cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-5 of 5 results.

A196087 Sum of all parts minus the total numbers of parts of all partitions of n.

Original entry on oeis.org

0, 1, 3, 8, 15, 31, 51, 90, 142, 228, 341, 525, 757, 1110, 1572, 2233, 3084, 4286, 5812, 7910, 10580, 14145, 18659, 24626, 32099, 41814, 53976, 69559, 88932, 113557, 143967, 182241, 229353, 288078, 360029, 449158, 557757, 691369, 853628, 1051974
Offset: 1

Views

Author

Omar E. Pol, Nov 10 2011

Keywords

Comments

Also sum of parts of all partitions of n except the largest parts of the partitions. - Omar E. Pol, Feb 16 2012
Equals column 1 of A161224. - Omar E. Pol, Feb 26 2012
Partial sums of A207035. - Omar E. Pol, Apr 22 2012

Examples

			For n = 4 the five partitions of 4 are: 4, 3+1, 2+2, 2+1+1, 1+1+1+1. The sum of all parts is 4+3+1+2+2+2+1+1+1+1+1+1 = 20. The sum of all parts is also the product n*p(n) = 4*5 = 20, where p(n) = A000041(n) is the number of partitions of n. On the other hand the number of parts in all partitions of 4 is equal to 12, so a(4) = 20-12 = 8.
		

Crossrefs

Programs

  • Maple
    b:= proc(n, i) option remember; local f, g;
          if n=0 then [1, 0]
        elif i<1 then [0, 0]
        elif i>n then b(n, i-1)
        else f:= b(n, i-1); g:= b(n-i, i);
             [f[1]+g[1], f[2]+g[2] +g[1]*(i-1)]
          fi
        end:
    a:= n-> b(n, n)[2]:
    seq(a(n), n=1..50);  # Alois P. Heinz, Feb 20 2012
  • Mathematica
    b[n_, i_] := b[n, i] = Module[{f, g}, Which[n==0, {1, 0}, i<1, {0, 0}, i>n, b[n, i-1], True, f = b[n, i-1]; g = b[n-i, i]; {f[[1]] + g[[1]], f[[2]] + g[[2]] + g[[1]]*(i-1)}]]; a[n_] := b[n, n][[2]]; Table[a[n], {n, 1, 50}] (* Jean-François Alcover, Oct 22 2015, after Alois P. Heinz *)
  • PARI
    a(n) = n*numbpart(n) - sum(m=1, n, numdiv(m)*numbpart(n-m)); \\ Michel Marcus, Oct 22 2015

Formula

a(n) = n*A000041(n) - A006128(n) = A066186(n) - A006128(n).
a(n) = A207038(A000041(n)). - Omar E. Pol, Apr 21 2012
a(n) ~ exp(Pi*sqrt(2*n/3))/(4*sqrt(3)) * (1 - (3 + 6*gamma + Pi^2/24 + 3*log(6*n/Pi^2))/(Pi*sqrt(6*n))), where gamma is the Euler-Mascheroni constant A001620. - Vaclav Kotesovec, Oct 24 2016
G.f.: Sum_{k>=1} x^(2*k)/(1 - x^k)^2 / Product_{j>=1} (1 - x^j). - Ilya Gutkovskiy, Mar 05 2021
a(n) = Sum_{k=1..n-1} p(n+j,j), where p(n,j) is the number of partitions of j having exactly j parts. E.g., a(4) = p(5,1) + p(6,2) + p(7,3) = 1+3+4 = 8. - Gregory L. Simay, Aug 19 2022

A207034 Sum of all parts minus the number of parts of the n-th partition in the list of colexicographically ordered partitions of j, if 1<=n<=A000041(j).

Original entry on oeis.org

0, 1, 2, 2, 3, 3, 4, 3, 4, 4, 5, 4, 5, 5, 6, 4, 5, 5, 6, 6, 6, 7, 5, 6, 6, 7, 6, 7, 7, 8, 5, 6, 6, 7, 7, 7, 8, 7, 8, 8, 8, 9, 6, 7, 7, 8, 7, 8, 8, 9, 8, 8, 9, 9, 9, 10, 6, 7, 7, 8, 8, 8, 9, 8, 9, 9, 9, 10, 8, 9, 9, 10, 9, 10, 10, 10, 11, 7, 8, 8, 9, 8, 9
Offset: 1

Views

Author

Omar E. Pol, Feb 20 2012

Keywords

Comments

a(n) is also the column number in which is located the part of size 1 in the n-th zone of the tail of the last section of the set of partitions of k in colexicographic order, minus the column number in which is located the part of size 1 in the first row of the same tail, when k -> infinity (see example). For the definition of "section" see A135010.

Examples

			Illustration of initial terms, n = 1..15. Consider the last 15 rows of the tail of the last section of the set of partitions in colexicographic order of any integer >= 8. The tail contains at least A000041(8-1) = 15 parts of size 1. a(n) is also the number of dots in the n-th row of the diagram.
----------------------------------
n      Tail                  a(n)
----------------------------------
15        1 . . . . . .       6
14          1 . . . . .       5
13          1 . . . . .       5
12            1 . . . .       4
11          1 . . . . .       5
10            1 . . . .       4
9             1 . . . .       4
8               1 . . .       3
7             1 . . . .       4
6               1 . . .       3
5               1 . . .       3
4                 1 . .       2
3                 1 . .       2
2                   1 .       1
1                     1       0
----------------------------------
Written as a triangle:
0;
1;
2;
2,3;
3,4;
3,4,4,5;
4,5,5,6;
4,5,5,6,6,6,7;
5,6,6,7,6,7,7,8;
5,6,6,7,7,7,8,7,8,8,8,9;
6,7,7,8,7,8,8,9,8,8,9,9,9,10;
6,7,7,8,8,8,9,8,9,9,9,10,8,9,9,10,9,10,10,10,11;
...
Consider a matrix [j X A000041(j)] in which the rows represent the partitions of j in colexicographic order (see A211992). Every part of every partition is located in a cell of the matrix. We can see that a(n) is the number of empty cells in row n for any integer j, if A000041(j) >= n. The number of empty cells in row n equals the sum of all parts minus the number of parts in the n-th partition of j.
Illustration of initial terms. The smallest part of every partition is located in the last column of the matrix.
---------------------------------------------------------
.   j: 1    2       3         4           5             6
n a(n)
---------------------------------------------------------
1  0 | 1  1 1   1 1 1   1 1 1 1   1 1 1 1 1   1 1 1 1 1 1
2  1 |    . 2   . 2 1   . 2 1 1   . 2 1 1 1   . 2 1 1 1 1
3  2 |          . . 3   . . 3 1   . . 3 1 1   . . 3 1 1 1
4  2 |                  . . 2 2   . . 2 2 1   . . 2 2 1 1
5  3 |                  . . . 4   . . . 4 1   . . . 4 1 1
6  3 |                            . . . 3 2   . . . 3 2 1
7  4 |                            . . . . 5   . . . . 5 1
8  3 |                                        . . . 2 2 2
9  4 |                                        . . . . 4 2
10 4 |                                        . . . . 3 3
11 5 |                                        . . . . . 6
...
Illustration of initial terms. In this case the largest part of every partition is located in the first column of the matrix.
---------------------------------------------------------
.   j: 1    2       3         4           5             6
n a(n)
---------------------------------------------------------
1  0 | 1  1 1   1 1 1   1 1 1 1   1 1 1 1 1   1 1 1 1 1 1
2  1 |    2 .   2 1 .   2 1 1 .   2 1 1 1 .   2 1 1 1 1 .
3  2 |          3 . .   3 1 . .   3 1 1 . .   3 1 1 1 . .
4  2 |                  2 2 . .   2 2 1 . .   2 2 1 1 . .
5  3 |                  4 . . .   4 1 . . .   4 1 1 . . .
6  3 |                            3 2 . . .   3 2 1 . . .
7  4 |                            5 . . . .   5 1 . . . .
8  3 |                                        2 2 2 . . .
9  4 |                                        4 2 . . . .
10 4 |                                        3 3 . . . .
11 5 |                                        6 . . . . .
...
		

Crossrefs

Row r has length A187219(r). Partial sums give A207038. Row sums give A207035. Right border gives A001477. Where records occur give A000041 without repetitions.

Formula

a(n) = t(n) - A194548(n), if n >= 2, where t(n) is the n-th element of the following sequence: triangle read by rows in which row n lists n repeated k times, where k = A187219(n).
a(n) = A000120(A194602(n-1)) = A000120(A228354(n)-1).
a(n) = i - A193173(i,n), i >= 1, 1<=n<=A000041(i).

A207038 Partial sums of A207034.

Original entry on oeis.org

0, 1, 3, 5, 8, 11, 15, 18, 22, 26, 31, 35, 40, 45, 51, 55, 60, 65, 71, 77, 83, 90, 95, 101, 107, 114, 120, 127, 134, 142, 147, 153, 159, 166, 173, 180, 188, 195, 203, 211, 219, 228, 234, 241, 248, 256, 263, 271, 279, 288, 296, 304, 313, 322, 331, 341
Offset: 1

Views

Author

Omar E. Pol, Feb 21 2012

Keywords

Examples

			Written as a triangle:
0;
1;
3;
5,8;
11,15;
18,22,26,31;
35,40,45,51;
55,60,65,71,77,83,90;
95,101,107,114,120,127,134,142;
147,153,159,166,173,180,188,195,203,211,219,228;
234,241,248,256,263,271,279,288,296,304,313,322,331,341;
		

Crossrefs

Row n has length A187219(n). Right border gives A196087.

A182276 Sum of all parts minus the total number of parts of the shell model of partitions with n regions.

Original entry on oeis.org

0, 1, 3, 4, 8, 10, 15, 16, 20, 22, 31, 33, 38, 41, 51, 52, 56, 58, 67, 71, 74, 90, 92, 97, 100, 110, 112, 119, 123, 142, 143, 147, 149, 158, 162, 165, 181, 184, 192, 197, 201, 228, 230, 235, 238, 248, 250, 257, 261, 280, 284, 287, 299, 305, 310, 341
Offset: 1

Views

Author

Omar E. Pol, Apr 23 2012

Keywords

Comments

For the definition of "region of n" see A206437.

Examples

			Written has a triangle:
0,
1,
3,
4,    8;
10,  15;
16,  20, 22, 31;
33,  38, 41, 51;
52,  56, 58, 67, 71, 74, 90;
92,  97,100,110,112,119,123,142;
143,147,149,158,162,165,181,184,192,197,201,228;
230,235,238,248,250,257,261,280,284,287,299,305,310,341;
		

Crossrefs

Row j has length A187219(j). Right border gives A196087.

Programs

  • Mathematica
    lex[n_]:=DeleteCases[Sort@PadRight[Reverse /@ IntegerPartitions@n], x_ /; x==0,2];
    reg = {}; l = {};
    For[j = 1, j <= 56, j++,
      mx = Max@lex[j][[j]]; AppendTo[l, mx];
      For[i = j, i > 0, i--, If[l[[i]] > mx, Break[]]];
      t = Take[Reverse[First /@ lex[mx]], j - i];
      AppendTo[reg, Total@t - Length@t]
      ];
    Accumulate@reg  (* Robert Price, Jul 25 2020 *)

Formula

a(n) = A182244(n) - A182181(n).
a(A000041(n)) = A196087(n).

A210946 Triangle read by rows: T(n,k) = sum of parts in the k-th column of the mirror of the last section of the set of partitions of n with its parts aligned to the right margin.

Original entry on oeis.org

1, 3, 5, 9, 2, 12, 3, 20, 9, 2, 25, 11, 3, 38, 22, 9, 2, 49, 28, 14, 3, 69, 44, 26, 9, 2, 87, 55, 37, 14, 3, 123, 83, 62, 29, 9, 2, 152
Offset: 1

Views

Author

Omar E. Pol, Apr 21 2012

Keywords

Comments

Row n lists the positive terms of the n-th row of triangle A210953 in decreasing order.

Examples

			For n = 7 the illustration shows two arrangements of the last section of the set of 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)
.                 ---------
.                  25,11,3
.
The left hand picture shows the last section of 7 with its parts aligned to the right margin. In the right hand picture (the mirror) we can see that the sum of all parts of the columns 1..3 are 25, 11, 3 therefore row 7 lists 25, 11, 3.
Written as a triangle begins:
1;
3;
5;
9,    2;
12,   3;
20,   9,  2;
25,  11,  3;
38,  22,  9,  2;
49,  28, 14,  3;
69,  44, 26,  9,  2;
87,  55, 37, 14,  3,
123, 83, 62, 29,  9,  2;
		

Crossrefs

Showing 1-5 of 5 results.