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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. 1
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 (list; table; graph; refs; listen; history; text; internal format)
OFFSET

1,3

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.

LINKS

Alois P. Heinz, Rows n = 1..200, flattened

FORMULA

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

EXAMPLE

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].

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

CROSSREFS

Column 1 is A000041.

Leading diagonal gives A000290, n >= 1.

Second diagonal gives A000007.

Row sums give A206440.

Cf. A066183, A135010, A206438, A299768.

Sequence in context: A058546 A196774 A219245 * A091435 A118441 A244131

Adjacent sequences:  A299766 A299767 A299768 * A299772 A299773 A299774

KEYWORD

nonn,tabl

AUTHOR

Omar E. Pol, Mar 20 2018

STATUS

approved

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Last modified January 16 15:53 EST 2019. Contains 319195 sequences. (Running on oeis4.)