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.

A299769 Triangle read by rows: T(n,k) is the sum of all squares of the parts k in the last section of the set of partitions of n, with n >= 1, 1 <= k <= n.

Original entry on oeis.org

1, 1, 4, 2, 0, 9, 3, 8, 0, 16, 5, 4, 9, 0, 25, 7, 16, 18, 16, 0, 36, 11, 12, 18, 16, 25, 0, 49, 15, 32, 27, 48, 25, 36, 0, 64, 22, 28, 54, 32, 50, 36, 49, 0, 81, 30, 60, 54, 80, 75, 72, 49, 64, 0, 100, 42, 60, 90, 80, 100, 72, 98, 64, 81, 0, 121, 56, 108, 126, 160, 125, 180, 98, 128, 81, 100, 0, 144
Offset: 1

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Author

Omar E. Pol, Mar 20 2018

Keywords

Comments

The partial sums of the k-th column of this triangle give the k-th column of triangle A299768.
Note that the last section of the set of partitions of n is also the n-th section of the set of partitions of any positive integer >= n.

Examples

			Triangle begins:
   1;
   1,   4;
   2,   0,   9;
   3,   8,   0,  16;
   5,   4,   9,   0,  25;
   7,  16,  18,  16,   0,  36;
  11,  12,  18,  16,  25,   0,  49;
  15,  32,  27,  48,  25,  36,   0,  64;
  22,  28,  54,  32,  50,  36,  49,   0,  81;
  30,  60,  54,  80,  75,  72,  49,  64,   0, 100;
  42,  60,  90,  80, 100,  72,  98,  64,  81,   0, 121;
  56, 108, 126, 160, 125, 180,  98, 128,  81, 100,   0, 144;
  ...
Illustration for the 4th row of triangle:
.
.                                  Last section of the set
.        Partitions of 4.          of the partitions of 4.
.       _ _ _ _                              _
.      |_| | | |  [1,1,1,1]                 | |  [1]
.      |_ _| | |  [2,1,1]                   | |  [1]
.      |_ _ _| |  [3,1]                _ _ _| |  [1]
.      |_ _|   |  [2,2]               |_ _|   |  [2,2]
.      |_ _ _ _|  [4]                 |_ _ _ _|  [4]
.
For n = 4 the last section of the set of partitions of 4 is [4], [2, 2], [1], [1], [1], so the squares of the parts are respectively [16], [4, 4], [1], [1], [1]. The sum of the squares of the parts 1 is 1 + 1 + 1 = 3. The sum of the squares of the parts 2 is 4 + 4 = 8. The sum of the squares of the parts 3 is 0 because there are no parts 3. The sum of the squares of the parts 4 is 16. So the fourth row of triangle is [3, 8, 0, 16].
		

Crossrefs

Column 1 is A000041.
Leading diagonal gives A000290, n >= 1.
Second diagonal gives A000007.
Row sums give A206440.

Programs

  • Maple
    b:= proc(n, i) option remember; `if`(n=0 or i=1, 1+n*x, b(n, i-1)+
          (p-> p+(coeff(p, x, 0)*i^2)*x^i)(b(n-i, min(n-i, i))))
        end:
    T:= n-> (p-> seq(coeff(p, x, i), i=1..n))(b(n$2)-b(n-1$2)):
    seq(T(n), n=1..14);  # Alois P. Heinz, Jul 23 2018
  • Mathematica
    b[n_, i_] := b[n, i] = If[n==0 || i==1, 1 + n*x, b[n, i-1] + Function[p, p + (Coefficient[p, x, 0]*i^2)*x^i][b[n-i, Min[n-i, i]]]];
    T[n_] := Function[p, Table[Coefficient[p, x, i], {i, 1, n}]][b[n, n] - b[n-1, n-1]];
    T /@ Range[14] // Flatten (* Jean-François Alcover, Dec 10 2019, after Alois P.heinz *)

Formula

T(n,k) = A299768(n,k) - A299768(n-1,k). - Alois P. Heinz, Jul 23 2018