanley-Elder theorems, Integers, 16 (2016), A24.
- L. Babel, A. Brandstädt, and V. B. Le, Recognizing the P4-structure of bipartite graphs, Discrete Appl. Math. 93 (1999), 157-168.
- David Benson, Radha Kessar, and Markus Linckelmann, Hochschild cohomology of symmetric groups in low degrees, arXiv:2204.09970 [math.GR], 2022.
-
b:= proc(n, i) option remember; local f, g;
if n=0 or i=1 then [1, 0]
else f:= b(n, i-1); g:= `if`(i>n, [0$2], b(n-i, i));
[f[1]+g[1], f[2]+g[2]+`if`(i=4, g[1], 0)]
fi
end:
a:= n-> b(n, n)[2]:
seq(a(n), n=1..100); # Alois P. Heinz, Oct 27 2012
-
Table[ Count[ Flatten[ IntegerPartitions[n]], 4], {n, 1, 50} ]
(* second program: *)
b[n_, i_] := b[n, i] = Module[{g}, If[n == 0 || i == 1, {1, 0}, g = If[i > n, {0, 0}, b[n - i, i]]; b[n, i - 1] + g + {0, If[i == 4, g[[1]], 0]}]]; a[n_] := b[n, n][[2]]; Table[a[n], {n, 1, 100}] (* Jean-François Alcover, Oct 09 2015, after Alois P. Heinz *)
A024787
Number of 3's in all partitions of n.
Original entry on oeis.org
0, 0, 1, 1, 2, 4, 6, 9, 15, 21, 31, 45, 63, 87, 122, 164, 222, 298, 395, 519, 683, 885, 1146, 1475, 1887, 2401, 3050, 3845, 4837, 6060, 7563, 9402, 11664, 14405, 17751, 21807, 26715, 32634, 39784, 48352, 58649, 70969, 85690, 103232, 124143, 148951, 178407, 213277, 254509
Offset: 1
From _Omar E. Pol_, Oct 25 2012: (Start)
For n = 7 we have:
--------------------------------------
. Number
Partitions of 7 of 3's
--------------------------------------
7 .............................. 0
4 + 3 .......................... 1
5 + 2 .......................... 0
3 + 2 + 2 ...................... 1
6 + 1 .......................... 0
3 + 3 + 1 ...................... 2
4 + 2 + 1 ...................... 0
2 + 2 + 2 + 1 .................. 0
5 + 1 + 1 ...................... 0
3 + 2 + 1 + 1 .................. 1
4 + 1 + 1 + 1 .................. 0
2 + 2 + 1 + 1 + 1 .............. 0
3 + 1 + 1 + 1 + 1 .............. 1
2 + 1 + 1 + 1 + 1 + 1 .......... 0
1 + 1 + 1 + 1 + 1 + 1 + 1 ...... 0
------------------------------------
. 13 - 7 = 6
The difference between the sum of the third column and the sum of the fourth column of the set of partitions of 7 is 13 - 7 = 6 and equals the number of 3's in all partitions of 7, so a(7) = 6.
(End)
- Alois P. Heinz, Table of n, a(n) for n = 1..5000
- David Benson, Radha Kessar, and Markus Linckelmann, Hochschild cohomology of symmetric groups in low degrees, arXiv:2204.09970 [math.GR], 2022.
- Philip Cuthbertson, Fixed hooks in arbitrary columns of partitions, Integers (2025) Vol. 25, Art. No. A28. See p. 3.
- Eric Weisstein's World of Mathematics, q-Pochhammer Symbol
-
b:= proc(n, i) option remember; local g;
if n=0 or i=1 then [1, 0]
else g:= `if`(i>n, [0$2], b(n-i, i));
b(n, i-1) +g +[0, `if`(i=3, g[1], 0)]
fi
end:
a:= n-> b(n, n)[2]:
seq(a(n), n=1..100); # Alois P. Heinz, Oct 27 2012
-
Table[ Count[ Flatten[ IntegerPartitions[n]], 3], {n, 1, 50} ]
b[n_, i_] := b[n, i] = Module[{g}, If[n==0 || i==1, {1, 0}, g = If[i>n, {0, 0}, b[n-i, i]]; b[n, i-1] + g + {0, If[i==3, g[[1]], 0]}]]; a[n_] := b[n, n][[2]]; Table[a[n], {n, 1, 100}] (* Jean-François Alcover, Oct 09 2015, after Alois P. Heinz *)
Join[{0, 0}, (1/((1 - x^3) QPochhammer[x]) + O[x]^50)[[3]]] (* Vladimir Reshetnikov, Nov 22 2016 *)
A302246
Irregular triangle read by rows in which row n lists all parts of all partitions of n, in nonincreasing order.
Original entry on oeis.org
1, 2, 1, 1, 3, 2, 1, 1, 1, 1, 4, 3, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 5, 4, 3, 3, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 6, 5, 4, 4, 3, 3, 3, 3, 2, 2, 2, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 7, 6, 5, 5, 4, 4, 4, 3, 3, 3, 3, 3, 3, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 1, 1
Offset: 1
Triangle begins:
1;
2,1,1;
3,2,1,1,1,1;
4,3,2,2,2,1,1,1,1,1,1,1;
5,4,3,3,2,2,2,2,1,1,1,1,1,1,1,1,1,1,1,1;
6,5,4,4,3,3,3,3,2,2,2,2,2,2,2,2,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1;
...
For n = 4 the partitions of 4 are [4], [2, 2], [3, 1], [2, 1, 1], [1, 1, 1, 1]. There is only one 4, only one 3, three 2's and seven 1's, so the 4th row of this triangle is [4, 3, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1].
On the other hand for n = 4 the 4th row of A176206 is [4, 3, 2, 2, 1, 1, 1] and the divisors of these terms are [1, 2, 4], [1, 3], [1, 2], [1, 2], [1], [1], [1] the same as the 4th row of A338156. These divisors listed in nonincreasing order give the 4th row of this triangle. - _Omar E. Pol_, Jun 16 2022
The number of parts k in row n is
A066633(n,k).
The sum of all parts k in row n is
A138785(n,k).
The number of parts >= k in row n is
A181187(n,k).
The sum of all parts >= k in row n is
A206561(n,k).
The number of parts <= k in row n is
A210947(n,k).
The sum of all parts <= k in row n is
A210948(n,k).
-
nrows=10;Array[ReverseSort[Flatten[IntegerPartitions[#]]]&,nrows] (* Paolo Xausa, Jun 16 2022 *)
-
row(n) = my(list = List()); forpart(p=n, for (k=1, #p, listput(list, p[k]));); vecsort(Vec(list), , 4); \\ Michel Marcus, Jun 16 2022
A302247
Irregular triangle read by rows in which row n lists all parts of all partitions of n, in nondecreasing order.
Original entry on oeis.org
1, 1, 1, 2, 1, 1, 1, 1, 2, 3, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 3, 4, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 4, 5, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 4, 4, 5, 6, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1
Offset: 1
Triangle begins:
1;
1,1,2;
1,1,1,1,2,3;
1,1,1,1,1,1,1,2,2,2,3,4;
1,1,1,1,1,1,1,1,1,1,1,1,2,2,2,2,3,3,4,5;
1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,2,2,2,2,2,2,2,2,3,3,3,3,4,4,5,6;
...
For n = 4 the partitions of 4 are [4], [2, 2], [3, 1], [2, 1, 1], [1, 1, 1, 1]. There are seven 1's, three 2's, only one 3 and only one 4, so the 4th row of this triangle is [1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 3, 4].
On the other hand for n = 4 the 4th row of A176206 is [4, 3, 2, 2, 1, 1, 1] and the divisors of these terms are [1, 2, 4], [1, 3], [1, 2], [1, 2], [1], [1], [1] the same as the 4th row of A338156. These divisors listed in nondecreasing order give the 4th row of this triangle. - _Omar E. Pol_, Jun 16 2022
The number of parts k in row n is
A066633(n,k).
The sum of all parts k in row n is
A138785(n,k).
The number of parts >= k in row n is
A181187(n,k).
The sum of all parts >= k in row n is
A206561(n,k).
The number of parts <= k in row n is
A210947(n,k).
The sum of all parts <= k in row n is
A210948(n,k).
-
nrows=10; Array[Sort[Flatten[IntegerPartitions[#]]]&,nrows] (* Paolo Xausa, Jun 16 2022 *)
-
row(n) = my(list = List()); forpart(p=n, for (k=1, #p, listput(list, p[k]));); vecsort(Vec(list)); \\ Michel Marcus, Jun 16 2022
A024789
Number of 5's in all partitions of n.
Original entry on oeis.org
0, 0, 0, 0, 1, 1, 2, 3, 5, 8, 12, 17, 25, 35, 50, 68, 94, 126, 170, 226, 299, 391, 511, 660, 853, 1091, 1393, 1766, 2235, 2811, 3527, 4403, 5484, 6800, 8415, 10369, 12752, 15627, 19110, 23298, 28346, 34389, 41642, 50295, 60636, 72929, 87563, 104903, 125470
Offset: 1
From _Omar E. Pol_, Oct 25 2012: (Start)
For n = 8 we have:
--------------------------------------
. Number
Partitions of 8 of 5's
--------------------------------------
8 .............................. 0
4 + 4 .......................... 0
5 + 3 .......................... 1
6 + 2 .......................... 0
3 + 3 + 2 ...................... 0
4 + 2 + 2 ...................... 0
2 + 2 + 2 + 2 .................. 0
7 + 1 .......................... 0
4 + 3 + 1 ...................... 0
5 + 2 + 1 ...................... 1
3 + 2 + 2 + 1 .................. 0
6 + 1 + 1 ...................... 0
3 + 3 + 1 + 1 .................. 0
4 + 2 + 1 + 1 .................. 0
2 + 2 + 2 + 1 + 1 .............. 0
5 + 1 + 1 + 1 .................. 1
3 + 2 + 1 + 1 + 1 .............. 0
4 + 1 + 1 + 1 + 1 .............. 0
2 + 2 + 1 + 1 + 1 + 1 .......... 0
3 + 1 + 1 + 1 + 1 + 1 .......... 0
2 + 1 + 1 + 1 + 1 + 1 + 1 ...... 0
1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 .. 0
------------------------------------
. 7 - 4 = 3
The difference between the sum of the fifth column and the sum of the sixth column of the set of partitions of 8 is 7 - 4 = 3 and equals the number of 5's in all partitions of 8, so a(8) = 3.
(End)
-
b:= proc(n, i) option remember; local g;
if n=0 or i=1 then [1, 0]
else g:= `if`(i>n, [0$2], b(n-i, i));
b(n, i-1) +g +[0, `if`(i=5, g[1], 0)]
fi
end:
a:= n-> b(n, n)[2]:
seq(a(n), n=1..100); # Alois P. Heinz, Oct 27 2012
-
Table[ Count[ Flatten[ IntegerPartitions[n]], 5], {n, 1, 50} ]
(* second program: *)
b[n_, i_] := b[n, i] = Module[{g}, If[n == 0 || i == 1, {1, 0}, g = If[i > n, {0, 0}, b[n - i, i]]; b[n, i - 1] + g + {0, If[i == 5, g[[1]], 0]}]]; a[n_] := b[n, n][[2]]; Table[a[n], {n, 1, 100}] (* Jean-François Alcover, Oct 09 2015, after Alois P. Heinz *)
-
x='x+O('x^50); concat([0, 0, 0, 0], Vec(x^5/(1 - x^5) * prod(k=1, 50, 1/(1 - x^k)))) \\ Indranil Ghosh, Apr 06 2017
A024790
Number of 6's in all partitions of n.
Original entry on oeis.org
0, 0, 0, 0, 0, 1, 1, 2, 3, 5, 7, 12, 16, 24, 33, 47, 63, 89, 117, 159, 209, 278, 360, 474, 607, 786, 1001, 1280, 1615, 2049, 2565, 3222, 4011, 4998, 6180, 7653, 9407, 11571, 14154, 17308, 21063, 25630, 31044, 37586, 45339, 54646, 65646, 78804, 94305, 112761, 134473
Offset: 1
-
b:= proc(n, i) option remember; local g;
if n=0 or i=1 then [1, 0]
else g:= `if`(i>n, [0$2], b(n-i, i));
b(n, i-1) +g +[0, `if`(i=6, g[1], 0)]
fi
end:
a:= n-> b(n, n)[2]:
seq(a(n), n=1..100); # Alois P. Heinz, Oct 27 2012
-
Table[ Count[ Flatten[ IntegerPartitions[n]], 6], {n, 1, 52} ]
b[n_, i_] := b[n, i] = Module[{g}, If [n == 0 || i == 1, {1, 0}, g = If[i > n, {0, 0}, b[n - i, i]]; b[n, i - 1] + g + {0, If[i == 6, g[[1]], 0]}]]; a[n_] := b[n, n][[2]]; Table[a[n], {n, 1, 100}] (* Jean-François Alcover, Oct 09 2015, after Alois P. Heinz *)
A024794
Number of 10's in all partitions of n.
Original entry on oeis.org
0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 2, 3, 5, 7, 11, 15, 22, 30, 43, 57, 79, 104, 140, 183, 242, 312, 407, 520, 670, 849, 1081, 1359, 1715, 2141, 2678, 3322, 4125, 5085, 6274, 7691, 9430, 11502, 14025, 17024, 20655, 24959, 30140, 36270, 43612, 52274, 62604, 74763
Offset: 1
- Alois P. Heinz, Table of n, a(n) for n = 1..1000
- David Benson, Radha Kessar, and Markus Linckelmann, Hochschild cohomology of symmetric groups in low degrees, arXiv:2204.09970 [math.GR], 2022.
- Joseph Vandehey, Digital problems in the theory of partitions, Integers (2024) Vol. 24A, Art. No. A18. See p. 3.
Cf.
A066633,
A000070(n-1),
A024786,
A024787,
A024788,
A024789,
A024790,
A024791,
A024792,
A024793,
A000041.
-
b:= proc(n, i) option remember; local g;
if n=0 or i=1 then [1, 0]
else g:= `if`(i>n, [0$2], b(n-i, i));
b(n, i-1) +g +[0, `if`(i=10, g[1], 0)]
fi
end:
a:= n-> b(n, n)[2]:
seq(a(n), n=1..100); # Alois P. Heinz, Oct 27 2012
-
Table[ Count[ Flatten[ IntegerPartitions[n]], 10], {n, 1, 55} ]
b[n_, i_] := b[n, i] = Module[{g}, If[n == 0 || i == 1, {1, 0}, g = If[i > n, {0, 0}, b[n - i, i]]; b[n, i - 1] + g + {0, If[i == 10, g[[1]], 0]}]]; a[n_] := b[n, n][[2]]; Table[a[n], {n, 1, 100}] (* Jean-François Alcover, Oct 09 2015, after Alois P. Heinz *)
A207032
Triangle read by rows: T(n,k) = number of odd/even parts >= k in the last section of the set of partitions of n, if k is odd/even.
Original entry on oeis.org
1, 1, 1, 3, 0, 1, 3, 3, 0, 1, 7, 1, 2, 0, 1, 9, 6, 2, 2, 0, 1, 15, 4, 4, 1, 2, 0, 1, 19, 13, 4, 5, 1, 2, 0, 1, 32, 10, 10, 3, 4, 1, 2, 0, 1, 40, 24, 10, 9, 4, 4, 1, 2, 0, 1, 60, 23, 18, 8, 8, 3, 4, 1, 2, 0, 1, 78, 46, 22, 19, 8, 9, 3, 4, 1, 2, 0, 1
Offset: 1
Triangle begins:
1;
1, 1;
3, 0, 1;
3, 3, 0, 1;
7, 1, 2, 0, 1;
9, 6, 2, 2, 0, 1;
15, 4, 4, 1, 2, 0, 1;
19, 13, 4, 5, 1, 2, 0, 1;
32, 10, 10, 3, 4, 1, 2, 0, 1;
40, 24, 10, 9, 4, 4, 1, 2, 0, 1;
60, 23, 18, 8, 8, 3, 4, 1, 2, 0, 1;
78, 46, 22, 19, 8, 9, 3, 4, 1, 2, 0, 1;
Cf.
A006128,
A066897,
A066898,
A135010,
A138121,
A138135,
A138137,
A141285,
A181187,
A182703,
A206563,
A207031.
A024791
Number of 7's in all partitions of n.
Original entry on oeis.org
0, 0, 0, 0, 0, 0, 1, 1, 2, 3, 5, 7, 11, 16, 23, 32, 45, 61, 84, 112, 151, 199, 263, 342, 446, 574, 739, 943, 1201, 1518, 1917, 2404, 3010, 3749, 4661, 5766, 7122, 8759, 10753, 13153, 16059, 19544, 23743, 28759, 34774, 41938, 50491, 60642, 72718, 87004, 103934, 123908
Offset: 1
-
b:= proc(n, i) option remember; local g;
if n=0 or i=1 then [1, 0]
else g:= `if`(i>n, [0$2], b(n-i, i));
b(n, i-1) +g +[0, `if`(i=7, g[1], 0)]
fi
end:
a:= n-> b(n, n)[2]:
seq(a(n), n=1..100); # Alois P. Heinz, Oct 27 2012
-
<< DiscreteMath`Combinatorica`; Table[ Count[ Flatten[ Partitions[n]], 7], {n, 1, 52} ]
Table[Count[Flatten[IntegerPartitions[n]],7],{n,55}] (* Harvey P. Dale, Feb 26 2015 *)
b[n_, i_] := b[n, i] = Module[{g}, If[n == 0 || i == 1, {1, 0}, g = If[i > n, {0, 0}, b[n - i, i]]; b[n, i - 1] + g + {0, If[i == 7, g[[1]], 0]}]]; a[n_] := b[n, n][[2]]; Table[a[n], {n, 1, 100}] (* Jean-François Alcover, Oct 09 2015, after Alois P. Heinz *)
-
x='x+O('x^50); concat([0, 0, 0, 0, 0, 0], Vec(x^7/(1 - x^7) * prod(k=1, 50, 1/(1 - x^k)))) \\ Indranil Ghosh, Apr 06 2017
Comments