A206283 -id:A206283 - cp's OEIS frontend

A181187 Triangle read by rows: T(n,k) = sum of k-th largest elements in all partitions of n.

1, 3, 1, 6, 2, 1, 12, 5, 2, 1, 20, 8, 4, 2, 1, 35, 16, 8, 4, 2, 1, 54, 24, 13, 7, 4, 2, 1, 86, 41, 22, 13, 7, 4, 2, 1, 128, 61, 35, 20, 12, 7, 4, 2, 1, 192, 95, 54, 33, 20, 12, 7, 4, 2, 1, 275, 136, 80, 49, 31, 19, 12, 7, 4, 2, 1, 399, 204, 121, 76, 48, 31, 19, 12, 7, 4, 2, 1, 556, 284

Offset: 1

Wouter Meeussen, Oct 09 2010

For the connection with A066897 and A066898 see A206563. - Omar E. Pol, Feb 13 2012
T(n,k) is also the total number of parts >= k in all partitions of n. - Omar E. Pol, Feb 14 2012
The first differences of row n together with 1 give the row n of triangle A066633. - Omar E. Pol, Feb 26 2012
We define the k-th rank of a partition as the k-th part minus the number of parts >= k. Since the first part of a partition is also the largest part of the same partition so the Dyson's rank of a partition is the case for k = 1. It appears that the sum of the k-th ranks of all partitions of n is equal to zero. - Omar E. Pol, Mar 04 2012
T(n,k) is also the total number of divisors >= k of all positive integers in a sequence with n blocks where the m-th block consists of A000041(n-m) copies of m, with 1 <= m <= n. - Omar E. Pol, Feb 05 2021

			From _Omar E. Pol_, Feb 13 2012: (Start)
Illustration of initial terms. First five rows of triangle as sums of columns from the partitions of the first five positive integers:
.
.                            5
.                            3+2
.                  4         4+1
.                  2+2       2+2+1
.          3       3+1       3+1+1
.     2    2+1     2+1+1     2+1+1+1
.  1  1+1  1+1+1   1+1+1+1   1+1+1+1+1
. -------------------------------------
.  1, 3,1, 6,2,1, 12,5,2,1, 20,8,4,2,1 --> This triangle
.  |  |/|  |/|/|   |/|/|/|   |/|/|/|/|
.  1, 2,1, 4,1,1,  7,3,1,1, 12,4,2,1,1 --> A066633
.
For more information see A207031 and A206563.
...
Triangle begins:
    1;
    3,   1;
    6,   2,   1;
   12,   5,   2,  1;
   20,   8,   4,  2,  1;
   35,  16,   8,  4,  2,  1;
   54,  24,  13,  7,  4,  2,  1;
   86,  41,  22, 13,  7,  4,  2,  1;
  128,  61,  35, 20, 12,  7,  4,  2, 1;
  192,  95,  54, 33, 20, 12,  7,  4, 2, 1;
  275, 136,  80, 49, 31, 19, 12,  7, 4, 2, 1;
  399, 204, 121, 76, 48, 31, 19, 12, 7, 4, 2, 1;
(End)

Alois P. Heinz, Rows n = 1..141, flattened

Row sums are A066186. First column is A006128. Reverse of each row converges to A000070.
Columns 2-3: A096541, A207033. - Omar E. Pol, Feb 18 2012
T(2n,n) gives A216053(n+1).
Cf. A206283.

p:= (f, g)-> zip((x, y)-> x+y, f, g, 0):
b:= proc(n, i) option remember; local f, g;
      if n=0 or i=1 then [1, n]
    else f:= b(n, i-1); g:= `if`(i>n, [0], b(n-i, i));
         p(p(f, g), [0$i, g[1]])
      fi
    end:
T:= proc(n) local j, l, r, t;
      l, r, t:= b(n, n), 1, 1;
      for j from n to 2 by -1 do t:= t+l[j]; r:=r, t od;
      seq([r][1+n-j], j=1..n)
    end:
seq(T(n), n=1..14); # Alois P. Heinz, Apr 05 2012

Table[Plus @@ (PadRight[ #,n]& /@ IntegerPartitions[n]),{n,16}]
(* Second program: *)
T[n_, n_] = 1; T[n_, k_] /; k, ] = 0; Table[Table[T[n, k], {k, n, 1, -1}] // Accumulate // Reverse, {n, 1, 16}] // Flatten (* Jean-François Alcover, Oct 10 2015, after Omar E. Pol *)

T(n,k) = Sum_{j=1..n} A207031(j,k). - Omar E. Pol, May 02 2012

Better definition from Omar E. Pol, Feb 13 2012

A210952 Triangle read by rows: T(n,k) = sum of all parts of the k-th column of the partitions of n but with the partitions aligned to the right margin.

Original entry on oeis.org

1, 1, 3, 1, 3, 5, 1, 3, 7, 9, 1, 3, 7, 12, 12, 1, 3, 7, 14, 21, 20, 1, 3, 7, 14, 24, 31, 25, 1, 3, 7, 14, 26, 40, 47, 38, 1, 3, 7, 14, 26, 43, 61, 66, 49, 1, 3, 7, 14, 26, 45, 70, 92, 93, 69, 1, 3, 7, 14, 26, 45, 73, 106, 130, 124, 87, 1, 3, 7, 14

Offset: 1

Omar E. Pol, Apr 22 2012

			For n = 6 the illustration shows the partitions of 6 aligned to the right margin and below the sums of the 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
-------------------------
.  1,  3,  7, 14, 21, 20
.
So row 6 lists 1, 3, 7, 14, 21, 20.
Triangle begins:
1;
1, 3;
1, 3, 5;
1, 3, 7,  9;
1, 3, 7, 12, 12;
1, 3, 7, 14, 21, 20;
1, 3, 7, 14, 24, 31, 25;
1, 3, 7, 14, 26, 40, 47, 38;
1, 3, 7, 14, 26, 43, 61, 66, 49;
1, 3, 7, 14, 26, 45, 70, 92, 93, 69:

Mirror of triangle A206283. Rows sums give A066186. Rows converge to A014153. Right border gives A046746, >= 1.

Cf. A135010, A138121, A181187, A210763, A210950, A210951, A210953, A210961, A210970.

T(n,k) = Sum_{j=1..n} A210953(j,k). - Omar E. Pol, May 26 2012

A207381 Total sum of the odd-indexed parts of all partitions of n.

Original entry on oeis.org

1, 3, 7, 14, 25, 45, 72, 117, 180, 275, 403, 596, 846, 1206, 1681, 2335, 3183, 4342, 5820, 7799, 10321, 13622, 17798, 23221, 30009, 38706, 49567, 63316, 80366, 101805, 128211, 161134, 201537, 251495, 312508, 387535, 478674, 590072, 724920, 888795, 1086324

Offset: 1

Omar E. Pol, Feb 17 2012

nonn

For more information see A206563.

			For n = 5, write the partitions of 5 and below write the sums of their odd-indexed parts:
.    5
.    3+2
.    4+1
.    2+2+1
.    3+1+1
.    2+1+1+1
.    1+1+1+1+1
.  ------------
.   20 + 4 + 1 = 25
The total sum of the odd-indexed parts is 25 so a(5) = 25.

Alois P. Heinz, Table of n, a(n) for n = 1..1000

Cf. A066186, A066897, A066898, A181187, A194714, A206283, A206563, A207031, A207032, A207382.

b:= proc(n, i) option remember; local g, h;
      if n=0 then [1, 0$2]
    elif i<1 then [0$3]
    else g:= b(n, i-1); h:= `if`(i>n, [0$3], b(n-i, i));
         [g[1]+h[1], g[2]+h[3], g[3]+h[2]+i*h[1]]
      fi
    end:
a:= n-> b(n,n)[3]:
seq(a(n), n=1..50); # Alois P. Heinz, Mar 12 2012

b[n_, i_] := b[n, i] = Module[{g, h}, If[n == 0 , {1, 0, 0}, If[i < 1, {0, 0, 0},  g = b[n, i - 1]; h = If[i > n, {0, 0, 0}, b[n - i, i]]; {g[[1]] + h[[1]], g[[2]] + h[[3]], g[[3]] + h[[2]] + i*h[[1]]}]]]; a[n_] := b[n, n][[3]]; Table [a[n], {n, 1, 50}] (* Jean-François Alcover, Dec 09 2016 after Alois P. Heinz *)

a(n) = A066186(n) - A207382(n) = A066897(n) + A207382(n).

More terms from Alois P. Heinz, Mar 12 2012

A207382 Sum of the even-indexed parts of all partitions of n.

Original entry on oeis.org

0, 1, 2, 6, 10, 21, 33, 59, 90, 145, 213, 328, 467, 684, 959, 1361, 1866, 2588, 3490, 4741, 6311, 8422, 11067, 14579, 18941, 24630, 31703, 40788, 52019, 66315, 83891, 106034, 133182, 167045, 208397, 259637, 321895, 398498, 491295, 604725, 741579, 908008

Offset: 1

Omar E. Pol, Feb 17 2012

nonn

Also the sum of the floors of half the parts of all partitions of n, because the sum of one kind for a partition equals the sum of the other kind for the conjugate partition. Furthermore, this generalizes to taking m-th indices and dividing by m. - George Beck, Apr 15 2017

			For n = 5, write the partitions of 5 and below write the sums of their even-indexed parts:
. 5
. 3+2
. 4+1
. 2+2+1
. 3+1+1
. 2+1+1+1
. 1+1+1+1+1
------------
.   8 + 2   = 10
The sum of the even-indexed parts is 10, so a(5) = 10.
From _George Beck_, Apr 15 2017: (Start)
Alternatively, sum the floors of the parts divided by 2:
. 2
. 1+1
. 2+0
. 1+1+0
. 1+0+0
. 1+0+0+0
. 0+0+0+0+0
The sum is 10, so a(5) = 10. (End)

Alois P. Heinz, Table of n, a(n) for n = 1..1000

For more information see A206563.

Cf. A066186, A066897, A066898, A181187, A194714, A206283, A207031, A207032, A207381.

b:= proc(n, i) option remember; local g, h;
      if n=0 then [1, 0$2]
    elif i<1 then [0$3]
    else g:= b(n, i-1); h:= `if`(i>n, [0$3], b(n-i, i));
         [g[1]+h[1], g[2]+h[3], g[3]+h[2]+i*h[1]]
      fi
    end:
a:= n-> b(n,n)[2]:
seq (a(n), n=1..50); # Alois P. Heinz, Mar 12 2012

b[n_, i_] := b[n, i] = Module[{g, h}, Which[n==0, {1, 0, 0}, i<1, {0, 0, 0}, True, g = b[n, i-1]; h = If[i>n, {0, 0, 0}, b[n-i, i]]; {g[[1]] + h[[1]], g[[2]] + h[[3]], g[[3]] + h[[2]] + i*h[[1]]}]]; a[n_] := b[n, n][[2]]; Table [a[n], {n, 1, 50}] (* Jean-François Alcover, Feb 03 2017, after Alois P. Heinz *)
a[n_]:= Total@Flatten@Quotient[IntegerPartitions[n], 2];
Table [a[n], {n, 1, 50}] (* George Beck, Apr 15 2017 *)

a(n) = A066186(n) - A207381(n) = A207381(n) - A066897(n).

More terms from Alois P. Heinz, Mar 12 2012

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A181187 Triangle read by rows: T(n,k) = sum of k-th largest elements in all partitions of n.

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A210952 Triangle read by rows: T(n,k) = sum of all parts of the k-th column of the partitions of n but with the partitions aligned to the right margin.

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A207381 Total sum of the odd-indexed parts of all partitions of n.

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A207382 Sum of the even-indexed parts of all partitions of n.

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